3.3 Calculate boiler heat input and condenser heat rejection | 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 3.3

A boiler and a condenser transfer heat across the boundary of a flowing-fluid control volume. For either component, identify the working fluid, inlet and outlet states, and whether heat enters or leaves. In the state numbering used here, state 2 is the boiler inlet and state 3 its outlet; state 4 is the condenser inlet and state 1 its outlet. These state numbers are a common arrangement in a simple vapour power cycle, but the calculation method applies whenever inlet and outlet data are known. We assume steady operation, negligible kinetic- and potential-energy changes, and no shaft work in either component. Enthalpies in the examples are supplied directly; no table values are assumed.

What you will learn

Set up a steady-flow energy balance for a boiler or condenser.
Calculate heat-transfer rate from mass flow rate and inlet and outlet enthalpies.
Use a consistent sign convention and distinguish heat added from heat rejected.
Check component heat-transfer results using units and physical direction.

1. Define the component and sign convention

Draw a control-volume boundary around the boiler alone or the condenser alone. Fluid crosses that boundary at an inlet and an outlet, and heat crosses the component wall. The boiler adds heat to the working fluid. The condenser removes heat from it.

Use heat transfer into the control volume as positive. With this convention, boiler heat transfer is positive, while condenser heat transfer is negative. It is often clearer to report condenser heat rejection as a positive magnitude, labelled

\dot Q_{out}

, rather than as negative heat into the control volume.

For steady flow, mass does not build up inside the component. With one inlet and one outlet, the mass flow rates are equal: \dot m_{in}=\dot m_{out}=\dot m. Enthalpy,

h

, accounts for energy carried with each unit mass of flowing fluid.

State the component boundary and identify its inlet and outlet before using an equation.
Use one sign convention consistently: heat into the control volume is positive.
A heat-rejection magnitude is positive even though heat transfer into the condenser control volume is negative.

2. Apply the steady-flow energy balance

The general energy balance includes heat transfer, work, energy carried by mass, and changes in kinetic and potential energy. For a boiler or condenser with one inlet and one outlet, no shaft work, and negligible kinetic- and potential-energy changes, it reduces to heat transfer equalling mass flow rate multiplied by the outlet-to-inlet enthalpy change.

For the boiler, heat added raises the fluid enthalpy, so heat transfer into the control volume is positive. For the condenser, the fluid usually leaves with lower enthalpy, so signed heat transfer into the control volume is negative. The positive rejection rate is the inlet enthalpy minus the outlet enthalpy, multiplied by mass flow rate.

If the question asks for energy rather than a rate, multiply the enthalpy difference by the mass that passes through during the stated interval. Keep the time basis consistent: mass flow rate in kilograms per second and enthalpy in kilojoules per kilogram give a heat-transfer rate in kilowatts.

\dot Q=\dot m(h_{out}-h_{in})

Use enthalpies at the component inlet and outlet, not at unrelated cycle states.
Use supplied property data or values obtained from the property source specified in the problem.
Check whether the requested result is a rate or a total amount of heat.

3. Interpret and check the answer

A boiler heat-input rate is the rate at which energy enters the working fluid. A condenser heat-rejection rate is the positive magnitude of energy leaving the working fluid. These are component heat transfers; do not confuse either one with net cycle work or cycle efficiency.

Check the enthalpy change against the expected physical direction. For positive boiler heat input under the stated assumptions, outlet enthalpy exceeds inlet enthalpy. For positive condenser heat rejection, inlet enthalpy exceeds outlet enthalpy.

Check units explicitly:

\mathrm{kg/s}

multiplied by

\mathrm{kJ/kg}

gives

\mathrm{kJ/s}

, which is

\mathrm{kW}

. If the sign or units do not fit the physical process, review the state assignment and subtraction order.

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

A negative signed condenser heat transfer means heat leaves the control volume under the chosen convention.
Report the requested heat-rejection magnitude as a positive value.
No property-table reading is needed when all required enthalpies are supplied.

