3.2 Account for turbine and pump isentropic efficiencies | University of Alberta MEC E 340: Applied Thermodynamics

University of Alberta MEC E 340: Applied Thermodynamics

Vapour Power Cycles

MEC E 340 Applied Thermodynamics — Study topic 3.2

Consider a steady-flow control volume containing a turbine or pump, with one inlet at state 1 and one outlet at state 2. The working fluid may be a vapour or a liquid. For each device, compare the actual process with an ideal isentropic process that has the same inlet state and outlet pressure. The actual outlet is state 2a; the ideal outlet is state 2s. The subscript s means isentropic, not a separate physical outlet. The comparison measures how well the real device approaches the ideal work transfer. This lesson assumes steady operation, negligible changes in kinetic and potential energy, and negligible heat transfer unless stated otherwise. Numerical property data used below are supplied directly in each example; no table values are presumed.

What you will learn

Explain why actual turbine and pump work differs from ideal isentropic work.
Use the correct isentropic-efficiency definition for a turbine or a pump.
Find actual outlet enthalpy and specific work from supplied state data.
Check work direction, units, and physical plausibility.

1. Set up the device and the comparison

For a one-inlet, one-outlet steady-flow device, mass conservation gives the same mass flow rate at inlet and outlet. With heat transfer and changes in kinetic and potential energy neglected, the steady-flow energy balance reduces to a relation between enthalpy change and shaft work. For a turbine, the fluid loses enthalpy and the device delivers work. For a pump, the fluid gains enthalpy and the device requires work input.

To define the ideal comparison, preserve the actual inlet state and the specified outlet pressure. The ideal outlet state is found by setting its entropy equal to the inlet entropy. Then use suitable supplied data or property tables to find its enthalpy. This is an ideal reference calculation, not a claim that the actual outlet entropy remains constant.

Use positive magnitudes for work output from a turbine and work input to a pump in the efficiency definitions. This avoids sign confusion. Specific work and enthalpy differences have units of kJ/kg; multiplying specific work by mass flow rate gives power.

\dot{m}_1=\dot{m}_2,\quad q\approx 0,\quad \Delta ke\approx\Delta pe\approx 0

State 1 is the inlet; 2a is the actual outlet; 2s is the ideal isentropic outlet.
The comparison uses the same inlet state and outlet pressure.
With the stated assumptions, turbine work output equals the enthalpy drop; pump work input equals the enthalpy rise.

2. Turbine isentropic efficiency

An ideal turbine extracts the maximum work for the specified inlet state and outlet pressure. Its isentropic outlet enthalpy is therefore lower than the actual outlet enthalpy. The actual turbine produces less work than the ideal reference.

Turbine isentropic efficiency is actual work output divided by isentropic work output. Under the assumptions above, the work terms can be replaced by enthalpy drops. Once efficiency and the ideal outlet enthalpy are known, solve for the actual outlet enthalpy. A lower actual enthalpy than the isentropic outlet would imply more work than the ideal reference and signals a setup or data error.

To locate state 2s from property data, use the inlet entropy and the specified outlet pressure. For example, in a vapour-table problem, the appropriate table or stated property model must supply the enthalpy at that pressure and entropy. If the problem does not give enough information to determine it, do not invent a value.

\eta_t=\frac{h_1-h_{2a}}{h_1-h_{2s}}

For a turbine, the actual enthalpy drop is smaller than the ideal enthalpy drop.
Efficiency is a ratio of work magnitudes and is dimensionless.
Use supplied efficiency data; do not assume a standard efficiency.

3. Pump isentropic efficiency

A pump raises the fluid pressure and requires work input. In the ideal comparison, the pump process is isentropic between the specified inlet state and outlet pressure. The ideal enthalpy rise is less than the actual enthalpy rise because a real pump needs additional work.

Pump isentropic efficiency is ideal work input divided by actual work input. With negligible heat, kinetic-energy, and potential-energy changes, use the corresponding enthalpy rises. Rearranging gives the actual outlet enthalpy when the ideal enthalpy rise and efficiency are known.

For a liquid, a problem may supply the approximation that specific volume is constant, allowing ideal pump work to be estimated as specific volume times pressure rise. This approximation is appropriate only when it is stated or clearly permitted by the supplied data. Convert pressure units so that the product has units of energy per mass:

\mathrm{m^3/kg}\cdot\mathrm{kPa}=\mathrm{kJ/kg}

.

\eta_p=\frac{h_{2s}-h_1}{h_{2a}-h_1}

For a pump, actual work input and enthalpy rise exceed their ideal counterparts.
The pump efficiency ratio is the inverse arrangement of the turbine ratio.
If using the liquid approximation, state it and keep pressure units consistent.

4. A reliable calculation and reasonableness check

Write down the device, working fluid, inlet state, actual outlet pressure, and assumptions before calculating. Identify whether the question asks for ideal or actual outlet properties, specific work, or power. Find the isentropic outlet property first, using the inlet entropy and outlet pressure with only the data or property source available in the problem.

