1.3 Apply steady-flow mass and energy balances to cycle devices | University of Alberta MEC E 340: Applied Thermodynamics

University of Alberta MEC E 340: Applied Thermodynamics

Cycle Analysis Foundations

Applying control-volume balances to turbines, compressors, and heaters

A thermodynamic cycle contains devices through which working fluid flows. To analyze one device, draw a control volume around it and track mass and energy crossing its boundary. The fluid entering and leaving has energy that includes enthalpy, kinetic energy, and potential energy; heat and shaft work may also cross the boundary. In many cycle-device problems, changes in kinetic and potential energy are small enough to neglect, but that is an assumption to state, not an automatic rule. This lesson reviews the steady-flow balances needed to connect a device’s inlet and outlet states to its heat transfer or work.

What you will learn

Choose a control volume and identify its inlets, outlets, and energy interactions.
Apply steady-flow mass and energy balances to a cycle device.
Use supplied enthalpy data and consistent sign conventions to calculate a device’s heat or work transfer.
Check that the result agrees with the stated assumptions and energy-flow directions.

1. Define the control volume and the sign convention

A control volume is a chosen region in space through which mass may flow. For a turbine, it can enclose the turbine casing; for a compressor, it can enclose the compressor. Number the inlet and outlet states on a sketch before using property data. State 1 commonly denotes an inlet and state 2 an outlet, but the numbering must match the problem and diagram.

Use the working fluid and all supplied state information. Enthalpy is especially useful for flowing streams because it accounts for internal energy and the flow work needed to push fluid across the control-volume boundary. Obtain enthalpies from the data supplied in the problem or an identified property source; do not assume values that are not given.

For this lesson, take heat transfer into the control volume as positive and work delivered by the control volume as positive. Thus, a turbine delivering shaft work has positive work, while a compressor requiring shaft work has negative work under this convention. Some engineering calculations instead report work input as a positive magnitude; identify the convention and convert carefully.

Steady operation means the amount of mass and energy stored inside the control volume does not change with time. It does not mean that mass or energy transfer is zero: streams can continuously enter and leave, and heat and work can cross the boundary.

\dot{m}_{in}=\dot{m}_{

\mathrm{out}

}

Label states, flow directions, and heat and work directions before calculating.
Use a consistent sign convention and supplied property data.
Steady flow means no accumulation inside the control volume.

2. Apply the mass and energy balances

The steady-flow mass balance says that total mass flow entering equals total mass flow leaving when there is no accumulation. In a device with one inlet and one outlet, the mass flow rates are equal. With multiple streams, add the rates on each side; do not equate just one inlet to one outlet unless the device has only those streams.

The steady-flow energy balance accounts for heat, work, and the energy carried by each stream. For a single-inlet, single-outlet device, the stream contribution per unit mass is enthalpy plus kinetic energy and gravitational potential energy. If the inlet and outlet speeds or elevations are not supplied, a problem may direct you to neglect their changes. State that simplification explicitly.

For a one-inlet, one-outlet device with negligible changes in kinetic and potential energy, the energy balance becomes especially direct: heat transfer minus work output equals mass flow rate times the enthalpy rise. This same equation covers different devices. A turbine often has an enthalpy decrease that supports work output; a compressor often has an enthalpy increase that requires work input. A heater with no shaft work has heat input associated with an enthalpy increase.

For a cycle, the working fluid returns to its initial state after all devices and processes. The cycle’s net energy change is zero, but an individual device generally has different inlet and outlet enthalpies. Apply a control-volume balance to each device as needed; do not set each device’s enthalpy change to zero just because the complete cycle returns to its starting state.

\dot{Q}-\dot{W}_{\mathrm{out}}=\dot{m}(h_2-h_1)

For one inlet and one outlet at steady state, mass flow rates are equal.
Keep kinetic and potential energy terms unless the problem permits neglecting them.
Use the energy balance for the device being analyzed, not the whole cycle.

3. Use properties, solve, and check

Begin with the information provided: mass flow rate, inlet and outlet states, enthalpies, heat-transfer rate, or work rate. If a state is described by pressure and temperature, use the appropriate supplied property data to obtain enthalpy. If enthalpy is given directly, no extra table lookup is needed. Do not infer a phase or a property value from a schematic plot.

