4.2 Balance a heat-recovery steam generator | University of Alberta MEC E 340: Applied Thermodynamics

University of Alberta MEC E 340: Applied Thermodynamics

Combined Power and Cogeneration

Energy accounting for hot exhaust gas and a water–steam stream

An HRSG uses energy from a hot gas stream to heat water and produce steam. For the balance, treat the gas side and the water–steam side together as one steady-flow control volume. Number the gas states 1 (inlet) and 2 (outlet), and the water–steam states 3 (inlet) and 4 (outlet). Unless a problem says otherwise, take the HRSG as steady, with no shaft work and negligible changes in kinetic and potential energy. The gas and water–steam streams do not mix. Heat crossing the outside boundary is positive into the control volume. The main task is to account for how much energy the gas loses, how much the water–steam stream gains, and whether any energy leaves through the HRSG boundary.

What you will learn

Identify the control volume and the two fluid streams in a heat-recovery steam generator (HRSG).
Apply steady-flow mass and energy balances using consistent state numbers and sign conventions.
Calculate a steam flow rate, a gas outlet temperature, or an external heat loss from supplied data.
Check that the calculated heat transfers and state changes are physically consistent.

1. Set up the two-stream control volume

The HRSG control volume contains the heat-transfer equipment and the fluid passages on both sides. The hot gas enters at state 1 and leaves at state 2. Feedwater or a water–steam mixture enters at state 3 and leaves at state 4. These labels describe stream locations; they do not imply particular temperatures, pressures, or phases. Use the states and property values given in the problem.

For a steady HRSG with no leakage, each stream has the same mass flow rate at its inlet and outlet. If there is no mixing between streams, the gas mass flow rate need not equal the water–steam mass flow rate. They are separate streams, so write a mass balance for each one.

For the energy balance, use enthalpy at each inlet and outlet. Enthalpy accounts for the energy carried by a flowing stream, including its flow work. Define the heat loss to the surroundings as a positive quantity when it leaves the HRSG. With this convention, the gas-side energy decrease supplies the water–steam energy increase and any external heat loss.

\dot m_{g,1}=\dot m_{g,2},\quad \dot m_{w,3}=\dot m_{w,4}

Gas: state 1 in, state 2 out. Water–steam: state 3 in, state 4 out.
At steady state, each stream’s inlet mass flow equals its outlet mass flow.
Do not equate the two streams’ mass flow rates unless the problem gives a reason.

2. Apply the steady-flow energy balance

The general energy balance includes heat transfer, work, enthalpy flow, and changes in kinetic and potential energy. For the usual HRSG model, there is no shaft work and the kinetic- and potential-energy changes are negligible. Heat transfer across the outer boundary may also be negligible if the problem states that the HRSG is insulated.

Writing the balance with heat into the control volume positive gives a useful form: the gas enthalpy decrease equals the water–steam enthalpy increase plus heat lost to the surroundings. If external heat loss is neglected, the gas enthalpy decrease equals the water–steam enthalpy increase. This is energy conservation, not an assumption that the two mass flow rates are equal.

For a gas treated with constant specific heat over the stated temperature range, its enthalpy change can be represented by the supplied specific heat times the temperature change. Use this only when the problem permits that approximation. For water or steam, take the inlet and outlet enthalpies from the data supplied in the problem or from the specified property source. Do not infer a phase or invent a table value.

\dot m_g(h_1-h_2)=\dot m_w(h_4-h_3)+\dot Q_{\mathrm{loss}}

State the sign convention before substituting values.
A heat loss reduces the energy available to raise the water–steam enthalpy.
Use property data and idealizations explicitly supplied or named.

3. Choose properties, solve, and check

First list the known mass flow rates, temperatures, and enthalpies, with units. If a problem gives water–steam enthalpies directly, use them. If it instead specifies states for property-table lookup, use the named table and confirm that the phase assumptions match the stated conditions. Enthalpy differences are often enough; an arbitrary reference value cancels when the same reference is used consistently.

Rearrange the balance for the requested unknown only after the sign convention and assumptions are clear. For example, with negligible heat loss and a constant gas specific heat, a known gas inlet temperature and water–steam enthalpy rise can determine the gas outlet temperature. A known gas-side energy decrease and water–steam enthalpy rise can determine the steam production rate.