Worked example

Boiler heat-input rate

A steady-flow boiler receives water at state 2 and discharges steam at state 3. The supplied data are

\dot m=2.40\ \mathrm{kg/s}

,

h_2=210\ \mathrm{kJ/kg}

, and

h_3=3220\ \mathrm{kJ/kg}

. Assume no shaft work and negligible kinetic- and potential-energy changes. Find the heat-input rate.

Boiler control volume

Steady-flow boiler with one inlet and one outlet; schematic device, not to scale. Heat enters the working fluid, and no shaft work is assumed.

Set the boundary and states
Take the boiler as the control volume. State 2 is the inlet and state 3 is the outlet; the given enthalpies are the only property data needed.
Use the steady-flow balance
With no shaft work and negligible kinetic- and potential-energy changes, heat input equals mass flow rate times the increase in enthalpy. \dot Q_{in}=\dot m(h_3-h_2)
Calculate and check
The outlet enthalpy is greater than the inlet enthalpy, so positive heat input is physically consistent. \dot Q_{in}=(2.40\ kg/s)(3220-210\ kJ/kg)=7224\ kW

Answer: The boiler heat-input rate is

7224\ \mathrm{kW}

, or

7.224\ \mathrm{MW}

.

Check: The enthalpy rise is

3010\ \mathrm{kJ/kg}

. Multiplying by the positive mass flow rate gives positive heat input, as expected.

Worked example

Condenser heat-rejection rate

A steady-flow condenser receives vapour at state 4 and discharges liquid at state 1. Supplied data are

\dot m=1.80\ \mathrm{kg/s}

,

h_4=2400\ \mathrm{kJ/kg}

, and

h_1=190\ \mathrm{kJ/kg}

. Assume no shaft work and negligible kinetic- and potential-energy changes. Find the positive heat-rejection rate.

Condenser control volume

Steady-flow condenser with one inlet and one outlet; schematic device, not to scale. Heat leaves the working fluid, and no shaft work is assumed.

Identify the heat direction
Take the condenser as the control volume. State 4 is the inlet and state 1 is the outlet. Since heat leaves the fluid, calculate the positive rejection magnitude using inlet enthalpy minus outlet enthalpy.
Apply the simplified balance
The signed heat-transfer rate into the control volume is negative; its magnitude is the requested heat-rejection rate.
$\dot Q_{out}=\dot m(h_4-h_1)$
Calculate and check
The inlet enthalpy exceeds the outlet enthalpy, consistent with heat leaving the working fluid.
$\dot Q_{out}=(1.80\ \mathrm{kg/s})(2400-190\ \mathrm{kJ/kg})=3978\ \mathrm{kW}$

Answer: The condenser heat-rejection rate is

3978\ \mathrm{kW}

, or

3.978\ \mathrm{MW}

. With heat into the control volume defined as positive, the signed heat-transfer rate is

\dot Q=-3978\ \mathrm{kW}

.

Check: The positive rejection magnitude agrees with the enthalpy decrease. The signed heat transfer is negative because energy leaves the condenser control volume.

Worked example

Calculate both component duties

For a steady-flow vapour cycle, a boiler takes the working fluid from state 2 to state 3, and a condenser takes it from state 4 to state 1. Supplied enthalpies are

h_2=250\ \mathrm{kJ/kg}

,

h_3=3300\ \mathrm{kJ/kg}

,

h_4=2500\ \mathrm{kJ/kg}

, and

h_1=200\ \mathrm{kJ/kg}

. The mass flow rate is

5.00\ \mathrm{kg/s}

. Find the boiler heat-input rate and condenser heat-rejection rate. Assume steady operation, no shaft work in either component, and negligible kinetic- and potential-energy changes.