Keep units consistent. A mass flow rate in kilograms per second multiplied by an enthalpy difference in kilojoules per kilogram gives kilojoules per second, equivalent to kilowatts. When energy rates use watts instead, convert consistently. Write whether a calculated work rate is output or input; a negative result under the stated sign convention means work input.

Finally, check the result against the physical description. An adiabatic turbine with a positive enthalpy drop should deliver work under the stated approximations. A compressor with an enthalpy rise and heat loss needs more work input than the enthalpy rise alone would require. A heater with no work interaction should supply heat when the fluid’s enthalpy increases. These checks catch sign errors, but they do not replace the balance.

1\ \mathrm{kJ/s}=1\ \mathrm{kW}

Read only the properties provided or obtained from a named source.
Check dimensions: mass flow rate times specific energy is an energy rate.
Compare the sign and direction of the result with the device description.

Worked example

Adiabatic turbine work output

Steam flows steadily through a turbine at

2.5\ \mathrm{kg/s}

. Supplied inlet and outlet enthalpies are

3400\ \mathrm{kJ/kg}

and

2800\ \mathrm{kJ/kg}

, respectively. Assume one inlet and one outlet, adiabatic operation, and negligible changes in kinetic and potential energy. Find the turbine work output rate.

Steady-flow turbine

Control volume around the turbine; one inlet at state 1 and one outlet at state 2. Adiabatic, steady operation; kinetic and potential energy changes are neglected.

Set the control volume and directions
Enclose the turbine. State 1 is the inlet and state 2 is the outlet. Heat transfer is zero, and positive work denotes work delivered by the turbine. The stated assumptions allow the enthalpy-only form of the energy balance.
$\dot{Q}=0$
Apply mass and energy balances
Steady operation with one inlet and one outlet gives the same mass flow rate through both. Solve the energy balance for work output.
$\dot{W}_{\mathrm{out}}=\dot{m}(h_1-h_2)$
Substitute the supplied data
The enthalpy decreases across the turbine, so the calculated work output should be positive.
$\dot{W}_{\mathrm{out}}=(2.5\ \mathrm{kg/s})(3400-2800\ \mathrm{kJ/kg})=1500\ \mathrm{kW}$

Answer: The turbine delivers

1500\ \mathrm{kW}

of work.

Check: The outlet enthalpy is lower than the inlet enthalpy, and the adiabatic balance predicts positive work output. The units reduce to kilojoules per second, or kilowatts.

Worked example

Compressor work input with heat loss

Air flows steadily through a compressor at

0.80\ \mathrm{kg/s}

. Its enthalpy rises from

300\ \mathrm{kJ/kg}

at state 1 to

430\ \mathrm{kJ/kg}

at state 2. Heat is lost from the compressor at

10\ \mathrm{kW}

. Assume one inlet and one outlet and negligible changes in kinetic and potential energy. Find the compressor work input rate.

Compressor with heat loss

Control volume around the compressor; one inlet at state 1 and one outlet at state 2. Heat leaves, shaft work enters, and kinetic and potential energy changes are neglected.

Assign signs
Heat into the control volume is positive, so the stated heat loss gives a negative heat-transfer rate. Work delivered out is positive, so compressor work input will appear as a negative work output.
$\dot{Q}=-10\ \mathrm{kW}$
Apply the energy balance
Use the supplied inlet and outlet enthalpies and solve for work output. The enthalpy increase requires energy, while heat loss removes energy from the control volume.
$\dot{W}_{\mathrm{out}}=\dot{Q}-\dot{m}(h_2-h_1)$
Calculate work input
The negative result is work input under the chosen convention. Report the input as a positive magnitude.
$\dot{W}_{\mathrm{out}}=-10-(0.80)(430-300)=-114\ \mathrm{kW}$

Answer: The compressor requires

114\ \mathrm{kW}

of work input.

Check: The enthalpy increase alone accounts for

104\ \mathrm{kW}

, and the additional

10\ \mathrm{kW}

offsets heat loss. Thus

114\ \mathrm{kW}

input is consistent with the stated energy flows.

