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so i firgured this might not be the right place exactly to ask for help about a thermodynamics type of question. but i found some electrical questions present so i decided to post it here too for some help.

so basically i have been doing an experiment on steam powersteam power diagram

and have measured and calculated quite a bit of data:

voltage, load resistance, current, power, rpm of turbine, angular speed, k constant, torque, heater's power, boiler's temp and pressure, time taken for water mass to reduce by 100g

and now im stuck at this question which im not sure of the "correct way" to go about it.

The question was to find out the overall efficiency and plot it against resistance, so resistance is pretty simple, but the overall efficiency is puzzling me at the moment.

my guess would be that:

overall efficiency is = (power generated(calculated from voltage and current values) + mechanical power generated from the turbine (Pmech= torque * angular speed)) / (the total net power introduced into the system which is the heater's power).

i would like to know what you guys think of what i thought of to solve the question?

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  • \$\begingroup\$ \$\dfrac{Useful\space power\space out}{power\space in}\$ \$\endgroup\$
    – Andy aka
    Oct 18, 2014 at 9:49

2 Answers 2

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$$Eff = \frac{P_{out}(overall \; system)}{P_{in}(overall \; system)} = \frac{V \times A}{W} $$

The result is always < 1 due to losses. The numerator of your equation is NOT the sum of power generated(calculated from voltage and current values) plus mechanical power from the turbine. The generator converts mechanical power to electrical power (think of it as a large transducer which is a device that converts one form of energy into another form). \$P_{gen} = P_{turbine} - energy \; losses\$ translating mechanical to electrical power. Don't add \$P_{gen} + P_{turbine}\$ in the numerator.

Stepping through the process:

  1. Heat is generated to boil the water by means of an electrical supply \$(P_{in} = W)\$. Heat input is regulated via a varistat. Not all heat can be used to raise the temperature of the water (some is lost to the environment because the boiler’s insulation is not perfect). This step has \$Eff_1\$ (<1)

  2. Water heats up and produces steam. The steam's pressure and temperature can be measured with \$P\$ and \$T\$. Steam exits the nozzle which drives the turbine wheel. Not all heat energy from the steam acts on the turbine because some steam escapes and not all the steam's heat energy can be converted into mechanical work (thermodynamic limitations prevent this). This step has \$Eff_2\$ (<1).

  3. The turbine's mechanical power (shaft angular speed x torque) is converted into electrical power by the generator. Due to system losses (bearing friction, generator windage, eddy currents, \$I^2 \times R\$ heat losses, etc), not all mechanical power can be converted to electrical power. This step has \$Eff_3\$ (<1).

Overall efficiency = \$ Eff_1 \times Eff_2 \times Eff_3 \$ = efficiency equation above.

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If the heat from your steam turbine comes entirely from the 240V AC supply, and the output power is defined to be the power delivered to the load potentiometer, then you can ignore the turbine in your efficiency calculation, right? If you want the overall efficiency, you don't need to calculate the power transferred and lost in each stage. Just look at the input electrical power and the output electrical power. Make sure your output power is given in terms of the resistance (I^2 * R or V^2 / R) to make it easy to plot.

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