Finally, check units and direction. A gas that supplies heat should have a lower outlet enthalpy than inlet enthalpy. The water–steam stream should gain enthalpy when it is heated or converted to a higher-enthalpy outlet state. A calculated heat loss should be nonnegative under the stated loss convention. If the balance implies the opposite without an input of heat, recheck the state labels, units, and signs.

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

Keep mass flow in kg/s, enthalpy in kJ/kg, and heat-transfer rate in kW.
Use the gas model only over the range and conditions for which it is given.
A balance checks conservation; it does not by itself establish detailed exchanger performance.

Worked example

Example 1: Find the gas outlet temperature

An insulated HRSG operates steadily. Gas enters at state 1 at 500 °C with a mass flow rate of 8.0 kg/s. The supplied gas model is constant specific heat,

c_p=1.10\ \mathrm{kJ/(kg\cdot K)}

. Water–steam enters at state 3 at 300 kg/s, and supplied property data give

h_3=500\ \mathrm{kJ/kg}

and

h_4=2500\ \mathrm{kJ/kg}

. Find the gas outlet temperature at state 2. Neglect kinetic and potential energy changes and shaft work.

HRSG control volume

Schematic two-stream device representation, not to scale. The gas and water–steam streams pass through separate passages; no external heat or work is assumed.

Define the balance
The control volume includes both HRSG streams. It is steady and insulated, so external heat transfer is zero. The gas enthalpy decrease equals the water–steam enthalpy increase.
$\dot m_g c_p(T_1-T_2)=\dot m_w(h_4-h_3)$
Substitute the supplied data
The water–steam enthalpy rise is 2000 kJ/kg. Dividing its energy gain rate by the gas heat-capacity rate gives the gas temperature drop.
$T_1-T_2=\frac{300(2500-500)}{8.0(1.10)}=6818\ \mathrm{K}$

Answer: The arithmetic gives

T_2=500-6818=-6318\ ^\circ\mathrm{C}

, which is physically implausible for the stated gas stream and HRSG. The supplied data are mutually inconsistent with the assumed steady, insulated balance and constant-

c_p

model. This is the result of the balance check, not a credible predicted outlet temperature.

Check: The water–steam stream would gain

300(2000)=600{,}000\ \mathrm{kW}

. The gas heat-capacity rate is only

8.0(1.10)=8.8\ \mathrm{kW/K}

, so the required temperature drop is far greater than the 500 °C inlet temperature allows in this model. A physical calculation needs consistent flow rates, enthalpies, and gas conditions.

Worked example

Example 2: Find water–steam mass flow rate

A steady, insulated HRSG receives gas at 12.0 kg/s. The supplied gas model has

c_p=1.05\ \mathrm{kJ/(kg\cdot K)}

, and the gas cools from 450 °C at state 1 to 200 °C at state 2. Water enters at state 3 with

h_3=420\ \mathrm{kJ/kg}

and leaves at state 4 with

h_4=2820\ \mathrm{kJ/kg}

, using property data supplied with the problem. Find the water–steam mass flow rate. Neglect shaft work and kinetic and potential energy changes.

HRSG control volume

Schematic two-stream device representation, not to scale. Separate gas and water–steam streams exchange energy internally; external heat transfer and shaft work are neglected.

Write the insulated balance
With no external heat transfer, gas energy loss equals water–steam energy gain. The enthalpy change for the gas follows the stated constant-specific-heat model.
$\dot m_g c_p(T_1-T_2)=\dot m_w(h_4-h_3)$
Calculate the gas energy decrease
The temperature difference is 250 K; a temperature difference in kelvins has the same numerical size as in degrees Celsius.
$\dot Q_{g,\mathrm{decrease}}=12.0(1.05)(450-200)=3150\ \mathrm{kW}$
Solve for the water–steam flow
Each kilogram of water–steam gains 2400 kJ, so divide the gas energy-decrease rate by that enthalpy rise.
$\dot m_w=\frac{3150}{2820-420}=1.3125\ \mathrm{kg/s}$

Answer: The water–steam mass flow rate is approximately

1.31\ \mathrm{kg/s}

.

Check: The water–steam gain rate is

1.3125(2400)=3150\ \mathrm{kW}

, equal to the gas energy decrease. The gas cools and the water–steam enthalpy rises, consistent with heat recovery.

