Force

Problem

A thermometer having a time constant of 0. min is placed in a temperature bath and after the thermometer comes to equilibrium with the bath, the temperature of the bath is increased linearly with time at the rate of I deg C / min what is the difference between the indicated temperature and bath temperature (a) 0. 1 min (b) 10. min after the change in temperature begins. (c) What is the maximum deviation between the indicated temperature and bath temperature and when does it occurs. (d) plot the forcing function and the response on the same graph. After the long enough time buy how many minutes does the response lag the input?

Solution

Consider thermometer to be in equilibrium with temperature bath at temperature (a) the difference between the indicated temperature and bath temperature

Problem

Determine the transfer function H(s)/Q(s) for the liquid level shown in figure below Resistance R1 and R2 are linear. The flow rate from tank 3 is maintained constant at b by means of a pump; the flow rate from tank 3 is independent of head h. The tanks are non-interacting.

Solution

Balance on tank 3 gives writing the steady state equation Subtracting and writing in terms of deviation Taking Laplace transforms

We have three equations and 4 unknowns (Q(s),H(s),H1(s) and H2(s). So we can express one in terms of other. From (1) Combining equation 4,5,6

Problem

Heat transfer equipment shown in fig. consists of tow tanks, one nested inside the other. Heat is transferred by convection through the wall of inner tank.

1. Hold up volume of each tank is 1 ft3
2. The cross sectional area for heat transfer is 1 ft2
3. The overall heat transfer coefficient for the flow of heat between the tanks is 10 Btu/(hr)(ft2)(oF)
4. Heat capacity of fluid in each tank is 2 Btu/(lb)(oF)

5.  Density of each fluid is 50 lb/ft3 Initially the temp of feed stream to the outer tank and the contents of the outer tank are equal to 100 oF.

Contents of inner tank are initially at 100 oF. the flow of heat to the inner tank (Q) changed according to a step change from 0 to 500 Btu/hr. (a) Obtain an expression for the laplace transform of the temperature of inner tank T(s). (b) Invert T(s) and obtain T for t= 0,5,10, U

Solution

(a)For outer tank Substituting numerical values Now Ti(s) = 0, since there is no change in temp of feed stream to outer tank. Which gives For inner tank

Problem

The thermal system shown in fig P 13. 6 is controlled by PD controller. Data; w = 250 lb/min; ? = 62. 5 lb/ft3; V1 = 4 ft3, V2=5 ft3; V3=6ft3; C = 1 Btu/(lb)(°F) Change of 1 psi from the controller changes the flow rate of heat of by 500 Btu/min. the temperature of the inlet stream may vary. There is no lag in the measuring element. (a) Draw a block diagram of the control system with the appropriate transfer function in each block. Each transfer function should contain a numerical values of the parameters. b) From the block diagram, determine the overall transfer function relating the temperature in tank 3 to a change in set point. (c ) Find the offset for a unit steo change in inlet temperature if the controller gain KC is 3psi/°F of temperature error and the derivative time is 0. 5 min. Fig. (1) (b)

Problem

For the control shown, the characteristics equation is s 4 +4 s3 +6 s 2 +4 s +(1 + k) =0 (a) Determine value of k above which the system is unstable. (b) Determine the value of k for which the two of the roots are on the imaginary axis.

Solution

s 4 +4 s3 +6 s 2 +4 s +(1 + k) =0 For the system to be unstable.

Discussion

The objective of the experiment is to use the capillary rise method to determine the surface tension of the working solution – in this case, the increasing concentrations of n-butanol solution.

In a solution, molecules experience intermolecular forces with each other. However, the molecules in the surface of the solution experience less intermolecular force because part of it is exposed to a different phase. Therefore, there is a tendency for the “bulk” solution to pull the molecules from the surface towards them. This concept is applied in a rain droplet, where because of the pull by the bulk part of the solution, it shapes into a form of a sphere – a shape with the least surface area. The rise of the solution in the capillary tube is the result of cohesion and adhesion.

Cohesion is the attraction of molecules within the same phase while adhesion refers to the attraction of molecules of different phases; say the n-butanol solution and the walls of the capillary tube. If the adhesion force is stronger than the cohesion force, the walls of the capillary tube will be wet, which in turn attracts molecules from the bulk of the solution upward until the pressure exerted from outside (environment) of the capillary tube is equal to the forces that lifts the solution upwards.

This equilibrium point will be used to determine the height of the rise of the solution, which is a factor in determining the surface tension of the solution. Different concentrations of n-butanol solution were prepared in volumetric flasks. Then the radius of the capillary tube was identified by performing the capillary rise method using deionized water. With the given surface tension of water at 30 degrees Celsius, the radius was calculated: r=2? pgh Surfactants are molecules that have a nonpolar tail and a polar end.

It lowers the surface tension from two different phases because of its ability to “pull” the molecules toward the molecules in the surface area. N-butanol is a surfactant therefore, it is hypothesized that the surface tension will decrease as the concentration of this surfactant in the solution increases. However, in the experiment, the results stated otherwise, that the more concentrated the n-butanol in the solution is, the higher the surface tension it manifests.

The equation used to compute for the surface tension is: ? = pghr2 where p is the density, g is the acceleration due to gravity, h is the height of the solution and r is the radius of the capillary. Sources of error can come from the deviating temperatures of the balance room and the laboratory. The capillary rise method should have been performed immediately right after identifying the density of the solution since a little change in temperature could greatly affect the behavior of the solution.

Moreover, the long duration of time in performing the capillary rise method could possibly turn the solution back again to two layered phases, instead of a homogenous mixture therefore, what is measured is the height of the crude deionized wated rather than the solution. Also, another source of error could be the prolonged “stagnant” state of the other solutions in the volumetric flask where the alcohols present in the solution could possibly be turned into vapor state inside the flasks.

The behavior of the n-butanol is to converge to each other and replace the water molecules at the surface. The concentration of this surfactant becomes bigger than the molecules in the bulk which gives the excess of concentration denoted by: ? = -sRT where s is the slope of the best-fit line by plotting the surface tension against ln C (concentration in mol m-3), R is the ideal gas constant, and the T is the temperature in Kelvin. The value of ? , is used to calculate the value of the cross-sectional molecule of n-butanol, as well as the molecular radius of the chemical.

Conclusion

Although the result stated otherwise, the concept of surface tension and the relationship of the concentration of the surfactant were understood with further research of other related experiments. Despite this, the use of capillary rise method gave way to compute for the radius without directly measuring it, but instead by having a given surface tension and determining the other sufficient factors in the Laplace equation. Sample Calculations: radius of capillary= 2(0. 07118Nm)995. 67kgm39. 8ms2(0. 0320m)=0. 00046m urface tension= 910kgm39. 8ms20. 022m(0. 00046m)2=0. 045Nm excess concentration= -0. 0028. 314JKmol(303K)= 0. 00000079Jmol cross-sectional area= 10. 00000079Jmol10101m216. 022X1023molecules=2. 09X10^48A molecular radius= 2. 09X10^48Avalue of pi=8. 16X10^23A

Literature Cited

1. Chang, Raymond. Physical Chemistry for the Chemical and Biological Sciences. 3rd ed. Sausalito, CA: University Science Books, 2000. Print. Page 840. csustan.
2. Surface Tension and Soap Bubbles . 03 February 1999. 25 June 2012. <http://www. chem. csustan. du/chem2000/Exp5/Bkg. htm>.
3. Prpich, A. , et. al. Tension at the Surface: Which Phase Is More Important, Liquid or Vapor?. 2009 Value of density @ 30 degrees Celsius taken from: <http://van. physics. illinois. edu/qa/listing. php? id=2170>
4. Value of ideal gas constant taken from: < Mohr, Peter J. ; Taylor, Barry N. ; Newell, David B. (2008). “CODATA
5. Recommended Values of the Fundamental Physical Constants: 2006″. Rev. Mod. Phys. 80 (2): 633–730. >. Value of surface tension of water @ 30 taken from: < Lange, p. 1663>

TM-SV-08-1 UNIVERSITY OF CALIFORNIA – COOPERATIVE EXTENSION 2008 SAMPLE COSTS TO PRODUCE PROCESSING TOMATOES TRANSPLANTED IN THE SACRAMENTO VALLEY Prepared by: Gene Miyao Karen M. Klonsky Pete Livingston UC Cooperative Extension Farm Advisor, Yolo, Solano, & Sacramento Counties UC Cooperative Extension Specialist, Department of Agricultural and Resource Economics, UC Davis UC Cooperative Extension Staff Research Associate, Department of Agricultural and Resource Economics, UC Davis

UC COOPERATIVE EXTENSION SAMPLE COSTS TO PRODUCE PROCESSING TOMATOES TRANSPLANTED In the Sacramento Valley – 2008 CONTENTS INTRODUCTION ………………………………………………………………………………………………………………………………………………………… 2 ASSUMPTIONS …………………………………………………………………………………………………………………………………………………………… CULTURAL PRACTICES AND MATERIAL INPUTS …………………………………………………………………………………………………………………. 3 CASH OVERHEAD ……………………………………………………………………………………………………………………………………………………………. 5 NON-CASH OVERHEAD …………………………………………………………………………………………………………………………………………………… REFERENCES ……………………………………………………………………………………………………………………………………………………………… 8 TABLE 1. COSTS PER ACRE TO PRODUCE PROCESSING TOMATOES …………………………………………………………………………………….. 10 TABLE 2. COSTS AND RETURNS PER ACRE TO PRODUCE PROCESSING TOMATOES ……………………………………………………………….. 12 TABLE 3.

MONTHLY CASH COSTS PER ACRE TO PRODUCE PROCESSING TOMATOES ……………………………………………………………. 14 TABLE 4. WHOLE FARM ANNUAL EQUIPMENT, INVESTMENT, AND BUSINESS OVERHEAD COSTS ………………………………………… 15 TABLE 5. HOURLY EQUIPMENT COSTS ……………………………………………………………………………………………………………………………. 17 TABLE 6. RANGING ANALYSIS ……………………………………………………………………………………………………………………………………….. 8 TABLE 7. COSTS AND RETURNS/ BREAKEVEN ANALYSIS ………………………………………………………………………………………………….. 19 TABLE 8. DETAILS OF O PERATIONS ………………………………………………………………………………………………………………………………… 20 INTRODUCTION The sample costs to produce transplanted processing tomatoes in the Sacramento Valley is based on the 2007 cost and returns study practices using 2008 prices and are presented in this study.

The price adjustments are for fuel, fertilizers, pesticides, water, labor rates, interest rates, and some cash overhead costs. This study is intended as a guide only, and can be used to make production decisions, determine potential returns, prepare budgets and evaluate production loans. Practices described are based on production practices considered typical for the crop and area, but may not apply to every situation. Sample costs for labor, materials, equipment, and custom services are based on current figures.

Blank columns, “Your Costs”, in Tables 1 and 2 are provided to enter actual costs of an individual farm operation. The hypothetical farm operations, production practices, overhead, and calculations are described under the assumptions. For additional information or an explanation of the calculations used in the study, call the Department of Agricultural and Resource Economics, University of California, Davis, (530) 752-2414 or the local UC Cooperative Extension office.

Two additional cost of production study for processing tomatoes grown in this region are also available: “Sample Costs To Produce Processing Tomatoes, Direct Seeded, In the Sacramento Valley – 2007”, and “Sample Costs To Produce Processing Tomatoes, Transplanted, In the Sacramento Valley – 2007”. Sample Cost of Production Studies for many commodities are available and can be requested through the Department of Agricultural Economics, UC Davis, (530) 752-2414. Current studies can be downloaded from the department website http://coststudies. ucdavis. edu/ or obtained from selected county UC Cooperative Extension offices.

The University of California prohibits discrimination or harassment of any person on the basis of race, color, national origin, religion, sex, gender identity , pregnancy (including childbirth, and medical conditions related to pregnancy or childbirth), physical or mental disability , medical condition (cancer-related or genetic characteristics), ancestry, marital status, age, sexual orientation, citizenship, or service in the uniformed services (as defined by the Uniformed Services Employment and Reemployment Rights Act of 1994: service in the uniformed services includes membership, application for membership, performance of service, application for service, or obligation for service in the uniformed services) in any of its programs or activities. University policy also prohibits reprisal or retaliation against any person in any of its programs or activities for making a complaint of discrimination or sexual harassment or for using or participating in the investigation or resolution process of any such complaint. University policy is intended to be consistent with the provisions of applicable State and Federal laws.

Inquiries regarding the University’s nondiscrimination policies may be directed to the Affirmative Action/Equal Opportunity Director, University of California, Agriculture and Natural Resources, 1111 Franklin Street, 6th Floor, Oakland, CA 94607, (510) 987-0096. 2008 Transplanted Processing Tomato Cost and Returns Study Sacramento Valley UC Cooperative Extension 2 ASSUMPTIONS The following assumptions refer to tables 1 to 8 and pertain to sample costs and returns to produce transplanted processing tomatoes in the Sacramento Valley. Input prices and interest rates are based on 2008 values. However, production practices were not updated from the 2007 study. Practices described are not recommendations by the University of California, but represent production practices considered typical of a well-managed farm for this crop and area.

Some of the costs and practices listed may not be applicable to all situations nor used during every production year and/or additional ones not indicated may be needed. Processing tomato cultural practices and material input costs will vary by grower and region, and can be significant. The practices and inputs used in the cost study serve as a guide only. The costs are shown on an annual, per acre basis. The use of trade names in this report does not constitute an endorsement or recommendation by the University of California nor is any criticism implied by omission of other similar products. Farm. The hypothetical field and row-crop farm consists of 2,900 non-contiguous acres of rented land.

Tomatoes are transplanted on 630 acres (70% of the tomato acreage) and direct seeded on 270 acres (30% of the tomato acreage) for a total of 900 acres. Two thousand acres are planted to other rotational crops including alfalfa hay, field corn, safflower, sunflower, dry beans and/or wheat. For direct seeded tomato operations, please refer to the study titled, “Sample Costs to Produce Processing Tomatoes, Directed Seeded, in the Sacramento Valley – 2007”. The grower also owns various investments such as a shop and an equipment yard. In this report, practices completed on less than 100% of the acres are denoted as a percentage of the total tomato crop acreage.

CULTURAL PRACTICES AND MATERIAL INPUTS Land Preparation. Primary tillage which includes laser leveling, discing, rolling, subsoiling, land planing, and listing beds is done from August through early November in the year preceding transplanting. To maintain surface grade, 4% of the acres are laser leveled each year. Fields are stubbledisced and rolled (using a rice roller). Fields are subsoiled in two passes to a 30-inch depth and rolled. A medium-duty disk with a flat roller following is used. Ground is smoothed in two passes with a triplane. Beds on five-foot centers are made with a six-bed lister, and then shaped with a bed-shaper cultivator.

