Note : It is 36.7 Mj/kg (Mega) and not milli. Please help me. A farmer who raise

Question

Note: It is 36.7 Mj/kg (Mega) and not milli. Please help me.

A farmer who raises sunflowers has constructed an inexpensive extractor which provides 5000 gallons of sunflower seed oil per year. The farmer would like to construct a thermoelectric generator that would be powered by the sunflower seed oil. Selecting thermoelectric materials from Appendix C and assuming a hot shoe temperature of 550 degree C and a cold shoe temperature of 80 degree C, estimate the number of kilowatt-hours the farmer can reasonably expect to produce from the supply of sunflower seed oil. You may assume that the fuel has a heating value of 36.7 mj/kg and a density of 0.9 g/ml.

Explanation / Answer

Let us construct a carnot engine for extrac ting power generator

so efficiency of the generator = 1 - ( T_low) / T_high

T_high = hot temperature in kelovin = 550 + 273 = 823 K

T_low = low temp in kelvin = 80 + 273 = 353 K

so efficientgy of engine = 1 - (353 / 823 )=0.5710814 = 57.10814 %


now amount of seeds available = 5000 galloms = 18927.1 litres = 18.9271 m^3

density of oil = 900 kg / m^3

so mass of oil = 18.9271 * 900 = 17034.39 kg


so energy available in seeds = 17034.39 * 36.7 * 10^6 Joules = 6.25162113 * 10^11 Joules

efficiency =0.5710814

so maximum energy that can be produced = 0.5710814 * 6.25162113 * 10^11
= 3.57018453005754 * 10^11 Joules

1 kilo watt hour = 1000*3600 joules = 3.6 * 10^6 joules


so energy which can be produce in kwh = (3.57018453005754 * 10^11) / ( 3.6 * 10^6 )

= 9.917179 * 10^4 KWH = 99171.792 KWH

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