Next, use the efficiency definition for the correct device and solve for the requested actual quantity. Keep actual and ideal states distinct in every equation. If mass flow rate is provided, calculate power from mass flow rate times specific work, then convert units if needed.

Finally, check signs and trends. A turbine should deliver positive work, with

h_1>h_{2a}>h_{2s}

under the stated simplified model. A pump should take in positive work, with

h_{2a}>h_{2s}>h_1

. Both efficiencies should be positive and no greater than one for the comparison described here. A result outside these trends calls for a review of the state labels, work direction, efficiency rearrangement, or supplied data.

\dot{W}=\dot{m}\,w

Find the ideal outlet state before applying the efficiency.
Do not substitute the ideal enthalpy into the actual work calculation.
Check state ordering, work direction, dimensions, and efficiency range.

Worked example

Turbine: actual outlet enthalpy and work

A steady, adiabatic turbine handles vapour at a mass flow rate of

2.0\ \mathrm{kg/s}

. Supplied data give

h_1=3200\ \mathrm{kJ/kg}

and, at the outlet pressure with

s_{2s}=s_1

,

h_{2s}=2500\ \mathrm{kJ/kg}

. The turbine efficiency is

0.80

. Neglect changes in kinetic and potential energy. Find the actual outlet enthalpy and power output.

Steady turbine control volume

Schematic device view, not a property plot. Vapour enters at state 1 and exits at state 2a; the stated model is steady and adiabatic.

Identify the model
Use the turbine control volume with state 1 at the inlet and state 2a at the actual outlet. State 2s is the ideal outlet at the same outlet pressure. The problem supplies both inlet and isentropic-outlet enthalpies, as well as efficiency.
Apply the turbine efficiency
The ideal enthalpy drop is $700\ \mathrm{kJ/kg}$ . The actual drop is the stated fraction of that ideal drop. This gives the actual outlet enthalpy by subtracting the actual drop from the inlet enthalpy.
$h_1-h_{2a}=0.80(h_1-h_{2s})=560\ \mathrm{kJ/kg}$
Calculate power
For the stated adiabatic, steady model, the specific work output equals the actual enthalpy drop. Multiply by mass flow rate to obtain power.
$\dot{W}_{out}=2.0\ \mathrm{kg/s}\times560\ \mathrm{kJ/kg}=1120\ \mathrm{kW}$

Answer:

h_{2a}=2640\ \mathrm{kJ/kg}

and

\dot{W}_{out}=1120\ \mathrm{kW}

.

Check: The actual enthalpy drop,

560\ \mathrm{kJ/kg}

, is less than the ideal drop,

700\ \mathrm{kJ/kg}

. The actual outlet enthalpy is above the ideal outlet enthalpy, as expected for a turbine.

Worked example

Pump: estimate actual work for a liquid

A steady pump handles liquid water. The supplied model treats the liquid specific volume as constant at

0.00100\ \mathrm{m^3/kg}

. The pressure rise is

500\ \mathrm{kPa}

, the pump efficiency is

0.75

, and the inlet enthalpy is

100\ \mathrm{kJ/kg}

. Neglect heat transfer and changes in kinetic and potential energy. Estimate ideal pump work, actual specific work input, and actual outlet enthalpy.

Steady liquid pump control volume

Schematic device view, not a property plot. Liquid enters at state 1 and exits at state 2a; the supplied assumptions are steady operation and negligible heat transfer.

Estimate ideal pump work
For the stated constant-specific-volume liquid approximation, multiply specific volume by pressure rise. The given units produce kJ/kg directly. w_{s,in}=v\Delta P=(0.00100\ $\mathrm{m^3/kg}$ )(500\ $\mathrm{kPa}$ )=0.500\ $\mathrm{kJ/kg}$
Use pump efficiency
Pump efficiency is ideal work input divided by actual work input. Therefore, the actual input is the ideal input divided by efficiency. w_{a,in}= $\frac{0.500}{0.75}$ =0.667\ $\mathrm{kJ/kg}$
Find the outlet enthalpy
With the stated energy-balance assumptions, actual pump work input equals the actual enthalpy rise. Add this rise to the inlet enthalpy.
$h_{2a}=100+0.667=100.667\ \mathrm{kJ/kg}$

Answer: The ideal work input is

0.500\ \mathrm{kJ/kg}

, the actual work input is approximately

0.667\ \mathrm{kJ/kg}

, and

h_{2a}\approx100.667\ \mathrm{kJ/kg}

.

Check: Actual work input exceeds ideal work input, and the outlet enthalpy exceeds the inlet enthalpy. The approximation is based only on the specific volume and pressure rise supplied in the problem.