Transplanting. Planting is spread over a three-month period (late March through early June) to meet contracted weekly delivery schedules at harvest. The transplants are planted in a single line per bed. Direct seed is for the early season and precedes transplanting. All of the 630 acres are custom planted with greenhouse-grown transplants. Costs for extra seed (15%) purchased to allow for less than 100% germination and for non-plantable transplants are included in the respective categories in Table 2. Fertilization. In the fall, ahead of listing beds, a soil amendment, gypsum at 3. 0 tons per acre is custom broadcast spread on 20% of the acres.

After listing, as part of the bed shaping operation, 11-52-0 is shanked into the beds at 100 pounds per acre. Prior to planting, liquid starter fertilizer, 8-24-6 plus zinc, is banded below the seed line at 15 gallons of material per acre. Nitrogen fertilizer, UN-32 at 150 pounds of N per acre is sidedress-banded at layby. Additional N is applied under special needs on 20% of acres as CAN 17 at 100 pounds of product per acre as a sidedress. Irrigation. In this study, water is calculated to cost $31. 92 per acre-foot or $2. 66 per acre-inch and is a combination of 1/2 well water ($47. 67 per acre-foot) and 1/2 canal delivered surface water ($16. 17 per acre-foot).

The irrigation costs shown in Tables 1 and 3 include water, pumping, and labor charges. The transplants receive a single sprinkler irrigation after planting. Prior to initial furrow irrigation, fields are all chiseled to 12 inches deep in the furrow. Eight furrow irrigations are applied during the season. In 2008 Transplanted Processing Tomato Cost and Returns Study Sacramento Valley UC Cooperative Extension 3 this study 3. 5 acre-feet (42 acre-inches) is applied to the crop – 2. 0 acre-inches by sprinkler and 40 acreinches by furrow. Although sub-surface drip irrigation is gaining in popularity, it is not used in this study. Pest Management. The pesticides and rates mentioned in this cost study are listed n Integrated Pest Management for Tomatoes and UC Pest Management Guidelines, Tomato. For more information on other pesticides available, pest identification, monitoring, and management visit the UC IPM website at www. ipm. ucdavis. edu. Written recommendations are required for many pesticides and are made by licensed pest control advisors. For information and pesticide use permits, contact the local county agricultural commissioner’s office. Weeds. Beginning in January, Roundup plus Goal is sprayed on the fallow beds to control emerged weeds and repeated later with Roundup only. Before planting, the beds are cultivated twice to control weeds and to prepare the seedbed.

Wilcox Performer conditions bed and applies starter fertilizer. Trifluralin is broadcast sprayed at 1. 0 pint per acre and incorporated with a power mulcher. To control nutsedge, Dual Magnum at 1. 5 pints of product per acre is added to trifluralin as a tank-mix and applied to 30% or 189 acres. Matrix is applied to 80% or 504 acres in an 18-inch band at a rate of 2. 0 ounces of material per acre to control a range of weeds. A combination of hand weeding and mechanical cultivation is also used for weed control. The crop is mechanically cultivated with sled-mounted cultivators three times during the season. A contract labor crew hand removes weeds.

Insects and Diseases. The primary insect pests of seedlings included in this study are flea beetle, darkling ground beetle, and cutworm. Foliage and fruit feeders included are tomato fruitworm, various armyworm species, russet mite, stinkbug, and potato aphid. Diseases are primarily bacterial speck, late blight, and blackmold fruit rot. A Kocide and Dithane tank mix for bacterial speck is applied to 30% of the acres. All of the above applications are made by ground. The following applications are made by aircraft. Sulfur dust for russet mite control is applied to 70% of the acres. Asana for general insect control is applied to 40% of the acres.

Confirm for worm control is applied to 100% of the acres. Bravo is applied in June to 5% of the acres for late blight control and again in September as a fruit protectant fungicide on 15% of the acres. Fruit Ripener. Ethrel, a fruit ripening agent, is applied by ground before harvest to 5% of the acres at 4. 0 pints per acre. Harvest. The fruit is mechanically harvested using one primary harvester for 90% of the acres and one older harvester for special harvest situations and as a backup to the primary harvester. Typically growers with this acreage of processing tomatoes own tractors, trailer dollies, generator-light machines, and harvest support equipment.

Four manual sorters, a harvester driver, and two bulk-trailer tractor operators are used per harvester. A seasonal average of 1. 5 loads per hour at 25 tons per load are harvested with two (one day and one night) shifts of 10 hours each. Harvest efficiency includes down time, scheduled daily breaks, and transportation between fields. The processor pays the transportation cost of the tomatoes from the field to the processing plant. Costs for harvest operations are shown in Tables 1, 3 and 7; the equipment used is listed in Tables 4 and 5. If tomatoes are custom harvested, harvest expenses are subtracted from harvest costs in Tables 1 and 3, and the custom harvest charges added.

The equipment for harvest operations is then subtracted from investment costs in Table 4. Growers may choose to own harvesting equipment, purchased either new or 2008 Transplanted Processing Tomato Cost and Returns Study Sacramento Valley UC Cooperative Extension 4 used, or hire a custom harvester. Many factors are important in deciding which harvesting option a grower uses. The options are discussed in “Acquiring Alfalfa Hay Harvest Equipment: A Financial Analysis of Alternatives”. Yields. County average annual tomato crop yields in the Sacramento Valley over the past ten years ranged from 26. 34 to 43. 00 tons per acre. The reporting counties are Colusa, Sacramento, Solano, Sutter, Yolo, and sometimes Glenn counties.

Butte and Tehama are the only two Sacramento Valley counties that do not report processing tomatoes. The weighted average yields for the Sacramento Valley from 1997 to 2006 are shown in Table A. In this study, a yield of 35 tons per acre is used. Table A. Sacramento Valley Yield and Price † Tons $ Year per acre per ton 2006 35. 44 59. 28 2005 34. 30 49. 81 2004 40. 51 48. 06 2003 33. 74 48. 82 2002 37. 64 48. 37 2001 35. 23 48. 49 2000 34. 44 49. 54 1999 34. 58 58. 68 1998 29. 90 53. 68 1997 33. 24 50. 85 Average 34. 90 51. 56 Returns. Customarily, growers produce tomatoes under contract with various food processing companies. County † Source: California Agricultural Commissioner Crop Reports. verage prices in the Sacramento Valley ranged from $45. 66 to $62. 00 per ton over the last 10 years and the Valley-wide weighted averages are shown in Table A. A price of $70. 00 per ton is used in this study to reflect the return price growers are currently receiving. Assessments. Under a state marketing order a mandatory assessment fee is collected and administered by the Processing Tomato Advisory Board (PTAB). The assessment pays for inspecting and grading fruit, and varies between inspection stations. In Yolo County, inspection fees range from $6. 36 to $8. 90 per load with an average of $6. 75. Growers and processors share equally in the fee; growers pay $3. 38 per load in this study.

A truckload is assumed to be 25 tons. Tomato growers are also assessed a fee for the Curly Top Virus Control Program (CTVCP) administered by the California Department of Food and Agriculture (CDFA). Growers in Yolo County (District 111) are charged $0. 019 per ton. Additionally, several voluntary organizations assess member growers. California Tomato Growers Association (CTGA) represents growers’ interest in negotiating contract prices with processors. CTGA membership charges are $0. 17 per ton. The California Tomato Research Institute funds projects for crop improvement. CTRI membership charges are $0. 07 per ton. Labor. Basic hourly wages for workers are $11. 56 and $8. 0 per hour for machine operators and nonmachine (irrigators and manual laborers) workers, respectively. Adding 36% for the employer’s share of federal and state payroll taxes, insurance and other benefits raises the total labor costs to $15. 72 per hour for machine operators and $10. 88 per hour for non-machine labor. The labor for operations involving machinery is 20% higher than the field operation time, to account for equipment set up, moving, maintenance, and repair. The current minimum wage is $8. 00 per hour. CASH OVERHEAD Cash overhead consists of various cash expenses paid out during the year that are assigned to the whole farm and not to a particular operation.

These costs include property taxes, interest on operating capital, office expense, liability and property insurance, share rent, supervisors’ salaries, field sanitation, crop insurance, and investment repairs. Employee benefits, insurance, and payroll taxes are included in labor costs and not in overhead. Cash overhead costs are shown in Tables 1, 2, 3, and 4. Property Taxes. Counties charge a base property tax rate of 1% on the assessed value of the property. In some counties special assessment districts exist and charge additional taxes on property including equipment, buildings, and improvements. For this study, county taxes are calculated as 1% of the average value of the property. Average value equals new cost plus salvage value divided by 2 on a per acre basis. 008 Transplanted Processing Tomato Cost and Returns Study Sacramento Valley UC Cooperative Extension 5 Interest o n Operating Capital. Interest on operating capital is based on cash operating costs and is calculated monthly until harvest at a nominal rate of 6. 75% per year. A nominal interest rate is the typical market cost of borrowed funds. Insurance. Insurance for farm investments varies depending on the assets included and the amount of coverage. Property insurance provides coverage for property loss and is charged at 0. 740% of the average value of the assets over their useful life. Liability insurance covers accidents on the farm and costs $1,438 for the entire farm or $0. 50 per acre. Office Expense.

Office and business expenses are estimated to be $50,489 for the entire farm or $17. 41 per acre. These expenses include office supplies, telephones, bookkeeping, accounting, legal fees, road maintenance, office and shop utilities, and miscellaneous administrative expenses. Share Rent. Rent arrangements will vary. The tomato land in this study is leased on a share-rent basis with the landowner receiving 12% of the gross returns. The land rented includes developed wells and irrigation system. Field Supervisors’ Salary. Supervisor salaries for tomatoes, including insurance, payroll taxes, and benefits, and are $94,500 per year for two supervisors.

Two thirds of the supervisors’ time is allocated to tomatoes. The costs are $70. 00 per acre. Any returns above total costs are considered returns on risk and investment to management (or owners). Field Sanitation. Sanitation services provide portable toilet and washing facilities for the ranch during the crop season. The cost includes delivery and weekly service. Costs will vary depending upon the crops and number of portable units required. Crop Insurance. The insurance protects the grower from crop losses due to adverse weather conditions, fire, unusual diseases and/or insects, wildlife, earthquake, volcanic eruption, and failure of the irrigation system.

The grower can choose the protection level at 50% to 75% of production history or county yields. In this study, no level is chosen. The cost shown in the study is the average of the costs paid by the growers who reviewed this study. NON-CASH OVERHEAD Non-cash overhead is calculated as the capital recovery cost for equipment and other farm investments. Although farm equipment used for processing tomatoes may be purchased new or used, this study shows the current purchase price for new equipment. The new purchase price is adjusted to 60% to reflect a mix of new and used equipment. Annual ownership costs (equipment and investments) are shown in Tables 1, 2, and 5.

They represent the capital recovery cost for investments on an annual per acre basis. Capital Recovery Costs. Capital recovery cost is the annual depreciation and interest costs for a capital investment. It is the amount of money required each year to recover the difference between the purchase price and salvage value (unrecovered capital). It is equivalent to the annual payment on a loan for the investment with the down payment equal to the discounted salvage value. This is a more complex method of calculating ownership costs than straight-line depreciation and opportunity costs, but more accurately represents the annual costs of ownership because it takes the time value of money into account (Boehlje and Eidman).

The formula for the calculation of the annual capital recovery costs is; Capital *# && # * ,% Purchase ” Salvage( ) %Recovery(/ + ,Salvage ) Interest/ % ( Pr ice Value Value Rate + . ‘ $ , / ‘. Factor +$ 2008 Transplanted Processing Tomato Cost and Returns Study ! Sacramento Valley UC Cooperative Extension 6 Salvage Value. Salvage value is an estimate of the remaining value of an investment at the end of its useful life. For farm machinery the remaining value is a percentage of the new cost of the investment (Boehlje and Eidman). The percent remaining value is calculated from equations developed by the American Society of Agricultural Engineers (ASAE) based on equipment type and years of life. The life in years is estimated by dividing the wear out life, as given by ASAE by the annual hours of use in this operation.

For other investments including irrigation systems, buildings, and miscellaneous equipment, the value at the end of its useful life is zero. The salvage value for land is equal to the purchase price because land does not depreciate. The purchase price and salvage value for certain equipment and investments are shown in Table 5. Capital Recovery Factor. Capital recovery factor is the amortization factor or annual payment whose present value at compound interest is 1. The amortization factor is a table value that corresponds to the interest rate and the life of the equipment. Interest Rate. The interest rate of 4. 25% used to calculate capital recovery cost is the effective long-term interest rate in January 2008.

The interest rate is used to reflect the long-term realized rate of return to these specialized resources that can only be used effectively in the agricultural sector. Equipment Costs. Equipment costs are composed of three parts: non-cash overhead, cash overhead, and operating costs. Some of the cost factors have been discussed in previous sections. The operating costs consist of repairs, fuel, and lubrication. The fuel, lube, and repair cost per acre for each operation in Table 1 is determined by multiplying the total hourly operating cost in Table 5 for each piece of equipment used for the selected operation by the hours per acre. Tractor time is 10% higher than implement time for a given operation to account for setup, travel and down time. Repairs, Fuel and Lube.

Repair costs are based on purchase price, annual hours of use, total hours of life, and repair coefficients formulated by the ASAE. Fuel and lubrication costs are also determined by ASAE equations based on maximum Power-Take-Off horsepower, and fuel type. Prices for on-farm delivery of diesel and unleaded gasoline are $3. 54 and $3. 57 per gallon, respectively. Irrigation System. Irrigation equipment owned by the grower consists of main lines, hand moved sprinklers, portable pumps, V-ditchers, and siphon tubes. Risk. Risks associated with processing tomato production are not assigned a production cost. All acres are contracted prior to harvest and all tonnage-time delivery contracts are assumed to have been met. No excess acres are grown to fulfill contracts.

While this study makes an effort to model a production system based on typical, real world practices, it cannot fully represent financial, agronomic and market risks which affect the profitability and economic viability of processing tomato production. Table Values. Due to rounding the totals may be slightly different from the sum of the components. 2008 Transplanted Processing Tomato Cost and Returns Study Sacramento Valley UC Cooperative Extension 7 REFERENCES American Society of Agricultural Engineers. 2003. American Society of Agricultural Engineers Standards Yearbook. Russell H. Hahn and Evelyn E. Rosentreter (ed. ) St. Joseph, Missouri. 41st edition. Barker, Doug.

