WorksheetsSTEAM TURBINE EFFICIENCY WORKSHEET
Total questions: 15
Worksheet time: 2hrs 30mins
What does Thermal Efficiency (η_th) measure in a steam cycle?
How close the turbine performance is to ideal expansion
Losses in mechanical components
How effectively the steam cycle converts heat energy into work
The amount of fuel consumed
What is the formula for Thermal Efficiency (η_th)?
η_th = (W_turbine - W_pump) / Q_in
η_th = (h1 - h2_actual) / (h1 - h2s)
η_th = P_shaft / P_turbine
η_th = Q_in / (W_turbine - W_pump)
What does Mechanical Efficiency (η_mec) measure?
How effectively the steam cycle converts heat energy into work
Losses in mechanical components such as bearings and gears
The amount of heat supplied
The ideal expansion of steam
Example 1: Thermal Efficiency Given: W_turbine = 1200 kW, W_pump = 10 kW, Q_in = 3500 kW Calculate η_th = (W_turbine - W_pump) / Q_in = (1200 - 10) / 3500 = _____
0.34 or 34%
0.28 or 28%
0.45 or 45%
0.15 or 15%
Example 2: Isentropic Efficiency Given: h1 = 3200 kJ/kg, h2_actual = 2800 kJ/kg, h2s = 2700 kJ/kg Calculate η_is = (h1 - h2_actual) / (h1 - h2s) = (3200 - 2800) / (3200 - 2700) = _____
0.8 or 80%
0.5 or 50%
0.6 or 60%
0.9 or 90%
Example 3: Mechanical Efficiency Given: P_turbine = 1200 kW, P_shaft = 1100 kW Calculate η_mec = P_shaft / P_turbine = 1100 / 1200 = _____
0.9167 or 91.67%
0.8500 or 85.00%
0.9750 or 97.50%
0.8000 or 80.00%
A turbine produces 2400 kW, the pump uses 20 kW, and the heat supplied is 7000 kW. Find the thermal efficiency.
34%
28%
45%
52%
Problem 2: Isentropic Efficiency Given: h1 = 3500 kJ/kg, h2,actual = 2900 kJ/kg, h2s = 2750 kJ/kg. Find ηis.
80%
60%
50%
90%
Problem 3: Mechanical Efficiency Given: Pturbine = 600 kW, Pshaft = 560 kW. Find ηmec.
560/600 = 0.933 or 93.3%
600/560 = 1.071 or 107.1%
560/600 = 0.867 or 86.7%
600/560 = 0.933 or 93.3%
A low thermal efficiency in a power plant indicates:
that the plant wastes a large portion of input energy
that the plant produces more electricity than expected
that the plant operates at maximum possible output
that the plant uses less fuel for more output
The isentropic efficiency can never be 100% because:
irreversibilities and losses are always present in real processes.
ideal processes are always achievable in practice.
isentropic efficiency is always greater than 100%.
isentropic efficiency is not related to real processes.
Mechanical factors that can reduce turbine mechanical efficiency include:
Friction and misalignment
Increased fuel supply
Higher ambient temperature
Improved lubrication
What does a low thermal efficiency indicate about a power plant's performance?
Why can the isentropic efficiency never be 100%?
What mechanical factors can reduce turbine mechanical efficiency?