Worked example

Turbine: use efficiency to find power

A steady turbine receives a working fluid at

h_1=1450\ \mathrm{kJ/kg}

. Supplied property data at the specified outlet pressure and inlet entropy give

h_{2s}=1100\ \mathrm{kJ/kg}

. The turbine has an isentropic efficiency of

0.85

and a mass flow rate of

4.0\ \mathrm{kg/s}

. With heat transfer and changes in kinetic and potential energy neglected, determine actual specific work output and power output.

Steady turbine control volume

Schematic device view, not a property plot. The working fluid flows from state 1 to state 2a, and work leaves the turbine.

Calculate ideal work
The ideal turbine work output is the inlet enthalpy minus the supplied isentropic-outlet enthalpy.
$w_{s,out}=1450-1100=350\ \mathrm{kJ/kg}$
Calculate actual specific work
Actual turbine work output is efficiency times ideal turbine work output. This gives the actual enthalpy drop under the stated assumptions.
$w_{a,out}=0.85(350)=297.5\ \mathrm{kJ/kg}$
Calculate power
Multiply actual specific work output by the supplied mass flow rate. Since $1\ \mathrm{kJ/s}=1\ \mathrm{kW}$ , the result is directly in kilowatts.
$\dot{W}_{out}=(4.0)(297.5)=1190\ \mathrm{kW}$

Answer: The actual specific work output is

297.5\ \mathrm{kJ/kg}

, and power output is

1190\ \mathrm{kW}

.

Check: The actual work is below the ideal value of

350\ \mathrm{kJ/kg}

, consistent with an efficiency below one. The power units follow from multiplying kJ/kg by kg/s.

Common mistakes and how to avoid them

Using the pump efficiency definition for a turbine, or vice versa.

Correction: For a turbine, divide actual work output by ideal work output. For a pump, divide ideal work input by actual work input.

Treating the isentropic outlet as the actual outlet.

Correction: Label the reference state 2s and the real outlet 2a. Apply efficiency to find the actual work or enthalpy.

Using the same enthalpy ordering for both devices.

Correction: For a turbine, the ideal enthalpy drop is larger than the actual drop. For a pump, the actual enthalpy rise is larger than the ideal rise.

Assuming an efficiency value or property value not provided.

Correction: Use data supplied in the problem or a named property source. If required information is missing, state that rather than inventing it.

Lesson summary

Define state 1, actual outlet 2a, and isentropic reference outlet 2s at the same outlet pressure.
Turbine efficiency compares actual work output with ideal isentropic work output.
Pump efficiency compares ideal isentropic work input with actual work input.
Use a steady-flow energy balance to connect work with enthalpy change when heat transfer and kinetic- and potential-energy changes are negligible.
Check the work direction, enthalpy ordering, SI units, and whether efficiency lies between zero and one.

Check your understanding

Question 1

For a turbine with

h_1=900\ \mathrm{kJ/kg}

,

h_{2s}=600\ \mathrm{kJ/kg}

, and efficiency

0.80

, what is the actual specific work output?

$240\ \mathrm{kJ/kg}$
$300\ \mathrm{kJ/kg}$
$375\ \mathrm{kJ/kg}$
correctIndex: 0, "explanation": "The ideal work output is $900-600=300\ \mathrm{kJ/kg}$ . Turbine efficiency is actual divided by ideal, so actual work is $0.80(300)=240\ \mathrm{kJ/kg}$ ."}

Show answer and explanation

240\ \mathrm{kJ/kg}

The ideal work output is

900-600=300\ \mathrm{kJ/kg}

. Turbine efficiency is actual divided by ideal, so actual work is

0.80(300)=240\ \mathrm{kJ/kg}

.

Question 2

For an adiabatic pump under the simplified steady-flow assumptions, which enthalpy rise is larger?

The actual rise is larger than the isentropic rise.
The isentropic rise is larger than the actual rise.
The two rises must be equal for any pump.
correctIndex: 0, "explanation": "Pump efficiency is ideal work input divided by actual work input. Since actual input exceeds ideal input, the actual enthalpy rise is larger under the stated assumptions."}

Show answer and explanation

The actual rise is larger than the isentropic rise.

Pump efficiency is ideal work input divided by actual work input. Since actual input exceeds ideal input, the actual enthalpy rise is larger under the stated assumptions.

Key terms

Isentropic: A process with constant entropy; here it describes the ideal reference process from the inlet state to the specified outlet pressure.
Isentropic efficiency: A dimensionless comparison of actual device work with ideal isentropic work, arranged differently for turbines and pumps.
Specific work: Work transfer per unit mass of working fluid, commonly expressed in kJ/kg.

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3.2 · Account for turbine and pump isentropic efficiencies

University of Alberta MEC E 340: Applied Thermodynamics

MEC E 340 Applied Thermodynamics — Study topic 3.2

What you will learn

1. Set up the device and the comparison

2. Turbine isentropic efficiency

3. Pump isentropic efficiency

4. A reliable calculation and reasonableness check

Turbine: actual outlet enthalpy and work

Pump: estimate actual work for a liquid

Turbine: use efficiency to find power

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through MEC E 340

About this lesson and its review