California Workers’ Compensation Rating Data for Selected Agricultural Classifications as of January 2008. California Department of Insurance, Rate Regulation Branch. Boehlje, Michael D. , and Vernon R. Eidman. 1984. Farm Management. John Wiley and Sons. New York, NY. Blank, Steve, Karen Klonsky, Kim Norris, and Steve Orloff. 1992. Acquiring Alfalfa Hay Harvest Equipment: A Financial Analysis of Alternatives. University of California. Oakland, CA. Giannini Information Series No. 92-1. http://giannini. ucop. edu/InfoSeries/921-HayEquip. pdf. Internet accessed May, 2008. California State Automobile Association. 2008. Gas Price Averages 2007 – 2008.

AAA Press Room, San Francisco, CA. http://www. csaa. com/portal/site/CSAA/menuitem. 5313747aa611bd4e320cfad592278a0c/? vgnextoid= 8d642ce6cda97010VgnVCM1000002872a8c0RCRD. Internet accessed April, 2008. California State Board of equalization. Fuel Tax Division Tax Rates. http://www. boe. ca. gov/sptaxprog/spftdrates. htm. Internet accessed April, 2008. CDFA-California County Agricultural Commissioners, California Annual Agricultural Crop Reports. 1998 – 2007. California Department of Food and Agricultural, Sacramento, CA. http://www. nass. usda. gov/ca/bul/agcom/indexcac. htm. Internet accessed May, 2008. Energy Information Administration. 2008.

Weekly Retail on Highway http://tonto. eia. doe. gov/oog/info/gdu/gasdiesel. asp. Internet accessed April, 2008. Diesel Prices. Integrated Pest Management Education and Publications. 2008. “UC Pest Management Guidelines, Tomatoes. ” In M. L. Flint (ed. ) UC IPM Pest Management Guidelines. University of California. Division of Agriculture and Natural Resources. Oakland, CA. Publication 3339. http://www. ipm. ucdavis. edu/PMG/selectnewpest. tomatoes. html. Internet accessed May, 2008. Miyao, Gene, Karen M. Klonsky, and Pete Livingston. 2007. “Sample Costs To Produce Processing Tomatoes, Transplanted, In the Sacramento Valley – 2007”. University of California, Cooperative Extension.

Department of Agricultural and Resource Economics. Davis, CA. http://coststudies. ucdavis. edu/. Internet accessed April, 2008. Miyao, Gene, Karen M. Klonsky, and Pete Livingston. 2007. Sample Costs to Produce Processing Tomatoes, Direct Seeded, in the Sacramento Valley – 2007. University of California, Cooperative Extension. Department of Agricultural and Resource Economics. Davis, CA. http://coststudies. ucdavis. edu/. Internet accessed, April, 2008. 2008 Transplanted Processing Tomato Cost and Returns Study Sacramento Valley UC Cooperative Extension 8 Statewide Integrated Pest Management Project. 1998. Integrated Pest Management for Tomatoes. Fourth Edition. University of California.

Division of Agriculture and Natural Resources. Oakland, CA. Publication 3274. http://www. ipm. ucdavis. edu/PMG/selectnewpest. tomatoes. html. Internet accessed April, 2008. USDA-ERS. 2008. Farm Sector: Farm Financial Ratios. Agriculture and Rural Economics Division, ERS. USDA. Washington, DC. http://usda. mannlib. cornell. edu/reports/nassr/price/zapbb/agpran04. txt; Internet accessed January, 2008. ________________________ For information concerning the above or other University of California publications, contact UC DANR Communications Services at 800994-8849, online at http://anrcatalog. ucdavis. edu/InOrder/Shop/Shop. asp, or your local county UC Cooperative Extension office. 008 Transplanted Processing Tomato Cost and Returns Study Sacramento Valley UC Cooperative Extension 9 Table 1. UC COOPERATIVE EXTENSION COSTS PER ACRE TO PRODUCE TOMATOES SACRAMENTO VALLEY – 2008 TRANSPLANTED Labor Rate: $15. 72/hr. machine labor $10. 88/hr. non-machine labor Interest Rate: 6. 75% Yield per Acre: 35. 0 Ton Operation —————— Cash and Labor Costs per Acre —————–Time Labor Fuel, Lube Material Custom/ Total (Hrs/A) Cost & Repairs Cost Rent Cost 0. 00 0. 14 0. 42 0. 15 0. 36 0. 00 0. 10 0. 25 0. 08 0. 08 0. 26 1. 83 0. 17 0. 33 0. 00 0. 16 3. 00 0. 61 0. 33 0. 25 0. 25 0. 03 0. 04 10. 00 0. 00 0. 04 0. 00 0. 07 0. 00 0. 50 0. 00 0. 00 0. 0 0. 00 0. 32 0. 32 16. 42 0. 10 0. 93 0. 46 1. 49 0. 00 0. 00 0 3 8 3 7 0 2 5 1 1 10 39 3 6 0 3 33 12 6 5 5 1 1 109 0 1 0 1 0 9 0 0 0 0 12 6 212 2 58 32 92 0 0 344 0 18 53 10 22 0 6 12 3 3 19 145 7 13 0 6 0 21 13 15 12 1 2 0 0 2 0 3 0 17 0 0 0 0 8 0 122 4 177 34 215 0 0 482 0 0 0 0 0 79 0 42 12 13 0 146 36 13 354 9 18 0 112 0 0 5 0 107 1 0 15 20 0 0 5 4 27 2 0 0 727 0 0 0 0 14 14 887 7 0 0 0 0 1 0 0 0 0 0 8 0 0 165 0 0 0 0 0 0 0 0 0 0 0 6 0 50 0 3 1 6 0 0 0 231 0 0 0 0 0 0 239 7 20 61 13 29 81 8 59 16 17 28 338 46 33 519 19 51 32 131 20 17 6 3 216 1 3 21 24 50 27 7 4 33 2 20 6 1,292 6 235 66 308 14 14 66 2,017 1 17 0 25 70 294 6 4 6 423 2,440

Operation Preplant: Land Preparation – Laser Level – 4% of Acreage Land Preparation – Stubble Disc & Roll Land Preparation – Subsoil & Roll 2X Land Preparation – Disc & Roll Land Preparation – Triplane 2X Land Preparation – Apply Gypsum on 20% of Acreage Land Preparation – List Beds Land Preparation – Shape & Fertilize (11-52-0) Weed Control – Roundup & Goal Weed Control – Roundup Weed Control – Cultivate 2X TOTAL PREPLANT COSTS Cultural: Condition Bed & Starter Fertilizer Mulch Beds & Apply Treflan (& Dual on 30% of Acreage) Transplant Tomatoes Weed Control – Apply Matrix on 80% of Acreage Irrigate – Sprinklers 1X Weed Control – Cultivate 3X Fertilize – 150 Lbs N Sidedress Chisel Furrows Mulch Beds Disease Control – Bacterial Speck on 30% of Acreage Open Ditches Irrigate – Furrow 8X Disease Control – Late Blight on 5% of Acreage Close Ditches Mite Control – Sulfur on 70% of Acreage Fertilize – 20 Lbs N on 20% of Acreage Weed Control – Hand Hoe – Contract Train Vines Insect Control – Aphid on 40% of Acreage Disease Control – Fruit Rot on 15% of Acreage Insect Control – Worms Fruit Ripener – Ethrel on 5% of Acreage Pickup Truck Use (2 pickups) ATV Use TOTAL CULTURAL COSTS Harvest: Open Harvest Lane on 8% of Acreage Harvest In Field Hauling TOTAL HARVEST COSTS Assessment: Assessments/Fees TOTAL ASSESSMENT COSTS Interest on Operating Capital @ 6. 75% TOTAL OPERATING COSTS/ACRE CASH OVERHEAD: Liability Insurance Office Expense Field Sanitation Crop Insurance Field Supervisors’ Salary (2) Land Rent @ 12% of Gross Returns Property Taxes Property Insurance Investment Repairs TOTAL CASH OVERHEAD COSTS TOTAL CASH COSTS/ACRE Your Cost 008 Transplanted Processing Tomato Cost and Returns Study Sacramento Valley UC Cooperative Extension 10 UC COOPERATIVE EXTENSION Table 1 continued NON-CASH OVERHEAD: Investment Shop Building Storage Building Fuel Tanks & Pumps Shop Tools Booster Pumps Sprinkler Pipe Main Line Pipe – 10″ Semi Truck & Lowbed Trailer Pipe Trailers Truck-Service – 2 Ton Generators & Light Fuel Wagons Closed Mix System Siphon Tubes Implement Carrier Equipment TOTAL NON-CASH OVERHEAD COSTS TOTAL COSTS/ACRE Per producing Acre 25 10 8 5 21 52 28 12 12 13 3 1 2 4 3 755 953 — Annual Cost -Capital Recovery 2 1 1 0 2 6 3 1 1 3 1 0 0 0 0 94 116 2 1 1 0 2 6 3 1 1 3 1 0 0 0 0 94 116 2,555 008 Transplanted Processing Tomato Cost and Returns Study Sacramento Valley UC Cooperative Extension 11 Table 2. UC COOPERATIVE EXTENSION COSTS and RETURNS PER ACRE to PRODUCE TOMATOES SACRAMENTO VALLEY – 2008 TRANSPLANTED Labor Rate: $15. 72/hr. machine labor $10. 88/hr. non-machine labor Interest Rate: 6. 75% Yield per Acre: 35. 0 Ton Price or Value or Cost/Unit Cost/Acre 70. 00 2,450 2,450 Your Cost Quantity/Acre Unit GROSS RETURNS Processing Tomatoes 35. 00 TOTAL GROSS RETURNS FOR PROCESSING TOMATOES OPERATING COSTS Custom: Laser Level 0. 04 Gypsum Application 0. 20 Transplanting 8. 70 Air Application – Spray 10 Gal/Acre 1. 60 Air Application – Dust 28. 0 Fertilizer: Gypsum 0. 60 11-52-0 100. 00 8-24-6 15. 00 Zinc Chelate 6% 2. 00 UN-32 150. 00 CAN 17 118. 00 Herbicide: Roundup Ultra 2. 50 Goal 2XL 3. 00 Dual Magnum 0. 45 Treflan HFP 1. 00 Matrix DF 0. 48 Seed: Tomato Seed 10. 01 Transplant: Transplants – Growing 8. 70 Irrigation: Water 42. 00 Pump – Fuel, Lube, & Repairs 1. 00 Fungicide: Kocide 101 0. 60 Dithane DF 0. 60 Sulfur, Dust 98% 28. 00 Insecticide: Bravo Weatherstik 0. 60 Warrior T 1. 54 Confirm 12. 00 Contract: Contract Labor 5. 00 Growth Regulator: Ethrel 0. 03 Assessment: CDFA-CTVP 35. 00 CTGA 35. 00 CTRI 35. 00 PTAB 35. 00 Labor (machine) 9. 34 Labor (non-machine) 18. 08 Fuel – Gas 1. 5 Fuel – Diesel 77. 61 Lube Machinery repair Interest on Operating Capital @ 6. 75% TOTAL OPERATING COSTS/ACRE NET RETURNS ABOVE OPERATING COSTS/ACRE Ton Acre Ton Thou Acre Lb Ton Lb Lb Pint Lb N Lb Pint FlOz Pint Pint Oz Thou Thou AcIn Acre Lb Lb Lb Pint FlOz FlOz Hour Gal Ton Ton Ton Ton Hrs Hrs Gal Gal 165. 00 7. 00 19. 00 6. 25 0. 20 132. 00 0. 419 2. 28 0. 913 0. 745 0. 171 8. 59 1. 03 18. 63 4. 84 19. 25 11. 00 28. 00 2. 67 13. 00 3. 62 3. 89 0. 55 7. 85 3. 05 2. 23 9. 99 63. 00 0. 019 0. 17 0. 07 0. 135 15. 72 10. 88 3. 57 3. 54 7 1 165 10 6 79 42 34 2 112 20 21 3 8 5 9 110 244 112 13 2 2 15 5 5 27 50 2 1 6 2 5 147 197 7 275 42 159 66 2,017 406 008 Transplanted Processing Tomato Cost and Returns Study Sacramento Valley UC Cooperative Extension 12 UC COOPERATIVE EXTENSION Table 2 continued CASH OVERHEAD COSTS: Liability Insurance Office Expense Field Sanitation Crop Insurance Field Supervisors’ Salary (2) Land Rent @ 12% of Gross Returns Property Taxes Property Insurance Investment Repairs TOTAL CASH OVERHEAD COSTS/ACRE TOTAL CASH COSTS/ACRE NON-CASH OVERHEAD COSTS (CAPITAL RECOVERY): Shop Building Storage Building Fuel Tanks & Pumps Shop Tools Booster Pumps Sprinkler Pipe Main Line Pipe – 10″ Semi Truck & Lowbed Trailer Pipe Trailers Truck-Service – 2 Ton Generators & Light Fuel Wagons Closed Mix System

Siphon Tubes Implement Carrier Equipment TOTAL NON-CASH OVERHEAD COSTS/ACRE TOTAL COSTS/ACRE NET RETURNS ABOVE TOTAL COSTS/ACRE 1 17 0 25 70 294 6 4 6 423 2,440 2 1 1 0 2 6 3 1 1 3 1 0 0 0 0 94 116 2,555 -105 2008 Transplanted Processing Tomato Cost and Returns Study Sacramento Valley UC Cooperative Extension 13 Table 3. UC COOPERATIVE EXTENSION MONTHLY CASH COST PER ACRE TO PRODUCE TOMATOES SACRAMENTO VALLEY – 2008 TRANSPLANTED SEP 07 7 20 61 13 29 81 8 59 16 17 28 62 46 33 519 19 51 14 20 17 6 2 54 2 54 3 21 24 50 27 7 4 33 2 2 0 42 2 12 6 21 14 14 11 87 OCT 07 NOV 07 DEC 07 JAN 08 FEB MAR 08 08 APR MAY 08 08 JUN 08 JUL AUG 08 08 SEP 08 TOTAL

