A real turbine with a mass flow rate of 5.0 kg/s of steam has inlet conditions o

Question

A real turbine with a mass flow rate of 5.0 kg/s of steam has inlet conditions of 6.0 MPa and 700 oC, and exit conditions of saturated vapor at 50 kPa. If the power output of the turbine is 5.0 MW, determine the rate of heat loss from the turbine to 3 significant figures in kW. Assume the inlet/outlet kinetic and potential energy changes are negligible.

For problem (1) above, determine the increase in specific entropy of the steam as it flows through the turbine (units of kJ/kg K).

A small-scale pumped storage system pumps water from a lake to a higher elevation reservoir at times for use in a small hydroelectric turbine at times of high demand. The high-elevation reservoir is 45 meters above the height of the lake. For an energy storage capacity of 850 kWh what is the amount of water that needs to be stored at the high elevation reservoir in units of Mg (millions of grams)? Provide answer to 4 significant figures.

Explanation / Answer

Q2 . 70*3480For 6 MPa , 700 C

Specific entropy = 7.4247812760189 KJ/kg
specific enthalpy = 3894.4690345661 KJ/kg

now.. in the turbine.. the entropy will remain same for inlet as well as exit states...

for exit state
P = 50 Kpa.

saturated vapour

T_sat = 81.31673599664 C


Specific entropy_steam = 7.5929627746412 KJ/kgK

Specific enthalpy_steam = 2645.2132384179 KJ/kg


so change in enthalpy = 3894.4690345661 - 2645.2132384179 = 1249.2557961482 KJ/kg
mass flow rate = 5 kg/sec

so actual energy released = 5 * 1249.2557961482 = 6246.278980741 KJ/sec = 6246.27898 KW = 6.24627898 MW

actual energy produces = 5 MW
so... rate of heat loss = 6.24627898 - 5 = 1.24627898 MW


2) increase = 7.5929627746412 - 7.4247812760189 = 0.1681814986223 KJ/kgK

3) mass * g * height = 850 KWH
mass * 9.8 * 45 = 850 * 3600 * 1000

so mass = 6938775.510204 kg = 6938775510.204 grams = 6938775510.204 / ( 10^6 ) Mg = 6938.77551 Mg

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