Keep the component state pairs separate
Use states 2 and 3 for the boiler, and states 4 and 1 for the condenser. Each control volume has the same mass flow rate but a different enthalpy change.
Calculate the boiler duty
The boiler adds energy as the fluid moves from state 2 to state 3. \dot Q_{in}=5.00 (3300-250)=15250\ kW
Calculate the condenser duty
The condenser rejects energy as the fluid moves from state 4 to state 1. Report the positive rejection magnitude.
$\dot Q_{out}=5.00\,(2500-200)=11500\ \mathrm{kW}$

Answer: The boiler heat-input rate is

15.25\ \mathrm{MW}

, and the condenser heat-rejection rate is

11.50\ \mathrm{MW}

.

Check: Both rates are positive when reported as input and rejection magnitudes. The boiler enthalpy rise and condenser enthalpy drop have the expected directions. The difference between these component duties is not, by itself, a full cycle analysis.

Common mistakes and how to avoid them

Using outlet minus inlet enthalpy for a positive condenser heat-rejection magnitude.

Correction: That subtraction gives signed heat transfer into the condenser control volume. For positive rejection magnitude, use inlet enthalpy minus outlet enthalpy.

Using enthalpies from the wrong component states.

Correction: Match the enthalpy pair to the control volume: boiler inlet and outlet, or condenser inlet and outlet.

Reporting a heat-transfer rate in kilojoules rather than kilowatts.

Correction: With mass flow rate in kilograms per second and enthalpy in kilojoules per kilogram, the answer is kilojoules per second, equivalent to kilowatts.

Treating a supplied enthalpy difference as a heat rate without multiplying by mass flow rate.

Correction: An enthalpy difference is energy per unit mass. Multiply it by mass flow rate to obtain a rate.

Lesson summary

For the simplified steady-flow boiler or condenser balance, signed heat-transfer rate equals mass flow rate times outlet-minus-inlet enthalpy.
Boiler heat input is positive when outlet enthalpy exceeds inlet enthalpy.
Condenser heat rejection is reported as the positive magnitude of inlet-minus-outlet enthalpy times mass flow rate.
Use the correct state pair, consistent units, and the stated assumptions.

Check your understanding

Question 1

A condenser has

\dot m=2.00\ \mathrm{kg/s}

, inlet enthalpy

1800\ \mathrm{kJ/kg}

, and outlet enthalpy

300\ \mathrm{kJ/kg}

. What is its positive heat-rejection rate?

$3000\ \mathrm{kW}$
$-3000\ \mathrm{kW}$
$750\ \mathrm{kW}$
$4200\ \mathrm{kW}$

Show answer and explanation

3000\ \mathrm{kW}

The rejection magnitude is mass flow rate times inlet-minus-outlet enthalpy:

2.00(1800-300)=3000\ \mathrm{kW}

. Signed heat transfer into the condenser control volume would be negative.

Question 2

For a boiler, which enthalpy difference gives signed heat-transfer rate into the control volume under the convention used here?

\dot m(h_{out}-h_{in})
\dot m(h_{in}-h_{out})
h_{out}-h_{in}
\dot m(h_{out}+h_{in})

Show answer and explanation

\dot m(h_{out}-h_{in})

The simplified steady-flow balance gives heat into the control volume as mass flow rate multiplied by outlet-minus-inlet enthalpy.

Key terms

Control volume: A chosen region in space, such as the inside of a boiler or condenser, across whose boundary mass and energy may pass.
Enthalpy: A working-fluid property commonly used to account for energy carried by flowing fluid; here it is given per unit mass.
Heat-input rate: Energy transferred into a control volume per unit time.
Heat-rejection rate: The positive magnitude of energy transferred out of a control volume per unit time.
Steady flow: Operation in which conditions within the control volume do not change with time, so mass does not accumulate inside it.

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3.3 · Calculate boiler heat input and condenser heat rejection

University of Alberta MEC E 340: Applied Thermodynamics

MEC E 340 Applied Thermodynamics — study topic 3.3

What you will learn

1. Define the component and sign convention

2. Apply the steady-flow energy balance

3. Interpret and check the answer

Boiler heat-input rate

Condenser heat-rejection rate

Calculate both component duties

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