Beginning SEP 07 Ending SEP 08 Preplant: Laser Level – 4% of Acreage Land Prep – Stubble Disc & Roll Land Prep – Subsoil & Roll 2X Land Prep – Disc & Roll Land Prep – Triplane 2X Land Prep – Apply Gypsum on 20% of Acreage Land Prep – List Beds Land Prep – Shape Beds & Fertilize Weed Control – Roundup & Goal Weed Control – Roundup Weed Control – Cultivate 2X TOTAL PREPLANT COSTS Cultural: Condition Bed & Starter Fertilizer Mulch Beds & Apply Herbicide Transplant Tomatoes Weed Control – Apply Matrix on 80% of Acreage Irrigate – Sprinklers 1X Weed Control – Cultivate 2X Fertilize – 150 Lbs N – Sidedress Chisel Furrows Mulch Beds Disease Control – Bacterial Speck – 30% of Acreage Open Ditches Irrigate – Furrow 8X Disease Control – Late Blight 5% of Acreage Close Ditches Mite Control – Sulfur 70% of Acreage Fertilize – 20 Lb N 20% of Acreage Weed Control – Hand Hoe Train Vines Insect Control – Aphids 40% of Acreage Disease Control – Fruit Rot 15% of Acreage Insect Control – Worms – Confirm Fruit Ripener – Ethrel 5% of Acreage Pickup Truck Use (2 pickups) ATV Use TOTAL CULTURAL COSTS Harvest: Open Harvest Lane 8% of Acreage Harvest In Field Hauling TOTAL HARVEST COSTS Assessment: Assessments/Fees TOTAL ASSESSMENT COSTS Interest on Operating Capital @ 6. 5% TOTAL OPERATING COSTS/ACRE OVERHEAD: Liability Insurance Office Expense Field Sanitation Crop Insurance Field Supervisors’ Salary (2) Land Rent @ 12% of Gross Returns Property Taxes Property Insurance Investment Repairs TOTAL CASH OVERHEAD COSTS TOTAL CASH COSTS/ACRE 210 67 7 20 61 13 29 81 8 59 16 17 28 338 46 33 519 19 51 32 131 20 17 6 3 216 1 3 21 24 50 27 7 4 33 2 20 6 1,292 6 235 66 308 14 14 66 2,017 1 17 0 25 70 294 6 4 6 423 2,440 7 131 10 54 54 1 2 0 2 2 0 2 2 0 2 2 0 2 2 0 48 2 0 2 2 0 35 2 0 686 2 0 211 2 0 57 2 0 200 2 111 31 144 2 0 2 2 111 29 143 1 213 2 70 2 4 2 4 2 112 1 1 0 25 5 2 4 2 37 6 693 7 219 8 65 10 354 11 155 1 0 5 1 0 5 1 0 5 1 0 5 1 0 5 3 2 0 12 16 1 0 5 1 0 5 1 0 5 1 0 5 1 0 5 3 2 0 12 367 1 0 5 1 0 5 294 0 7 220 0 7 78 0 7 11 0 7 11 0 33 145 0 7 44 0 7 700 0 7 226 0 7 72 0 7 162 301 388 2008 Transplanted Processing Tomato Cost and Returns Study Sacramento Valley

UC Cooperative Extension 14 Table 4. UC COOPERATIVE EXTENSION WHOLE FARM ANNUAL EQUIPMENT, INVESTMENT, AND BUSINESS OVERHEAD COSTS SACRAMENTO VALLEY – 2008 TRANSPLANTED ANNUAL EQUIPMENT COSTS – Cash Overhead Insurance Taxes 318 430 331 448 477 645 828 1,118 1,060 1,433 211 285 17 24 58 78 45 60 22 30 132 178 58 79 22 29 245 330 195 263 36 49 209 283 1,265 1,710 99 134 91 123 72 97 72 97 9 12 62 83 62 83 35 47 10 14 10 14 10 14 10 14 9 12 175 236 6 8 6 8 6 8 6 8 97 131 70 94 20 26 6,465 8,737 3,879 5,242 Description 110 HP 2WD Tractor 130 HP 2WD Tractor 155 HP 2WD Tractor 200 HP Crawler 425 HP Crawler 92 HP 2WD Tractor ATV Bed Shaper – 3 Row Cultivator –

Alloway 3 Row Cultivator – Perfecta 3 Row Cultivator – Performer 3 Row Cultivator – 3 Row Cultivator – Sled 3 Row Disc – Stubble 18′ Disc – Finish 25′ Ditcher – V Harvester Tomato – Used Harvester -Tomato Lister – 3 Row Mulcher – 15′ Pickup Truck – 1/2 Ton Pickup Truck – 3/4 Ton Rear Blade – 8′ Rice Roller – 18′ Flat Roller – 18′ Ringroller – 30′ Saddle Tank – 300 Gallon Saddle Tank – 300 Gallon Saddle Tank – 300 Gallon Saddle Tank – 300 Gallon Spray Boom – 25′ Subsoiler – 16′ – 9 Shank Trailer Dolly Trailer Dolly Trailer Dolly Trailer Dolly Triplane – 16′ Vine Diverter Vine Trainer TOTAL 60% of New Cost * * Used to reflect a mix of new and used equipment. Yr 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 Price 66,445 69,163 99,594 172,650 221,197 44,015 4,017 13,292 10,236 5,100 30,281 11,868 4,980 49,847 44,743 8,631 46,108 331,980 20,176 20,507 17,655 17,655 2,269 14,139 14,139 7,952 2,374 2,374 2,374 2,374 1,781 35,605 1,451 1,451 1,451 1,451 22,253 16,046 4,800 1,444,424 866,654 Yrs Life 10 10 10 10 10 10 10 10 10 10 10 5 10 5 10 12 8 8 5 9 7 7 15 10 10 10 10 10 10 10 5 5 15 15 15 15 10 10 10

Salvage Value 19,627 20,430 29,418 50,998 65,338 13,001 710 2,351 1,810 902 5,355 3,866 881 16,237 7,912 1,195 10,411 10,000 6,572 4,098 1,766 1,766 218 2,500 2,500 1,406 420 420 420 420 580 11,598 139 139 139 139 3,935 2,838 480 302,935 181,761 Capital Recovery 6,678 6,952 10,010 17,353 22,233 4,424 443 1,466 1,129 562 3,339 1,974 549 8,293 4,934 855 5,799 48,743 3,357 2,406 2,747 2,747 197 1,559 1,559 877 262 262 262 262 296 5,923 126 126 126 126 2,454 1,769 560 173,739 104,243 Total 7,427 7,731 11,133 19,299 24,726 4,920 484 1,602 1,234 615 3,649 2,111 600 8,868 5,392 940 6,291 51,718 3,589 2,620 2,916 2,916 219 1,704 1,704 958 286 286 286 286 317 6,334 140 140 140 140 2,682 1,934 606 188,941 113,364 2008 Transplanted Processing Tomato Cost and Returns Study Sacramento Valley UC Cooperative Extension 15

UC COOPERATIVE EXTENSION Table 4 continued ANNUAL INVESTMENT COSTS —— Cash Overhead —–Insurance Taxes Repairs 243 18 89 9 31 40 328 132 147 294 59 45 614 118 157 2,325 329 24 121 12 42 54 444 178 199 397 80 61 830 160 212 3,142 1,643 221 439 44 210 487 2,219 700 531 722 145 313 4,152 586 3,860 16,272 Description INVESTMENT Booster Pumps Closed Mix System Fuel Tanks & Pumps Fuel Wagons Generators & Light Implement Carrier Main Line Pipe – 10″ Pipe Trailers Semi Truck & Lowbed Trailer Shop Building Shop Tools Siphon Tubes Sprinkler Pipe Storage Building Truck-Service – 2 Ton TOTAL INVESTMENT Price 59,757 4,412 21,949 2,186 7,620 9,742 80,676 35,000 36,170 72,168 14,465 11,066 150,980 29,112 38,600 573,903

Yrs Life 10 10 20 10 5 15 10 10 15 25 20 15 10 20 5 Salvage Value 5,976 441 2,195 219 762 974 8,068 700 3,617 7,217 1,447 1,107 15,098 2,911 3,860 54,592 Capital Recovery 6,967 514 1,579 255 1,584 844 9,407 4,311 3,133 4,575 1,041 958 17,604 2,095 8,022 62,889 Total 9,182 778 2,228 320 1,867 1,424 12,398 5,322 4,010 5,988 1,324 1,377 23,201 2,959 12,252 84,629 ANNUAL BUSINESS OVERHEAD Units/ Farm 900 2,900 900 900 2,900 2,900 Price/ Unit 25. 00 0. 48 70. 00 294. 00 0. 50 17. 41 Total Cost 22,500 1,392 63,000 264,600 1,450 50,489 Description Crop Insurance Field Sanitation Field Supervisors’ Salary (2) Land Rent @ 12% of Gross Returns Liability Insurance Office Expense

Unit Acre Acre Acre Acre Acre Acre 2008 Transplanted Processing Tomato Cost and Returns Study Sacramento Valley UC Cooperative Extension 16 Table 5. UC COOPERATIVE EXTENSION HOURLY EQUIPMENT COSTS SACRAMENTO VALLEY – 2008 TRANSPLANTED ——————- COSTS PER HOUR —————————- Cash Overhead ——– Operating ——-InsurFuel & Total Total ance Taxes Repairs Lube Oper. Costs/Hr. 0. 13 0. 18 3. 12 25. 99 29. 11 32. 20 0. 17 0. 22 3. 25 30. 71 33. 96 37. 82 0. 24 0. 32 4. 67 36. 62 41. 29 46. 86 0. 31 0. 42 4. 63 47. 25 51. 88 59. 12 0. 40 0. 54 5. 93 100. 40 106. 33 115. 61 0. 11 0. 14 2. 06 30. 71 32. 77 35. 24 0. 05 0. 07 1. 09 0. 0 1. 09 2. 54 0. 17 0. 24 2. 87 0. 00 2. 87 7. 69 0. 13 0. 18 2. 21 0. 00 2. 21 5. 92 0. 07 0. 09 1. 05 0. 00 1. 05 2. 90 0. 35 0. 47 6. 25 0. 00 6. 25 15. 98 0. 07 0. 09 2. 68 0. 00 2. 68 5. 05 0. 03 0. 05 1. 08 0. 00 1. 08 2. 03 0. 37 0. 50 8. 52 0. 00 8. 52 21. 85 0. 59 0. 79 7. 43 0. 00 7. 43 23. 64 0. 13 0. 18 2. 42 0. 00 2. 42 5. 84 0. 63 0. 85 2. 08 61. 07 63. 15 82. 07 1. 09 1. 47 124. 44 61. 07 185. 51 229. 90 0. 15 0. 21 4. 24 0. 00 4. 24 9. 76 0. 15 0. 20 2. 36 0. 00 2. 36 6. 67 0. 16 0. 22 1. 27 11. 97 13. 24 19. 81 0. 16 0. 22 1. 27 11. 97 13. 24 19. 81 0. 04 0. 06 0. 31 0. 00 0. 31 1. 30 0. 19 0. 25 1. 63 0. 00 1. 63 6. 76 0. 14 0. 9 1. 63 0. 00 1. 63 5. 52 0. 10 0. 14 0. 91 0. 00 0. 91 3. 79 0. 03 0. 04 0. 64 0. 00 0. 64 1. 47 0. 13 0. 17 0. 64 0. 00 0. 64 4. 14 0. 05 0. 07 0. 64 0. 00 0. 64 2. 00 0. 02 0. 02 0. 64 0. 00 0. 64 1. 07 0. 02 0. 02 0. 49 0. 00 0. 49 1. 12 0. 26 0. 35 8. 32 0. 00 8. 32 17. 83 0. 01 0. 01 0. 11 0. 00 0. 11 0. 28 0. 01 0. 01 0. 11 0. 00 0. 11 0. 28 0. 01 0. 01 0. 11 0. 00 0. 11 0. 28 0. 01 0. 01 0. 11 0. 00 0. 11 0. 28 0. 16 0. 21 3. 43 0. 00 3. 43 7. 74 0. 17 0. 23 2. 78 0. 00 2. 78 7. 57 0. 04 0. 05 2. 88 0. 00 2. 88 4. 03 Yr 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07 07

Description 110 HP 2WD Tractor 130 HP 2WD Tractor 155 HP 2WD Tractor 200 HP Crawler 425 HP Crawler 92 HP 2WD Tractor ATV Bed Shaper – 3 Row Cultivator – Alloway 3 Row Cultivator – Perfecta 3 Row Cultivator – Performer 3 Row Cultivator – 3 Row Cultivator – Sled 3 Row Disc – Stubble 18′ Disc – Finish 25′ Ditcher – V Harvester Tomato – Used Harvester -Tomato Lister – 9 Row Mulcher – 15′ Pickup Truck – 1/2 Ton Pickup Truck – 3/4 Ton Rear Blade – 8′ Rice Roller – 18′ Flat Roller – 18′ Ringroller – 30′ Saddle Tank – 300 Gallon Saddle Tank – 300 Gallon Saddle Tank – 300 Gallon Saddle Tank – 300 Gallon Spray Boom – 25′ Subsoiler – 16′ – 9 Shank Trailer Dolly Trailer Dolly Trailer Dolly Trailer Dolly Triplane – 16′ Vine Diverter Vine Trainer Actual Hours Capital Used Recovery 1,443. 2 2. 78 1,200. 0 3. 48 1,199. 3 5. 01 1,599. 4 6. 51 1,599. 8 8. 34 1,199. 2 2. 21 199. 5 1. 33 199. 5 4. 41 199. 8 3. 39 199. 8 1. 69 225. 1 8. 90 533. 0 2. 22 380. 0. 87 399. 2 12. 46 199. 5 14. 84 165. 2 3. 10 199. 4 17. 45 699. 0 41. 84 390. 0 5. 16 365. 4 3. 95 266. 5 6. 18 266. 5 6. 18 132. 2 0. 89 199. 2 4. 70 262. 5 3. 56 199. 5 2. 64 206. 6 0. 76 49. 1 3. 20 126. 0 1. 25 401. 9 0. 39 299. 4 0. 59 399. 5 8. 90 499. 6 0. 15 499. 7 0. 15 499. 3 0. 15 499. 7 0. 15 373. 8 3. 94 241. 9 4. 39 315. 0 1. 07 2008 Transplanted Processing Tomato Cost and Returns Study Sacramento Valley UC Cooperative Extension 17 Table 6. UC COOPERATIVE EXTENSION RANGING ANALYSIS SACRAMENTO VALLEY – 2008 TRANSPLANTED COSTS PER ACRE AT VARYING YIELDS FOR PROCESSING TOMATOES YIELD (TONS/ACRE) 26. 0 29. 0 32. 0 35. 0 38. 0 41. OPERATING COSTS/ACRE: Preplant Cost 338 338 338 338 338 338 Cultural Cost 1292 1,292 1,292 1,292 1,292 1,292 Harvest Cost 228 255 281 308 334 360 Assessment Cost 14 14 14 14 14 14 Interest on Operating Capital TOTAL OPERATING COSTS/ACRE TOTAL OPERATING COSTS/TON CASH OVERHEAD COSTS/ACRE TOTAL CASH COSTS/ACRE TOTAL CASH COSTS/TON NON-CASH OVERHEAD COSTS/ACRE TOTAL COSTS/ACRE TOTAL COSTS/TON 65 1937 74 422 2359 91 113 2472 95 65 1,964 68 422 2,386 82 114 2,500 86 65 1,990 62 423 2,413 75 115 2,528 79 66 2,017 58 423 2,440 70 116 2,555 73 66 2,044 54 423 2,466 65 117 2,583 68 66 2,071 51 423 2,493 61 117 2,611 64 44. 0 338 1,292 387 14 67 2,097 48 423 2,520 57 118 2,638 60