Worked example

Example 3: Determine external heat loss

A steady HRSG has a gas flow of 10.0 kg/s. The gas is modelled with the supplied constant

c_p=1.00\ \mathrm{kJ/(kg\cdot K)}

and cools from 400 °C at state 1 to 180 °C at state 2. Water–steam flows at 0.80 kg/s, with supplied enthalpies

h_3=600\ \mathrm{kJ/kg}

and

h_4=3000\ \mathrm{kJ/kg}

. Neglect shaft work and kinetic and potential energy changes. Find the heat-loss rate to the surroundings.

HRSG with heat loss

Schematic two-stream device representation, not to scale. Heat crosses outward to the surroundings; there is no shaft work.

Use the loss-positive balance
The gas enthalpy decrease is divided between the water–steam enthalpy increase and heat lost through the outer boundary.
$\dot Q_{\mathrm{loss}}=\dot m_g c_p(T_1-T_2)-\dot m_w(h_4-h_3)$
Evaluate both energy rates
The gas supplies 2200 kW. The water–steam stream gains 1920 kW. Their difference is the rate that leaves the HRSG boundary.
$\dot Q_{\mathrm{loss}}=10.0(1.00)(400-180)-0.80(3000-600)=280\ \mathrm{kW}$

Answer: The heat-loss rate is

280\ \mathrm{kW}

outward from the HRSG.

Check: The balance closes:

2200\ \mathrm{kW}=1920\ \mathrm{kW}+280\ \mathrm{kW}

. The positive loss agrees with the stated convention and is smaller than the gas energy decrease.

Common mistakes and how to avoid them

Setting gas mass flow equal to water–steam mass flow.

Correction: Write a separate steady mass balance for each non-mixing stream. Their flow rates can differ.

Treating heat lost to the surroundings as heat entering the HRSG.

Correction: Choose a convention first. Here, heat loss is positive outward and is added to the water–steam gain when matching the gas energy decrease.

Using a water or steam enthalpy without stated property data or a named property source.

Correction: Use only the supplied or specified property information, and check that the state description supports the property lookup.

Accepting any numerical result just because the algebra balances.

Correction: Check units and physical direction. An impossible outlet temperature or negative loss under a loss-positive convention signals inconsistent inputs, signs, or assumptions.

Lesson summary

Model the HRSG as one steady control volume with separate gas and water–steam streams.
Balance mass separately for each stream; do not assume their mass flow rates are equal.
For the usual model, gas energy decrease equals water–steam energy increase plus outward heat loss.
Use only stated gas models and supplied or named water–steam property data.
Verify the arithmetic, signs, units, and plausibility of the result.

Check your understanding

Question 1

A steady insulated HRSG has no shaft work and negligible kinetic and potential energy changes. Which statement matches its energy balance?

The gas enthalpy decrease equals the water–steam enthalpy increase.
The gas and water–steam mass flow rates must be equal.
The water–steam enthalpy increase equals zero.
The gas enthalpy decrease must be counted as heat entering from the surroundings.

Show answer and explanation

The gas enthalpy decrease equals the water–steam enthalpy increase.

With no external heat transfer, the gas-side energy decrease supplies the water–steam enthalpy increase. The two streams are separate, so their mass flow rates need not be equal.

Question 2

Gas supplies 900 kW to an HRSG and the water–steam stream gains 760 kW. With heat loss defined positive outward, what is the heat-loss rate?

140 kW outward
1660 kW outward
140 kW inward
760 kW outward

Show answer and explanation

140 kW outward

The balance gives loss as gas energy decrease minus water–steam energy gain: 900 kW − 760 kW = 140 kW outward.

Key terms

Heat-recovery steam generator (HRSG): A device that transfers energy from a hot gas stream to a water–steam stream to heat water or produce steam.
Control volume: A chosen region in space through whose boundary mass and energy may flow.
Enthalpy: A property used to account for energy carried by a flowing stream in a steady-flow balance.
Steady state: A condition in which properties and flow rates at fixed locations do not change with time.

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4.2 · Balance a heat-recovery steam generator

University of Alberta MEC E 340: Applied Thermodynamics

Energy accounting for hot exhaust gas and a water–steam stream

What you will learn

1. Set up the two-stream control volume

2. Apply the steady-flow energy balance

3. Choose properties, solve, and check

Example 1: Find the gas outlet temperature

Example 2: Find water–steam mass flow rate

Example 3: Determine external heat loss

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