# Isentropic Efficiency and Entropy Production Rate of Turbine

• AGiantGolden49er
In summary: Yes, the way our professor taught us, all the exit values under constant entropy are marked with subscripts.
AGiantGolden49er

## Homework Statement

Water vapor at 6 MPa, 600C enters a turbine operating at steady state and expands to 10 kPa. The mass flow rate is 2 kg/s, and the power developed is 2626 kW. Stray heat transfer and kinetic and potential energy effects are negligible. Determine (a) the isentropic turbine efficiency and (b) the rate of entropy production within the turbine, in kW/K.

## Homework Equations

dmcv / dt = ∑ mi - ∑ me
dEcv / dt = Qcv - Wcv + ∑ mi (hi + Vi2/2 + gzi) - ∑ me (he + Ve2/2 + gze)
dScv / dt = ∑ Qj / Tj + ∑ misi - ∑ mese + σcv

## The Attempt at a Solution

From the problem statement, I know that entry mass flow rate and exit mass flow rate are the same, because it is steady-state. I also know that, because of the steady-state operation, the energy balance equation simplifies to:
Wcv = m(h1 - h2)

Using the steam tables, I found the enthalpy and entropy values at state 1 to be 3658.4 kJ/kg and 7.1677 kJ/kg⋅K, respectively. I can use this enthalpy value, along with the given mass flow rate and work output values to find h2, but what I need to find is h2s. Once I find that, I feel like I can crack this problem wide open, but I'm having issues on how to do just that. Since the working fluid isn't an ideal gas, I can't use the relative pressure and volume relations, so I haven't the slightest clue on how to find h2s. If anyone can steer me in the right direction, that would be really appreciated.

AGiantGolden49er said:

## Homework Statement

Water vapor at 6 MPa, 600C enters a turbine operating at steady state and expands to 10 kPa. The mass flow rate is 2 kg/s, and the power developed is 2626 kW. Stray heat transfer and kinetic and potential energy effects are negligible. Determine (a) the isentropic turbine efficiency and (b) the rate of entropy production within the turbine, in kW/K.

## Homework Equations

dmcv / dt = ∑ mi - ∑ me
dEcv / dt = Qcv - Wcv + ∑ mi (hi + Vi2/2 + gzi) - ∑ me (he + Ve2/2 + gze)
dScv / dt = ∑ Qj / Tj + ∑ misi - ∑ mese + σcv

## The Attempt at a Solution

From the problem statement, I know that entry mass flow rate and exit mass flow rate are the same, because it is steady-state. I also know that, because of the steady-state operation, the energy balance equation simplifies to:
Wcv = m(h1 - h2)

Using the steam tables, I found the enthalpy and entropy values at state 1 to be 3658.4 kJ/kg and 7.1677 kJ/kg⋅K, respectively. I can use this enthalpy value, along with the given mass flow rate and work output values to find h2, but what I need to find is h2s. Once I find that, I feel like I can crack this problem wide open, but I'm having issues on how to do just that. Since the working fluid isn't an ideal gas, I can't use the relative pressure and volume relations, so I haven't the slightest clue on how to find h2s. If anyone can steer me in the right direction, that would be really appreciated.
##h_2s##?? That's the exit enthalpy value assuming constant entropy?

Chestermiller said:
##h_2s##?? That's the exit enthalpy value assuming constant entropy?
Yes, the way our professor taught us, all the exit values under constant entropy are marked with subscript s.

AGiantGolden49er said:
Yes, the way our professor taught us, all the exit values under constant entropy are marked with subscript s.
Then what you do is look up in your steam tables the state where the pressure is 10 kPa and the entropy is 7.1677. You might need to interpolate. You need to find the enthalpy for this state.

Chestermiller said:
Then what you do is look up in your steam tables the state where the pressure is 10 kPa and the entropy is 7.1677. You might need to interpolate. You need to find the enthalpy for this state.
So I think I found the enthalpy value, assuming constant entropy, which worked out to be around 2217.17 kJ/kg. I had to find quality under constant entropy first in order to do so. Consequently, I also found the isentropic turbine efficiency to be around 91.1%. Does this sound like I'm on track?

AGiantGolden49er said:
So I think I found the enthalpy value, assuming constant entropy, which worked out to be around 2217.17 kJ/kg. I had to find quality under constant entropy first in order to do so. Consequently, I also found the isentropic turbine efficiency to be around 91.1%. Does this sound like I'm on track?
Yes

## 1. What is isentropic efficiency of a turbine?

The isentropic efficiency of a turbine is a measure of how well the turbine is able to convert the available energy into useful work. It is the ratio of the actual work output of the turbine to the maximum work output that could be achieved if the process were isentropic (no heat transfer or friction losses).

## 2. What factors affect the isentropic efficiency of a turbine?

The isentropic efficiency of a turbine is affected by various factors such as the design and shape of the turbine blades, the quality of the working fluid, the inlet and outlet conditions, and the operating speed of the turbine.

## 3. How is isentropic efficiency calculated?

The isentropic efficiency of a turbine can be calculated by dividing the actual work output by the ideal work output (assuming isentropic conditions). The ideal work output can be calculated using the inlet and outlet pressure and temperature values.

## 4. What is entropy production rate of a turbine?

The entropy production rate of a turbine is a measure of the amount of entropy that is being generated in the turbine due to irreversibilities, such as friction and heat transfer. It is a key factor in determining the overall efficiency of the turbine.

## 5. How do you calculate the entropy production rate of a turbine?

The entropy production rate of a turbine can be calculated by multiplying the rate of change of entropy with respect to time by the mass flow rate through the turbine. This value can be obtained from the energy and mass balance equations for the turbine.

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