NET RETURNS PER ACRE ABOVE OPERATING COSTS FOR PROCESSING TOMATOES PRICE YIELD (DOLLARS/TON) (TONS/ACRE) Processing Tomatoes 26. 0 29. 0 32. 0 35. 0 38. 0 41. 0 44. 0 55. 00 -507 -369 -230 -92 46 184 323 60. 00 -377 -224 -70 83 236 389 543 65. 00 -247 -79 90 258 426 594 763 70. 00 -117 66 250 433 616 799 983 75. 00 13 211 410 608 806 1,004 1,203 80. 00 143 356 570 783 996 1,209 1,423 85. 00 273 501 730 958 1,186 1,414 1,643 NET RETURNS PER ACRE ABOVE CASH COSTS FOR PROCESSING TOMATOES PRICE YIELD (DOLLARS/TON) (TONS/ACRE) Processing Tomatoes 26. 0 29. 0 32. 0 35. 0 38. 0 41. 0 44. 0 55. 00 -929 -791 -653 -515 -376 -238 -100 60. 00 -799 -646 -493 -340 -186 -33 120 65. 0 -669 -501 -333 -165 4 172 340 70. 00 -539 -356 -173 10 194 377 560 75. 00 -409 -211 -13 185 384 582 780 80. 00 -279 -66 147 360 574 787 1,000 85. 00 -149 79 307 535 764 992 1,220 NET RETURNS PER ACRE ABOVE TOTAL COSTS FOR PROCESSING TOMATOES PRICE YIELD (DOLLARS/TON) (TONS/ACRE) Processing Tomatoes 26. 0 29. 0 32. 0 35. 0 38. 0 41. 0 44. 0 55. 00 -1,042 -905 -768 -630 -493 -356 -218 60. 00 -912 -760 -608 -455 -303 -151 2 65. 00 -782 -615 -448 -280 -113 54 222 70. 00 -652 -470 -288 -105 77 259 442 75. 00 -522 -325 -128 70 267 464 662 80. 00 -392 -180 32 245 457 669 882 85. 00 -262 -35 192 420 647 874 1,102 2008 Transplanted Processing Tomato Cost and Returns Study

Sacramento Valley UC Cooperative Extension 18 Table 7. UC COOPERATIVE EXTENSION COSTS AND RETURNS/ BREAKEVEN ANALYSIS SACRAMENTO VALLEY – 2008 TRANSPLANTED COSTS AND RETURNS – PER ACRE BASIS 1. Gross Returns Crop Processing Tomatoes 2,450 2,017 2. Operating Costs 3. Net Returns Above Oper. Costs (1-2) 433 4. Cash Costs 2,440 5. Net Returns Above Cash Costs (1-4) 10 6. Total Costs 2,555 7. Net Returns Above Total Costs (1-6) -105 COSTS AND RETURNS – TOTAL ACREAGE 1. Gross Returns Crop Processing Tomatoes 1,543,500 2. Operating Costs 1,270,748 3. Net Returns Above Oper. Costs (1-2) 272,752 4. Cash Costs 1,536,994 5. Net Returns Above Cash Costs (1-4) 6,506 6.

Total Costs 1,609,965 7. Net Returns Above Total Costs (1-6) -66,465 BREAKEVEN PRICES PER YIELD UNIT Base Yield (Units/Acre) 35. 0 Yield Units Ton ——– Breakeven Price To Cover ——-Operating Cash Total Costs Costs Costs ———— $ per Yield Unit ———–57. 63 69. 70 73. 01 CROP Processing Tomatoes BREAKEVEN YIELDS PER ACRE Yield Units Ton Base Price ($/Unit) 70. 00 ——– Breakeven Yield To Cover ——-Operating Cash Total Costs Costs Costs ———– Yield Units / Acre ———-28. 8 34. 9 36. 5 CROP Processing Tomatoes 2008 Transplanted Processing Tomato Cost and Returns Study Sacramento Valley UC Cooperative Extension 19 Table 8.

UC COOPERATIVE EXTENSION DETAILS OF OPERATIONS SACRAMENTO VALLEY – 2008 TRANSPLANTED Operation Laser Level – 4% Of Acreage Land Prep – Stubble Disc & Roll Land Prep – Subsoil & Roll 2X Land Prep – Disc & Roll Land Prep – Triplane 2X Land Prep – Apply Gypsum on 20% of Acreage Land Prep – List Beds Land Prep – Shape Beds & Fertilize Weed Control – Roundup & Goal Weed Control – Roundup Weed Control – Cultivate 2X Condition Beds & Apply Starter Fertilizer Power Mulch & Apply Herbicides – Treflan (& Dual on 30% of Acreage) Transplant Tomatoes Operation Month September September Tractor/ Power Unit Custom 425 HP Crawler Implement Laser Level Disc – Stubble 18′ Rice Roller – 18′ Subsoiler – 16′ – 9 Shank Disc – Finish 25′ Ringroller – 30′ Triplane – 16′ Broadcast Material Material Rate/Acre Unit 0. 04 Acre September 425 HP Crawler 200 HP Crawler September 200 HP Crawler September Gypsum Application October October January January January 200 HP Crawler 155 HP 2WD Tractor 130 HP 2WD Tractor 130 HP 2WD Tractor 110 HP 2WD Tractor 92 HP 2WD Tractor 110 HP 2WD Tractor 130 HP 2WD Tractor Custom

Gypsum Lister – 9 Row Bed Shaper – 3 Row Saddle Tank – 300 Gallon Saddle Tank – 300 Gallon Spray Boom – 25′ Saddle Tank – 300 Gallon Spray Boom – 25′ Cultivator – Alloway 3 Row Cultivator – Perfecta 3 Row Cultivator – Performer 3 Row Mulcher – 15′ Saddle Tank – 300 Gallon 0. 20 Ton 11-52-0 Zinc Chelate Roundup Ultra Goal 2 XL Roundup Ultra 100. 00 2. 00 1. 00 3. 00 1. 50 Lb Pint Pint FlOz Pint January March April Weed Control – Apply Matrix on 80% of Acreage Irrigate – Sprinklers 1X Weed Control – Cultivate 3X April April April April May May April May April April July April May June July June 130 HP 2WD Tractor Fertilize – 150 Lbs N Sidedress Chisel Furrows Mulch Beds Disease Control – Bacterial Speck – on 30% of Acreage Open Ditches Irrigate – Furrow 8X 10 HP 2WD Tractor 110 HP 2WD Tractor 110 HP 2WD Tractor 130 HP 2WD Tractor 200 HP Crawler 155 HP 2WD Tractor 130 HP 2WD Tractor 200 HP Crawler 200 HP Crawler Saddle Tank – 300 Gallon Cultivator – Sled 3 Row Labor Cultivator – Sled 3 Row Cultivator – Sled 3 Row Cultivator – 3 Row Cultivator – Sled 3 Row Saddle Tank – 300 Gallon Cultivator – 3 Row Cultivator – Sled 3 Row Saddle Tank – 300 Gallon Ditcher – V Ditcher – V Labor Labor Labor Labor 8-24-6 Treflan HFP Dual Magnum Tomato Seed Transplants – Growing Transplanting Matrix DF Water 15. 00 1. 00 0. 45 10. 44 8. 70 8. 70 0. 48 2. 00 Lb Pint Pint Thou Thou Thou Oz AcIn UN-32 150. 00 Lbs N Kocide 101 Dithane DF 0. 60 0. 60 Lb Lb Disease Control – Late Blight on 5% of Acreage Close Ditches

Air Application Spray 200 HP Crawler 200 HP Crawler Air Application Dust 130 HP 2WD Tractor Contract Labor 110 HP 2WD Tractor Air Application Spray Rear Blade – 8′ Rear Blade – 8′ Cultivator – Sled 3 Row Saddle Tank – 300 Gallon Vine Trainer Water Water Water Water Bravo Weatherstik 10. 00 10. 00 10. 00 10. 00 0. 15 AcIn AcIn AcIn AcIn Pint July July Mite Control – Sulfur on 70% of Acreage July Fertilize – 20 Lbs N on 20% of Acreage July Weed Control – Hand Hoe Train Vines Insect Control – Aphids on 40% of Acreage Disease Control – Fruit Rot on 15% of Acreage Insect Control – Worms Fruit Ripener – Ethrel on 5% of Acreage Open Harvest Lane on 8% of Acreage July July July Sulfur, Dust 98% CAN 17 Labor Warrior T Bravo Weatherstik Confirm 28. 00 118. 00 5. 00 1. 54 0. 45 12. 00 0. 03

GRAAD 12 NATIONAL SENIOR CERTIFICATE GRADE 12 PHYSICAL SCIENCES: PHYSICS (P1) FEBRUARY/MARCH 2012 MARKS: 150 TIME: 3 hours This question paper consists of 16 pages and 3 data sheets. Copyright reserved Please turn over Physical Sciences/P1 2 NSC DBE/Feb. –Mar. 2012 INSTRUCTIONS AND INFORMATION 1. 2. 3. Write your centre number and examination number in the appropriate spaces on the ANSWER BOOK. Answer ALL the questions in the ANSWER BOOK. This question paper consists of TWO sections: SECTION A (25) SECTION B (125) 4. 5. 6. 7. 8. 9. You may use a non-programmable calculator.

You may use appropriate mathematical instruments. Number the answers correctly according to the numbering system used in this question paper. YOU ARE ADVISED TO USE THE ATTACHED DATA SHEETS. Give brief motivations, discussions, et cetera where required. Round off your final numerical answers to a minimum of TWO decimal places. Copyright reserved Please turn over Physical Sciences/P1 3 NSC DBE/Feb. –Mar. 2012 SECTION A QUESTION 1: ONE-WORD ITEMS Give ONE word/term for each of the following descriptions. Write only the word/term next to the question number (1. 1–1. 5) in the ANSWER BOOK. 1. 1 1. 2 1. 3 1. 4 1. The type of energy an object has due to its motion The phenomenon which occurs when two light waves meet at a given point The unit of measurement of electrical resistance The basic principle on which electric generators function The type of line spectrum observed when electrons in an atom move from the excited state to the ground state (1) (1) (1) (1) (1) [5] QUESTION 2: MULTIPLE-CHOICE QUESTIONS Four options are provided as possible answers to the following questions. Each question has only ONE correct answer. Write only the letter (A–D) next to the question number (2. 1–2. 10) in the ANSWER BOOK. 2. A car of mass m collides head-on with a truck of mass 2m. If the car exerts a force of magnitude F on the truck during the collision, the magnitude of the force that the truck exerts on the car is … A B C D 1 F 2 F 2F 4F (2) 2. 2 An object moves in a straight line on a ROUGH horizontal surface. If the net work done on the object is zero, then … A B C D the object has zero kinetic energy. the object moves at constant speed. the object moves at constant acceleration. there is no frictional force acting on the object. (2) Copyright reserved Please turn over Physical Sciences/P1 4 NSC DBE/Feb. –Mar. 2012 2. 3

A ball is released from rest from a certain height above the floor and bounces off the floor a number of times. Ignore the effects of air resistance. Which ONE of the following velocity-time graphs best represents the motion of the ball? A velocity (m·s ) -1 B velocity (m·s ) -1 time (s) time (s) C velocity (m·s ) -1 D velocity (m·s ) -1 time (s) time (s) (2) 2. 4 The diagram below shows plane water waves that spread out after passing through a single slit. barrier with single slit plane water waves The wave phenomenon observed after the water waves pass through the slit is … A B C D reflection. diffraction. efraction. photoelectric effect. (2) Copyright reserved Please turn over Physical Sciences/P1 5 NSC DBE/Feb. –Mar. 2012 2. 5 Monochromatic light from a point source passes through a device X. A pattern is observed on a screen, as shown in the diagram below. Key: Colour band Dark band monochromatic light source X From the observation on the screen, it can be concluded that device X is a … A B C D 2. 6 prism. single slit. double slit. concave lens. (2) In the circuit diagram below, the internal resistance of the battery and the resistance of the conducting wires are negligible. The emf of the battery is E. E A V 2R R S

When switch S is closed, the reading on voltmeter V, in volts, is … A B C D 0 1 E 3 2 E 3 E screen (2) Copyright reserved Please turn over Physical Sciences/P1 6 NSC DBE/Feb. –Mar. 2012 2. 7 Two identical small metal spheres on insulated stands carry equal charges and are a distance d apart. Each sphere experiences an electrostatic force of magnitude F. The spheres are now placed a distance 1 2 d apart. The magnitude of the electrostatic force each sphere now experiences is … A B C D 2. 8 1 2 F F 2F 4F (2) A fully charged capacitor is connected in a circuit, as shown below. The capacitor discharges when switch S is closed.

R S V Which ONE of the following graphs correctly shows the change in the voltmeter reading with time when switch S is closed? potential difference (V) potential difference (V) A B time (s) time (s) potential difference (V) potential difference (V) C D time (s) time (s) Please turn over (2) Copyright reserved Physical Sciences/P1 7 NSC DBE/Feb. –Mar. 2012 2. 9 When light shines on a metal plate in a photocell, electrons are emitted. The graph below shows the relationship between the kinetic energy of the emitted photoelectrons and the frequency of the incoming light. D kinetic energy (J) 0 A B C requency (Hz) Which ONE of the points (A, B, C or D) on the graph represents the threshold frequency? A B C D 2. 10 A B C D (2) Overexposure to sunlight causes damage to plants and crops. Which ONE of the following types of electromagnetic radiation is responsible for this damage? A B C D Ultraviolet rays Radio waves Visible light X-rays (2) [20] TOTAL SECTION A: 25 Copyright reserved Please turn over Physical Sciences/P1 8 NSC DBE/Feb. –Mar. 2012 SECTION B INSTRUCTIONS AND INFORMATION 1. 2. 3. 4. Start EACH question on a NEW page. Leave ONE line between two subquestions, for example between QUESTION 3. and QUESTION 3. 2. Show the formulae and substitutions in ALL calculations. Round off your final numerical answers to a minimum of TWO decimal places. QUESTION 3 (Start on a new page. ) A stone is thrown vertically upward at a velocity of 10 m·s-1 from the top of a tower of height 50 m. After some time the stone passes the edge of the tower and strikes the ground below the tower. Ignore the effects of friction. vi = 10 m·s-1 1,5 m 50 m y1 3. 1 3. 2 Draw a labelled free-body diagram showing the force(s) acting on the stone during its motion. Calculate the: 3. 2. 1 3. 2. Time taken by the stone to reach its maximum height above the ground Maximum height that the stone reaches above the ground (1) (4) (4) (3) 3. 3 3. 4 USING THE GROUND AS REFERENCE (zero position), sketch a positiontime graph for the entire motion of the stone. On its way down, the stone takes 0,1 s to pass a window of length 1,5 m, as shown in the diagram above. Calculate the distance (y1) from the top of the window to the ground. (7) [19] Copyright reserved Please turn over Physical Sciences/P1 9 NSC DBE/Feb. –Mar. 2012 QUESTION 4 (Start on a new page. ) The bounce of a cricket ball is tested before it is used.