Worked example

Heat supplied in a steady-flow heater

A working fluid passes through a heater at

0.20\ \mathrm{kg/s}

. The supplied inlet enthalpy is

500\ \mathrm{kJ/kg}

and the outlet enthalpy is

750\ \mathrm{kJ/kg}

. Assume steady operation, one inlet and one outlet, no shaft work, and negligible changes in kinetic and potential energy. Find the heat-transfer rate into the fluid.

Steady-flow heater

Control volume around the heater; one inlet at state 1 and one outlet at state 2. Steady operation, no shaft work, and negligible kinetic and potential energy changes.

Identify the device interactions
Enclose the heater as the control volume. The stated no-shaft-work assumption sets work output to zero. The one-inlet, one-outlet steady mass balance gives a common mass flow rate.
$\dot{W}_{\mathrm{out}}=0$
Reduce the energy balance
With no shaft work and negligible kinetic and potential energy changes, heat transfer supplies the fluid’s enthalpy increase.
$\dot{Q}=\dot{m}(h_2-h_1)$
Calculate the heat rate
The positive result means heat enters the control volume, matching the description of a heater.
$\dot{Q}=(0.20\ \mathrm{kg/s})(750-500\ \mathrm{kJ/kg})=50\ \mathrm{kW}$

Answer: The heater supplies

50\ \mathrm{kW}

of heat to the fluid.

Check: The outlet enthalpy is higher than the inlet enthalpy, so positive heat input is consistent. The units are kilowatts.

Common mistakes and how to avoid them

Treating work input to a compressor as positive while using a convention where work output is positive.

Correction: Under the stated convention, compressor work output is negative. Report the positive magnitude separately as work input.

Using the complete-cycle energy change as the energy change across each individual device.

Correction: Apply the steady-flow balance to the specific device. Its inlet and outlet enthalpies can differ even though the working fluid eventually completes a cycle.

Dropping kinetic or potential energy terms without stating an assumption.

Correction: Retain the terms when relevant data are supplied; neglect them only when justified or directed by the problem.

Using an unsourced enthalpy value or assuming a phase from a schematic diagram.

Correction: Use property data supplied by the problem or an identified property source, and state the assumptions used.

Lesson summary

A steady-flow control volume has no mass or energy accumulation.
For one inlet and one outlet, the inlet and outlet mass flow rates are equal.
With negligible kinetic and potential energy changes, heat transfer minus work output equals mass flow rate times the enthalpy rise.
Sign convention, units, supplied properties, and a physical plausibility check are all part of a correct device analysis.

Check your understanding

Question 1

A steady adiabatic turbine has one inlet and one outlet, with negligible kinetic and potential energy changes. The supplied data show that outlet enthalpy is lower than inlet enthalpy. Under the convention that work output is positive, what does the balance predict?

Positive work output
Work input
Zero work because the device is adiabatic
Heat input equal to the enthalpy decrease

Show answer and explanation

Positive work output

With zero heat transfer, the decrease in stream enthalpy supports positive work output under the stated assumptions.

Question 2

A compressor has a mass flow rate of

0.50\ \mathrm{kg/s}

and an enthalpy rise of

80\ \mathrm{kJ/kg}

. It is adiabatic, and kinetic and potential energy changes are negligible. What is the work input rate?

$40\ \mathrm{kW}$
$160\ \mathrm{kW}$
$40\ \mathrm{kJ/kg}$
$-40\ \mathrm{kg/s}$

Show answer and explanation

40\ \mathrm{kW}

The required work input magnitude is mass flow rate times enthalpy rise:

(0.50)(80)=40\ \mathrm{kJ/s}=40\ \mathrm{kW}

.

Key terms

Control volume: A selected region in space across whose boundary mass and energy may flow.
Steady flow: Operation with no change over time in the mass and energy stored inside the control volume.
Enthalpy: A fluid property used in flow-energy calculations; its units here are kilojoules per kilogram.
Mass flow rate: The mass crossing a boundary per unit time, measured here in kilograms per second.

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1.3 · Apply steady-flow mass and energy balances to cycle devices

University of Alberta MEC E 340: Applied Thermodynamics

Applying control-volume balances to turbines, compressors, and heaters

What you will learn

1. Define the control volume and the sign convention

2. Apply the mass and energy balances

3. Use properties, solve, and check

Adiabatic turbine work output

Compressor work input with heat loss

Heat supplied in a steady-flow heater

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