The standard test is to drop a ball from a certain height onto a hard surface and then measure how high it bounces. During such a test, a cricket ball of mass 0,15 kg is dropped from rest from a certain height and it strikes the floor at a speed of 6,2 m·s-1. The ball bounces straight upwards at a velocity of 3,62 m·s-1 to a height of 0,65 m, as shown in the diagram below. The effects of air friction may be ignored. 0,15 kg 0,65 m 4. 1 4. 2 4. 3 Define the term impulse in words. Calculate the magnitude of the impulse of the net force applied to the ball during its collision with the floor.

To meet the requirements, a cricket ball must bounce to one third of the height that it is initially dropped from. Use ENERGY PRINCIPLES to determine whether this ball meets the minimum requirements. (2) (3) (5) [10] Copyright reserved Please turn over Physical Sciences/P1 10 NSC DBE/Feb. –Mar. 2012 QUESTION 5 (Start on a new page. ) A wooden block of mass 2 kg is released from rest at point P and slides down a curved slope from a vertical height of 2 m, as shown in the diagram below. It reaches its lowest position, point Q, at a speed of 5 m·s-1. P 2 kg 2m 9 kg Q 5. 1 5. 2 5. 3 5. 4

Define the term gravitational potential energy. Use the work-energy theorem to calculate the work done by the average frictional force on the wooden block when it reaches point Q. Is mechanical energy conserved while the wooden block slides down the slope? Give a reason for the answer. The wooden block collides with a stationary crate of mass 9 kg at point Q. After the collision, the crate moves to the right at 1 m·s-1. 5. 4. 1 5. 4. 2 Calculate the magnitude of the velocity of the wooden block immediately after the collision. The total kinetic energy of the system before the collision is 25 J.

Use a calculation to show that the collision between the wooden block and the crate is inelastic. (2) (6) (2) (4) (5) [19] QUESTION 6 (Start on a new page. ) An ambulance approaches an accident scene at constant velocity. The siren of the ambulance emits sound waves at a frequency of 980 Hz. A detector at the scene measures the frequency of the emitted sound waves as 1 050 Hz. 6. 1 6. 2 6. 3 Calculate the speed at which the ambulance approaches the accident scene. Use the speed of sound in air as 340 m·s-1. Explain why the measured frequency is higher than the frequency of the source.

The principle of the Doppler effect is applied in the Doppler flow meter. State ONE positive impact of the use of the Doppler flow meter on humans. (4) (2) (2) [8] Copyright reserved Please turn over Physical Sciences/P1 11 NSC DBE/Feb. –Mar. 2012 QUESTION 7 (Start on a new page. ) Learners investigate the change in the broadness of the central bright band formed when monochromatic light of different wavelengths passes through a single slit. They set up the apparatus, as shown in diagram below, and measure the broadness of the central bright band in the pattern observed on the screen.

The width of the slit is 5,6 x 10-7 m. screen first dark band monochromatic light ? midpoint of central bright band 0,033 m first dark band 0,45 m 7. 1 7. 2 7. 3 Write down an investigative question. Which TWO variables are kept constant? In one of their experiments, the distance from the midpoint of the central bright band to the first dark band is measured to be 0,033 m. Calculate the wavelength of the light used in this experiment. (2) (2) (5) 7. 4 How will the broadness of the central bright band of red light compare with that of blue light? Write down only GREATER THAN, SMALLER THAN or EQUAL TO.

Give a reason for the answer. (2) [11] Copyright reserved Please turn over Physical Sciences/P1 12 NSC DBE/Feb. –Mar. 2012 QUESTION 8 (Start on a new page. ) Two metal spheres, P and Q, on insulated stands, carrying charges of +5 x 10-9 C and +5 x 10-9 C respectively, are placed with their centres 20 mm apart. X is a point at a distance of 10 mm from sphere Q, as shown below. 20 mm +5 x 10-9 C P 10 mm +5 x 10-9 C Q X 8. 1 8. 2 8. 3 8. 4 Define the term electric field. Sketch the net electric field pattern for the two charges. Calculate the net electric field at point X due to the presence of P and Q.

Use your answer to QUESTION 8. 3 to calculate the magnitude of the electrostatic force that an electron will experience when placed at point X. (2) (3) (6) (3) [14] Copyright reserved Please turn over Physical Sciences/P1 13 NSC DBE/Feb. –Mar. 2012 QUESTION 9 (Start on a new page. ) 9. 1 Learners use Ohm’s law to determine which ONE of two resistors A and B has the greater resistance. For each resistor, they measure the current through the resistor for different potential differences across its ends. The graph below shows the results obtained in their investigation. A current (A) B potential difference (V) 9. 1. 1

The learners are supplied with the following apparatus: 6 V battery Voltmeter Ammeter Rheostat Resistors A and B Conducting wires Draw a circuit diagram to show how the learners must use the above apparatus to obtain each of the graphs shown above. (4) (1) 9. 1. 2 9. 1. 3 Write down ONE variable that must be kept constant during this investigation. Which ONE of A or B has the higher resistance? Give an explanation for the answer. (3) Copyright reserved Please turn over Physical Sciences/P1 14 NSC DBE/Feb. –Mar. 2012 9. 2 In the circuit diagram below, the battery has an emf of 12 V and an internal resistance of 0,8 ?.

The resistance of the ammeter and connecting wires may be ignored. 12 V 0,8 ? 2? 4? A 8? 8? V Calculate the: 9. 2. 1 9. 2. 2 9. 2. 3 Effective resistance of the circuit Reading on the ammeter Reading on the voltmeter (4) (3) (4) [19] Copyright reserved Please turn over Physical Sciences/P1 15 NSC DBE/Feb. –Mar. 2012 QUESTION 10 (Start on a new page. ) 10. 1 The essential components of a simplified DC motor are shown in the diagram below. coil B C N A D S brushes split-ring commutator When the motor is functioning, the coil rotates in a clockwise direction, as shown. 10. 1. Write down the function of each of the following components: (a) (b) 10. 1. 2 Split-ring commutator Brushes (1) (1) What is the direction of the conventional current in the part of the coil labelled AB? Write down only FROM A TO B or FROM B TO A. Will the coil experience a maximum or minimum turning effect (torque) if the coil is in the position as shown in the diagram above? State ONE way in which this turning effect (torque) can be increased. (1) 10. 1. 3 (1) (1) 10. 1. 4 10. 2 Alternating current (AC) is used for the long-distance transmission of electricity. 10. 2. 1 10. 2. Give a reason why AC is preferred over DC for long-distance transmission of electricity. An electric appliance with a power rating of 2 000 W is connected to a 230 V rms household mains supply. Calculate the: (a) (b) Peak (maximum) voltage rms current passing through the appliance (3) (3) [12] Please turn over (1) Copyright reserved Physical Sciences/P1 16 NSC DBE/Feb. –Mar. 2012 QUESTION 11 (Start on a new page. ) In the diagram shown below, electrons are released from a metal plate when light of a certain frequency is shone on its surface. incident light metal surface eA 11. 1 11. 2

Name the phenomenon described above. The frequency of the incident light on the metal plate is 6,16 x 1014 Hz and electrons are released with a kinetic energy of 5,6 x 10-20 J. Calculate the: 11. 2. 1 11. 2. 2 Energy of the incident photons Threshold frequency of the metal plate (1) (3) (5) 11. 3 The brightness of the incident light is now increased. What effect will this change have on the following: (Write down only INCREASES, DECREASES or REMAINS THE SAME. ) 11. 3. 1 11. 3. 2 The reading on the ammeter Explain the answer. The kinetic energy of the released photoelectrons Explain the answer. 2) (2) [13] 125 150 TOTAL SECTION B: GRAND TOTAL: Copyright reserved Physical Sciences/P1 1 NSC DBE/Feb. –Mar. 2012 DATA FOR PHYSICAL SCIENCES GRADE 12 PAPER 1 (PHYSICS) GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 12 VRAESTEL 1 (FISIKA) TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES NAME/NAAM Acceleration due to gravity Swaartekragversnelling Speed of light in a vacuum Spoed van lig in ‘n vakuum Planck’s constant Planck se konstante Coulomb’s constant Coulomb se konstante Charge on electron Lading op elektron Electron mass Elektronmassa Permittivity of free space Permittiwiteit van vry ruimte

SYMBOL/SIMBOOL g c h k e me ?0 VALUE/WAARDE 9,8 m·s-2 3,0 x 108 m·s-1 6,63 x 10-34 J·s 9,0 x 109 N·m2·C-2 -1,6 x 10-19 C 9,11 x 10-31 kg 8,85 x 10-12 F·m-1 Copyright reserved Please turn over Physical Sciences/P1 2 NSC DBE/Feb. –Mar. 2012 TABLE 2: FORMULAE/TABEL 2: FORMULES MOTION/BEWEGING v f = v i + a ? t 1 1 ? x = v i ? t + 2 a? t 2 or/of ? y = v i ? t + 2 a? t 2 2 2 v f = v i + 2a? x or/of v f = v i + 2a? y FORCE/KRAG 2 2 ? v +vf ? ? v +vf ? ?x = ? i ? ?t or/of ? y = ? i ? ?t ? 2 ? ? 2 ? Fnet = ma Fnet ? t = ? p ? p = mv f ? v i p = mv w = mg WORK, ENERGY AND POWER/ARBEID, ENERGIE EN DRYWING W = F? x cos ? 1 or/of K = mv 2 2 P= W ? t Ek = 1 mv 2 2 U = mgh Wnet = ? K ?K = K f ? K i or/of or/of or/of E P = mgh Wnet = ? Ek ?Ek = Ekf ? Eki P = Fv WAVES, SOUND AND LIGHT/GOLWE, KLANK EN LIG v=f? fL = v ± vL fs v ± vs or/of fL = v ± vL fb v ± vb 1 f E = hf T= E= h c ? E = Wo + Ek sin ? = m? a where/waar 1 E = hf and/en W0 = hf0 and/en Ek = mv 2 2 Copyright reserved Please turn over Physical Sciences/P1 3 NSC DBE/Feb. –Mar. 2012 ELECTROSTATICS/ELEKTROSTATIKA Q 1Q 2 r2 V E= d kQ Q U= 1 2 r Q C= V F= kQ r2 F E= q W V= q ? A C= 0 d E= ELECTRIC CIRCUITS/ELEKTRIESE STROOMBANE V R= I R s = R1 + R 2 + … 1 1 1 = + + … R p R1 R 2 W = Vq W = VI ? t W= I2R ? t V 2 ? t W= R emf ( ? ) = I(R + r) emk ( ? ) = I(R + r) q=I ? t W ? t P= P = VI P = I2R V2 P= R ALTERNATING CURRENT/WISSELSTROOM I rms = I max 2 Vmax 2 / I I wgk = maks 2 Vwgk = Vmaks 2 Paverage = Vrms I rms 2 Paverage = I rms R / / / Pgemiddeld = Vwgk I wgk Pgemiddeld = I 2 R wgk Vrms = / Paverage 2 Vrms = R Pgemiddeld = 2 Vwgk R Copyright reserved

Centripetal Force Lab Activity Analysis:

1. A) Average Percent Difference: 50g: (values expressed in newtons)

Step 1: Calculate the average value of the two variables Average Value= Value 1+ Value 2 /2 = 0. 49+ 0. 61/2 = 1. 1/2 = 0. 55

Step 2: Calculate the difference between the two variables Difference= Value 2- Value 1 = Fc- Fg = 0. 61- 0. 49 = 0. 12

Step 3: Calculate % difference % difference= difference of the variables / average of the variables x 100 = 0. 12/ 0. 55 x 100 = 21. 81% 100g: (values expressed in newtons)

Step 1: Calculate the average value of the two variables Average Value= Value 1+ Value 2 /2 = 0. 98+ 1. 84/2 = 2. 82/2 = 1. 41

Step 2: Calculate the difference between the two variables Difference= Value 2- Value 1 = Fc- Fg = 1. 84- 0. 98 = 0. 86

Step 3: Calculate % difference % difference= difference of the variables / average of the variables x 100 = 0. 86/ 1. 41 x 100 = 60. 99% 150g: (values expressed in newtons)

Step 1: Calculate the average value of the two variables Average Value= Value 1+ Value 2 /2 = 1. 47+ 2. 19/2 = 3. 66/2 = 1. 83

Step 2: Calculate the difference between the two variables Difference= Value 2- Value 1 = Fc- Fg = 2. 19- 1. 47 = 0. 72

Step 3: Calculate % difference % difference= difference of the variables / average of the variables x 100 = 0. 72/ 1. 83 x 100 = 39. 34% 200g: (values expressed in newtons)

Step 1: Calculate the average value of the two variables Average Value= Value 1+ Value 2 /2 = 1. 96+ 2. 66/2 = 4. 62/2 = 2. 31

Step 2: Calculate the difference between the two variables Difference= Value 2- Value 1 = Fc- Fg = 2. 66- 1. 96 = 0. 70

Step 3: Calculate % difference difference= difference of the variables / average of the variables x 100 = 0. 70/2. 31 x 100 = 30. 30% 250g: (values expressed in newtons)

Step 1: Calculate the average value of the two variables Average Value= Value 1+ Value 2 /2 = 2. 45+ 3. 57/2 = 6. 02/2 = 3. 01

Step 2: Calculate the difference between the two variables Difference= Value 2- Value 1 = Fc- Fg = 3. 57- 2. 45 = 1. 12

Step 3: Calculate % difference % difference= difference of the variables / average of the variables x 100 = 1. 12/ 3. 01 x 100 = 37. 20% Average % difference: = Sum of all 5 averages/5 21. 81+ 60. 99+ 39. 34+ 30. 30+ 37. 20/ 5 = 189. 64/ 5 = 37. 92% B) Slope Calculations (Graph is displayed on a separate sheet) 50g: Slope= Rise/Run = 0. 61/0. 49 = 1. 25 100g: Slope= Rise/Run = 1. 84/0. 98 = 1. 877 150g: Slope= Rise/Run = 2. 19/1. 47 = 1. 489 200g: Slope= Rise/Run = 2. 66/1. 96 = 1. 357 250g: Slope= Rise/Run = 3. 57/2. 45 = 1. 457 After calculating the slope of each section of the graph (each section corresponds to a certain mass used in the lab activity) it is evident that it varies from it’s expected value by a great amount.

The expected value of the slope was 1 as the rise and the run were supposed to be equal. However in our case the rise and the run varied greatly and therefore because they were different numbers the slope did not turn out to be 1 (the only way to get a slope of 1 is if both the numerator and denominator are equal, as a number divided by itself is always 1 and a number divided by a different number can never equal 1).

2. Yes the data collected did verify the equation Fc=42Rmf2. This is because the only varying value in this case “f”, had a direct relationship with the value of Fc.

The only other values that had to be determined in this lab was the radius and the mass of the rubber stopper but they were constant variables (constant at 0. 87m and 12. 4g respectively) meaning that they had no varying effect on the value of Fc. For there to be a relationship between Fc and 42Rmf2 when the value of any of the variables changes the value of Fc has to change as well Because the value of “f” had a direct relationship with the value of Fc, when the value of “f” changed the value of Fc changed as well. In this particular case when the value of “f” grew so did the value of Fc.

For example, during the 50g test the frequency was 1. 2Hz and the Fc was 0. 61N, and during the 100g test the frequency was 2. 08Hz and the Fc was 1. 84N. This shows that as the frequency increases so does the Fc acting on the system. This therefore shows the relationship between Fc and 42Rmf2.

3. A) When the string was pulled down and the stopper was still spinning, the stopper started spinning at a faster rate (took less time to complete 1 cycle around the trip)

B) This happens simply because the radius is being shortened.

Because the stopper on the end of the string is moving around the horizontal circle at a constant speed it is therefore being acted upon by a constant net-force. In this case the net-force acting upon it (the stopper) is Fc, therefore because it is Fc acting upon it, the force can be calculated by the formula 42Rmf2 as that is equal to Fc. In this case because the string with the stopper on the end was being pulled down this means that the radius of the entire circle was decreasing (less string= smaller distance= smaller radius).

In that formula if the radius is smaller that means that the centripetal force will be larger. In this case that larger the centripetal force acting on the rubber stopper, the faster the rubber stopper rotates around the horizontal circle.

C) The laws of conservation of energy state that the total energy in the system stays the same but simply takes on different forms (kinetic and potential being examples). Therefore this case is not contrary to the laws of conservation of energy simply because when the radius is decreasing the rubber stopper speeds up.

In the laws of conservation of energy when an object is speeding up the object is gaining kinetic energy. However in this case while the stopper is speeding up the hanging mass (along with some of the string) is falling to the ground. From a conservation of energy perspective when an object loses height it loses potential energy. Therefore in this case the object at the top gains kinetic energy while the mass loses potential energy. Because of this energy transfer no energy is lost in the system as hen the object is losing potential energy the other object in the same system is gaining kinetic energy, therefore the energy stays the same.

D) In figure skating the skaters do the exact same thing as what was done in this lab experiment. In order to spin faster they bend low (get low to the ground) and tuck their arms and legs in. This causes them to spin much faster than they were originally spinning and follows the same principles that the rubber stopper experiment followed. When they get low they lose potential energy but getting low causes them to tuck in (tuck in their legs and arms) and ultimately have a smaller radius.

This smaller radius causes them to have a much greater centripetal force and ultimately causes them to spin faster and causes them to gain kinetic energy. This follows the laws of conservation of energy as when they lose potential energy they gain kinetic energy (theoretically no energy lost- only transferred) Sources of Error: In this particular lab activity there were not very many potential sources of error simply because it was not as complicated an activity as many others. Therefore all errors that were made were simply human measurement errors.

The main source of error in this lab activity was measuring the period/frequency. This was a challenge simply because the person measuring had to do many different things in a very small amount of time. That one person was responsible for firstly choosing a spot along the path of the horizontal circle to begin the measurement from, then that same person had to start the watch during the very small time frame in which the rubber stopper passed by that specific point on the circle. From there the person had to count the stopper pass by 5 times and stop the watch when it passed by the 5th time.

This made it very difficult to get a completely accurate measurement for the period and the frequency, as it was very difficult to get an exact measurement of that time period. These slight miscalculations of the frequency caused the calculation of the centripetal force to be slightly wrong as well because the calculation of centripetal force depended on the frequency. This is evident because our “Fg” and “Fc” calculations are way off, as they were supposed to be nearly the same number as

Fg= Fc. – X-axis= Fc – Y-axis= Fg – point 1= 50g – point 2= 100g – point 3= 150g – point 4= 200g – point 5= 250g

Introduction

It may be recalled that a d. c. generator can be run as a d. c. motor. In like manner, an alternator may operate as a motor by connecting its armature winding to a 3-phase supply. It is then called a synchronous motor. As the name implies, a synchronous motor runs at synchronous speed (Ns = 120f/P) i. e. , in synchronism with the revolving field produced by the 3-phase supply. The speed of rotation is, therefore, tied to the frequency of the source.

Since the frequency is fixed, the motor speed stays constant irrespective of the load or voltage of 3phase supply. However, synchronous motors are not used so much because they run at constant speed (i. e. , synchronous speed) but because they possess other unique electrical properties. In this chapter, we shall discuss the working and characteristics of synchronous motors.

Construction

A synchronous motor is a machine that operates at synchronous speed and converts electrical energy into mechanical energy. It is fundamentally an alternator operated as a motor.

Like an alternator, a synchronous motor has the following two parts: a stator which houses 3-phase armature winding in the slots of the stator core and receives power from a 3-phase supply [See (Fig. (11. 1)]. a rotor that has a set of salient poles excited by direct current to form alternate N and S poles. The exciting coils are connected in series to two slip rings and direct current is fed into the winding from an external exciter mounted on the rotor shaft. The stator is wound for the same number of poles as the rotor poles.

As in the case of an induction motor, the number of poles determines the synchronous speed of the motor: Fig. (11. 1) 293 Synchronous speed, N s = where 120f P f = frequency of supply in Hz P = number of poles An important drawback of a synchronous motor is that it is not self-starting and auxiliary means have to be used for starting it.

Some Facts about Synchronous Motor

Some salient features of a synchronous motor are: A synchronous motor runs at synchronous speed or not at all. Its speed is constant (synchronous speed) at all loads. The only way to change its speed is to alter the supply frequency (Ns = 120 f/P). The outstanding characteristic of a synchronous motor is that it can be made to operate over a wide range of power factors (lagging, unity or leading) by adjustment of its field excitation. Therefore, a synchronous motor can be made to carry the mechanical load at constant speed and at the same time improve the power factor of the system. Synchronous motors are generally of the salient pole type. A synchronous motor is not self-starting and an auxiliary means has to be used for starting it. We use either induction motor principle or a separate starting motor for this purpose.

If the latter method is used, the machine must be run up to synchronous speed and synchronized as an alternator.

Operating Principle

The fact that a synchronous motor has no starting torque can be easily explained. Consider a 3-phase synchronous motor having two rotor poles NR and SR. Then the stator will also be wound for two poles NS and SS. The motor has direct voltage applied to the rotor winding and a 3-phase voltage applied to the stator winding. The stator winding produces a rotating field which revolves round the stator at synchronous speed Ns(= 120 f/P).

The direct (or zero frequency) current sets up a two-pole field which is stationary so long as the rotor is not turning. Thus, we have a situation in which there exists a pair of revolving armature poles (i. e. , NS ? SS) and a pair of stationary rotor poles (i. e. , NR ? SR). Suppose at any instant, the stator poles are at positions A and B as shown in Fig. (11. 2 (i)). It is clear that poles NS and NR repel each other and so do the poles SS and SR. Therefore, the rotor tends to move in the anticlockwise direction. After a period of half-cycle (or ? = 1/100 second), the polarities of the stator poles are reversed but the polarities of the rotor poles remain the same as shown in Fig. (11. 2 (ii)). Now SS and NR attract 294 each other and so do NS and SR. Therefore, the rotor tends to move in the clockwise direction. Since the stator poles change their polarities rapidly, they tend to pull the rotor first in one direction and then after a period of half-cycle in the other. Due to high inertia of the rotor, the motor fails to start. Fig. (10. 2) Hence, a synchronous motor has no self-starting torque i. e. , a synchronous motor cannot start by itself.

How to get continuous unidirectional torque? If the rotor poles are rotated by some external means at such a speed that they interchange their positions along with the stator poles, then the rotor will experience a continuous unidirectional torque. This can be understood from the following discussion: Suppose the stator field is rotating in the clockwise direction and the rotor is also rotated clockwise by some external means at such a speed that the rotor poles interchange their positions along with the stator poles. Suppose at any instant the stator and rotor poles are in the position shown in Fig. 11. 3 (i)). It is clear that torque on the rotor will be clockwise. After a period of half-cycle, the stator poles reverse their polarities and at the same time rotor poles also interchange their positions as shown in Fig. (11. 3 (ii)). The result is that again the torque on the rotor is clockwise. Hence a continuous unidirectional torque acts on the rotor and moves it in the clockwise direction. Under this condition, poles on the rotor always face poles of opposite polarity on the stator and a strong magnetic attraction is set up between them.

This mutual attraction locks the rotor and stator together and the rotor is virtually pulled into step with the speed of revolving flux (i. e. , synchronous speed). (iii) If now the external prime mover driving the rotor is removed, the rotor will continue to rotate at synchronous speed in the clockwise direction because the rotor poles are magnetically locked up with the stator poles. It is due to 295 this magnetic interlocking between stator and rotor poles that a synchronous motor runs at the speed of revolving flux i. e. , synchronous speed. Fig. (11. 3) 11. Making Synchronous Motor Self-Starting A synchronous motor cannot start by itself. In order to make the motor self-starting, a squirrel cage winding (also called damper winding) is provided on the rotor. The damper winding consists of copper bars embedded in the pole faces of the salient poles of the rotor as shown in Fig. (11. 4). The bars are short-circuited at the ends to form in effect a partial Fig. (11. 4) squirrel cage winding. The damper winding serves to start the motor. To start with, 3-phase supply is given to the stator winding while the rotor field winding is left unenergized.

The rotating stator field induces currents in the damper or squirrel cage winding and the motor starts as an induction motor. (ii) As the motor approaches the synchronous speed, the rotor is excited with direct current. Now the resulting poles on the rotor face poles of opposite polarity on the stator and a strong magnetic attraction is set up between them. The rotor poles lock in with the poles of rotating flux. Consequently, the rotor revolves at the same speed as the stator field i. e. , at synchronous speed. iii) Because the bars of squirrel cage portion of the rotor now rotate at the same speed as the rotating stator field, these bars do not cut any flux and, therefore, have no induced currents in them. Hence squirrel cage portion of the rotor is, in effect, removed from the operation of the motor. 296 It may be emphasized here that due to magnetic interlocking between the stator and rotor poles, a synchronous motor can only run at synchronous speed. At any other speed, this magnetic interlocking (i. e. , rotor poles facing opposite polarity stator poles) ceases and the average torque becomes zero.

Consequently, the motor comes to a halt with a severe disturbance on the line. Note: It is important to excite the rotor with direct current at the right moment. For example, if the d. c. excitation is applied when N-pole of the stator faces Npole of the rotor, the resulting magnetic repulsion will produce a violent mechanical shock. The motor will immediately slow down and the circuit breakers will trip. In practice, starters for synchronous motors arc designed to detect the precise moment when excitation should be applied.

Equivalent Circuit

Unlike the induction motor, the synchronous motor is connected to two electrical systems; a d. . source at the rotor terminals and an a. c. system at the stator terminals. 1. Under normal conditions of synchronous motor operation, no voltage is induced in the rotor by the stator field because the rotor winding is rotating at the same speed as the stator field. Only the impressed direct current is present in the rotor winding and ohmic resistance of this winding is the only opposition to it as shown in Fig. (11. 5 (i)). 2. In the stator winding, two effects are to be considered, the effect of stator field on the stator winding and the effect of the rotor field cutting the stator conductors at synchronous speed.

Fig. (11. 5) (i) The effect of stator field on the stator (or armature) conductors is accounted for by including an inductive reactance in the armature winding. This is called synchronous reactance Xs. A resistance Ra must be considered to be in series with this reactance to account for the copper losses in the stator or armature winding as shown in Fig. (11. 5 (i)). This 297 resistance combines with synchronous reactance and gives the synchronous impedance of the machine. The second effect is that a voltage is generated in the stator winding by the synchronously-revolving field of the rotor as shown in Fig. 11. 5 (i)). This generated e. m. f. EB is known as back e. m. f. and opposes the stator voltage V. The magnitude of Eb depends upon rotor speed and rotor flux ? per pole. Since rotor speed is constant; the value of Eb depends upon the rotor flux per pole i. e. exciting rotor current If. Fig. (11. 5 (i)) shows the schematic diagram for one phase of a star-connected synchronous motor while Fig. (11. 5 (ii)) shows its equivalent circuit. Referring to the equivalent circuit in Fig. (11. 5 (ii)). Net voltage/phase in stator winding is Er = V ? Eb Armature current/phase, I a = where 2 Zs = R 2 + X s a hasor difference Er Zs This equivalent circuit helps considerably in understanding the operation of a synchronous motor. A synchronous motor is said to be normally excited if the field excitation is such that Eb = If the field excitation is such that Eb < V, the motor is said to be under-excited. The motor is said to be over-excited if the field excitation is such that Eb > V. As we shall see, for both normal and under excitation, the motor has lagging power factor. However, for over-excitation, the motor has leading power factor.

Note: In a synchronous motor, the value of Xs is 10 to 100 times greater than Ra. Consequently, we can neglect Ra unless we are interested in efficiency or heating effects.

Motor on Load

In d. c. motors and induction motors, an addition of load causes the motor speed to decrease. The decrease in speed reduces the counter e. m. f. enough so that additional current is drawn from the source to carry the increased load at a reduced speed. This action cannot take place in a synchronous motor because it runs at a constant speed (i. e. , synchronous speed) at all loads.

What happens when we apply mechanical load to a synchronous motor? The rotor poles fall slightly behind the stator poles while continuing to run at 298 synchronous speed. The angular displacement between stator and rotor poles (called torque angle ? ) causes the phase of back e. m. f. Eb to change w. r. t. supply voltage V. This increases the net e. m. f. Er in the stator winding. Consequently, stator current Ia ( = Er/Zs) increases to carry the load. Fig. (11. 6) The following points may be noted in synchronous motor operation: (i) A synchronous motor runs at synchronous speed at all loads.

It meets the increased load not by a decrease in speed but by the relative shift between stator and rotor poles i. e. , by the adjustment of torque angle ?. (ii) If the load on the motor increases, the torque angle a also increases (i. e. , rotor poles lag behind the stator poles by a greater angle) but the motor continues to run at synchronous speed. The increase in torque angle ? causes a greater phase shift of back e. m. f. Eb w. r. t. supply voltage V. This increases the net voltage Er in the stator winding. Consequently, armature current Ia (= Er/Zs) increases to meet the load demand. iii) If the load on the motor decreases, the torque angle ? also decreases. This causes a smaller phase shift of Eb w. r. t. V. Consequently, the net voltage Er in the stator winding decreases and so does the armature current Ia (= Er/Zs).

Pull-Out Torque

There is a limit to the mechanical load that can be applied to a synchronous motor. As the load increases, the torque angle ? also increases so that a stage is reached when the rotor is pulled out of synchronism and the motor comes to a standstill. This load torque at which the motor pulls out of synchronism is called pull—out or breakdown torque.

Its value varies from 1. 5 to 3. 5 times the full— load torque. When a synchronous motor pulls out of synchronism, there is a major disturbance on the line and the circuit breakers immediately trip. This protects the motor because both squirrel cage and stator winding heat up rapidly when the machine ceases to run at synchronous speed. 299 11. 8 Motor Phasor Diagram Consider an under-excited ^tar-connected synchronous motor (Eb < V) supplied with fixed excitation i. e. , back e. m. f. Eb is constantLet V = supply voltage/phase Eb = back e. m. f. /phase Zs = synchronous impedance/phase (i) Motor on no load

When the motor is on no load, the torque angle ? is small as shown in Fig. (11. 7 (i)). Consequently, back e. m. f. Eb lags behind the supply voltage V by a small angle ? as shown in the phasor diagram in Fig. (11. 7 (iii)). The net voltage/phase in the stator winding, is Er. Armature current/phase, Ia = Er/Zs The armature current Ia lags behind Er by ? = tan-1 Xs/Ra. Since Xs >> Ra, Ia lags Er by nearly 90°. The phase angle between V and Ia is ? so that motor power factor is cos ?. Input power/phase = V Ia cos ? Fig. (11. 7) Thus at no load, the motor takes a small power VIa cos ? phase from the supply to meet the no-load losses while it continues to run at synchronous speed. Motor on load When load is applied to the motor, the torque angle a increases as shown in Fig. (11. 8 (i)). This causes Eb (its magnitude is constant as excitation is fixed) to lag behind V by a greater angle as shown in the phasor diagram in Fig. (11. 8 (ii)). The net voltage/phase Er in the stator winding increases. Consequently, the motor draws more armature current Ia (=Er/Zs) to meet the applied load. Again Ia lags Er by about 90° since Xs >> Ra. The power factor of the motor is cos ?. 300 Input power/phase, Pi = V Ia cos ?

Mechanical power developed by motor/phase Pm = Eb ? Ia ? cosine of angle between Eb and Ia = Eb Ia cos(? ? ? ) Fig. (11. 8) 11. 9 Effect of Changing Field Excitation at Constant Load In a d. c. motor, the armature current Ia is determined by dividing the difference between V and Eb by the armature resistance Ra. Similarly, in a synchronous motor, the stator current (Ia) is determined by dividing voltage-phasor resultant (Er) between V and Eb by the synchronous impedance Zs. One of the most important features of a synchronous motor is that by changing the field excitation, it can be made to operate from lagging to eading power factor. Consider a synchronous motor having a fixed supply voltage and driving a constant mechanical load. Since the mechanical load as well as the speed is constant, the power input to the motor (=3 VIa cos ? ) is also constant. This means that the in-phase component Ia cos ? drawn from the supply will remain constant. If the field excitation is changed, back e. m. f Eb also changes. This results in the change of phase position of Ia w. r. t. V and hence the power factor cos ? of the motor changes. Fig. (11. 9) shows the phasor diagram of the synchronous motor for different values of field excitation.

Note that extremities of current phasor Ia lie on the straight line AB. (i) Under excitation The motor is said to be under-excited if the field excitation is such that Eb < V. Under such conditions, the current Ia lags behind V so that motor power factor is lagging as shown in Fig. (11. 9 (i)). This can be easily explained. Since Eb < V, the net voltage Er is decreased and turns clockwise. As angle ? (= 90°) between Er and Ia is constant, therefore, phasor Ia also turns clockwise i. e. , current Ia lags behind the supply voltage. Consequently, the motor has a lagging power factor. 301 ii) Normal excitation The motor is said to be normally excited if the field excitation is such that Eb = V. This is shown in Fig. (11. 9 (ii)). Note that the effect of increasing excitation (i. e. , increasing Eb) is to turn the phasor Er and hence Ia in the anti-clockwise direction i. e. , Ia phasor has come closer to phasor V. Therefore, p. f. increases though still lagging. Since input power (=3 V Ia cos ? ) is unchanged, the stator current Ia must decrease with increase in p. f. Fig. (11. 9) Suppose the field excitation is increased until the current Ia is in phase with the applied voltage V, making the p. . of the synchronous motor unity [See Fig. (11. 9 (iii))]. For a given load, at unity p. f. the resultant Er and, therefore, Ia are minimum. (iii) Over excitation The motor is said to be overexcited if the field excitation is such that Eb > V. Under-such conditions, current Ia leads V and the motor power factor is leading as shown in Fig. (11. 9 (iv)). Note that Er and hence Ia further turn anti-clockwise from the normal excitation position. Consequently, Ia leads V. From the above discussion, it is concluded that if the synchronous motor is under-excited, it has a lagging power factor.

As the excitation is increased, the power factor improves till it becomes unity at normal excitation. Under such conditions, the current drawn from the supply is minimum. If the excitation is further increased (i. e. , over excitation), the motor power factor becomes leading. Note. The armature current (Ia) is minimum at unity p. f and increases as the power factor becomes poor, either leading or lagging. 302

Phasor Diagrams With Different Excitations

Fig. (11. 10) shows the phasor diagrams for different field excitations at constant load. Fig. (11. 10 (i)) shows the phasor diagram for normal excitation (Eb = V), whereas Fig. 11. 10 (ii)) shows the phasor diagram for under-excitation. In both cases, the motor has lagging power factor. Fig. (11. 10 (iii)) shows the phasor diagram when field excitation is adjusted for unity p. f. operation. Under this condition, the resultant voltage Er and, therefore, the stator current Ia are minimum. When the motor is overexcited, it has leading power factor as shown in Fig. (11. 10 (iv)). The following points may be remembered: (i) For a given load, the power factor is governed by the field excitation; a weak field produces the lagging armature current and a strong field produces a leading armature current. ii) The armature current (Ia) is minimum at unity p. f and increases as the p. f. becomes less either leading or lagging. Fig. (11. 10)

Power Relations

Consider an under-excited star-connected synchronous motor driving a mechanical load. Fig. (11. 11 (i)) shows the equivalent circuit for one phase, while Fig. (11. 11 (ii)) shows the phasor diagram. Fig. (11. 11) 303 (i) (ii) Input power/phase, Pi = V Ia cos ? Mechanical power developed by the motor/phase, Pm = Eb ? Ia ? cosine of angle between Eb and Ia = Eb Ia cos(? ? ? ) Armature Cu loss/phase = I 2 R a = Pi ? Pm a Output power/phasor, Pout = Pm ?

Iron, friction and excitation loss. Fig. (11. 12) shows the power flow diagram of the synchronous motor. Fig. (11. 12) 11. 12 Motor Torque Gross torque, Tg = 9. 55 where Pm N-m Ns Pm = Gross motor output in watts = Eb Ia cos(? ? ? ) Ns = Synchronous speed in r. p. m. Shaft torque, Tsh = 9. 55 Pout N-m Ns It may be seen that torque is directly proportional to the mechanical power because rotor speed (i. e. , Ns) is fixed. 11. 13 Mechanical Power Developed By Motor (Armature resistance neglected) Fig. (11. 13) shows the phasor diagram of an under-excited synchronous motor driving a mechanical load.

Since armature resistance Ra is assumed zero. tan? = Xs/Ra = ? and hence ? = 90°. Input power/phase = V Ia cos ? Fig. (11. 13) 304 Since Ra is assumed zero, stator Cu loss (I 2 R a ) will be zero. Hence input power a is equal to the mechanical power Pm developed by the motor. Mech. power developed/ phase, Pm = V Ia cos ? Referring to the phasor diagram in Fig. (11. 13), (i) AB = E r cos ? = I a X s cos ? Also AB = E b sin ? ? E b sin ? = I a X s cos ? or I a cos ? = E b sin ? Xs Substituting the value of Ia cos ? in exp. (i) above, Pm = = V Eb Xs VEb Xs per phase for 3-phase

It is clear from the above relation that mechanical power increases with torque angle (in electrical degrees) and its maximum value is reached when ? = 90° (electrical). Pmax = V Eb Xs per phase Under this condition, the poles of the rotor will be mid-way between N and S poles of the stator.

Power Factor of Synchronous Motors

In an induction motor, only one winding (i. e. , stator winding) produces the necessary flux in the machine. The stator winding must draw reactive power from the supply to set up the flux. Consequently, induction motor must operate at lagging power factor.

But in a synchronous motor, there are two possible sources of excitation; alternating current in the stator or direct current in the rotor. The required flux may be produced either by stator or rotor or both. If the rotor exciting current is of such magnitude that it produces all the required flux, then no magnetizing current or reactive power is needed in the stator. As a result, the motor will operate at unity power factor. 305.  If the rotor exciting current is less (i. e. , motor is under-excited), the deficit in flux is made up by the stator. Consequently, the motor draws reactive power to provide for the remaining flux.

Hence motor will operate at a lagging power factor. If the rotor exciting current is greater (i. e. , motor is over-excited), the excess flux must be counterbalanced in the stator. Now the stator, instead of absorbing reactive power, actually delivers reactive power to the 3-phase line. The motor then behaves like a source of reactive power, as if it were a capacitor. In other words, the motor operates at a leading power factor. To sum up, a synchronous motor absorbs reactive power when it is underexcited and delivers reactive power to source when it is over-excited. 11. 15 Synchronous Condenser

A synchronous motor takes a leading current when over-excited and, therefore, behaves as a capacitor. An over-excited synchronous motor running on no-load in known as synchronous condenser. When such a machine is connected in parallel with induction motors or other devices that operate at low lagging power factor, the leading kVAR supplied by the synchronous condenser partly neutralizes the lagging reactive kVAR of the loads. Consequently, the power factor of the system is improved. Fig. (11. 14) shows the power factor improvement by synchronous condenser method. The 3 ? ? load takes current IL at low lagging power factor cos ?

L. The synchronous condenser takes a current Im which leads the voltage by an angle ? m. The resultant current I is the vector sum of Im and IL and lags behind the voltage by an angle ?. It is clear that ? is less than ? L so that cos ? is greater than cos ? L. Thus the power factor is increased from cos ? L to cos ?. Synchronous condensers are generally used at major bulk supply substations for power factor improvement.

Advantages

By varying the field excitation, the magnitude of current drawn by the motor can be changed by any amount. This helps in achieving stepless control of power factor. ii) The motor windings have high thermal stability to short circuit currents. The faults can be removed easily. 306 Fig. (11. 14)

Disadvantages

There are considerable losses in the motor. The maintenance cost is high. It produces noise. Except in sizes above 500 RVA, the cost is greater than that of static capacitors of the same rating. (v) As a synchronous motor has no self-starting torque, then-fore, an auxiliary equipment has to be provided for this purpose. 11. 16 Applications of Synchronous Motors (i) Synchronous motors are particularly attractive for low speeds (< 300 r. . m. ) because the power factor can always be adjusted to unity and efficiency is high. (ii) Overexcited synchronous motors can be used to improve the power factor of a plant while carrying their rated loads. (iii) They are used to improve the voltage regulation of transmission lines. High-power electronic converters generating very low frequencies enable us to run synchronous motors at ultra-low speeds. Thus huge motors in the 10 MW range drive crushers, rotary kilns and variable-speed ball mills. 307

Comparison of Synchronous and Induction Motors

Speed 2. 3. 4. Power factor Excitation Economy 3-phase Induction Motor Remains constant (i. e. , Ns) from Decreases with load. no-load to full-load. Can be made to operate from Operates at lagging lagging to leading power factor. power factor. Requires d. c. excitation at the No excitation for the rotor. rotor. Economical fcr speeds below Economical for 300 r. p. m. speeds above 600 r. p. m. Self-starting No self-starting torque. Auxiliary means have to be provided for starting.

Andre Ampere biography Andre-Marie Ampere & Electromagnetism Andre-Marie Ampere was first, a Frenchman, second a physicist and third a mathematician. Andre was born on 20 January in the year 1775 at the Parish of St. Nizier, Lyon, France. During his childhood his father tried to teach him Latin, but he found that Andre’s interests and abilities lied in the study of mathematics. Certainly, Andre cherish the time that his father spent teaching him, for later, during the French Revolution, his father was captured and executed.

Andre met Julie Carron in 1796 and married her three years later. Around the same time, Andre tutored in mathematics, chemistry, and languages. He moved to Bourg-en-Bresse, to teach physics and chemistry in 1801. Unfortunately his wife died two years later leaving him with their infant son, Jean-Jacques Ampere. Andre was appointed of mathematics at the University of Lyon just one year later. In 1809, Andre Ampere was appointed professor of mathematics at the Polytechnic school in Paris. He was admitted as a member of the Institute in 1814 and in 1820, after H.

C. Orsted’s discovery that a magnetic needle is acted on by a voltaic current, Andre sent a paper of his own to the Academy that was much more detailed. He didn’t wait, on September 18, 1820, the very same day that he sent his paper, he presented a demonstration to the Academy that parallel wires with electric currents would pull or push at one another based on whether the electric currents was moving in the same or opposite directions. In demonstrating this experiment he laid the foundation of electrodynamics.

Andre Ampere is best known for the Ampere Circuital Law (Ampere’s Law), which states that for any closed loop path, the sum of the length elements times the magnetic field in the direction of the length element is equal to the permeability times the electric current enclosed in the loop. .Andre also invented the astatic needle and the ampere was named after him. Andre led an inquisitorial life, questioning things he did not fully understand, testing the things that he thought he understood, and proving not only his own theories but the series of many that came after him.

Andre Ampere practically invented the science of electromagnetism and he will always be remembered in years to come. Works Cited “Andre Marie Ampere. ” Columbia Electronic Encyclopedia, 6Th Edition (2011): 1. Academic Search Complete. Web. 25 Jan. 2012. Princeton University. “Ampere’s Theory. ” PrincetonUniversity. edu. Princeton University, 27 Oct. 2010. Web. . Princeton University. “Excerpts: Ampere’s Theory of Magnetism. ” PrincetonUniversity. edu. Princeton University, 27 Oct. 2010. Web. . Nave, C. R. “Ampere’s Law. ” Ampere’s Law. Hyperphysics – Georgia State University, 2011. Web. 25 Jan. 2012. .