7.5 Balance mass and energy for mixing moist-air streams | University of Alberta MEC E 340: Applied Thermodynamics

University of Alberta MEC E 340: Applied Thermodynamics

Psychrometry and Moist Air

MEC E 340 study topic 7.5

Consider a mixing junction with two or more moist-air inlets and one outlet. The working mixture is dry air plus water vapour. Number the inlet states 1 and 2, and the outlet state 3. Unless a problem says otherwise, treat the junction as steady, with no shaft work, negligible changes in kinetic and potential energy, and no heat transfer to the surroundings. Also state whether liquid water forms or enters; the examples below assume it does not. Moist-air mixing is easiest when balances use dry-air mass flow as the basis. The first law then balances the enthalpy carried by the dry air and vapour, while a separate water balance tracks the vapour content.

What you will learn

Choose a control volume and state the assumptions for a moist-air mixing problem.
Apply dry-air, water-vapour, and energy balances to find mixed-stream properties.
Keep mass-flow units and enthalpy bases consistent, then check the result.

1. Choose the basis and define the states

At each inlet and outlet, specify the dry-air mass flow rate, humidity ratio, and moist-air enthalpy. The humidity ratio

\omega

is kilograms of water vapour per kilogram of dry air. The specific enthalpy

h

is commonly reported per kilogram of dry air for moist-air calculations. These two quantities therefore share the same dry-air basis, even though one describes water content and the other energy.

Use state 1 and state 2 for the incoming streams and state 3 for the mixed stream. A mixing junction is an open control volume: mass enters and leaves while the process operates. At steady state, material does not accumulate inside the control volume. Do not treat the junction as a closed system or as a cycle.

\omega = m_v/m_{da}

Write every moist-air enthalpy basis explicitly: kJ/kg dry air.
Use dry-air flow rates in kg dry air/s when multiplying by humidity ratio or enthalpy.

2. Apply the mass balances

Dry air is conserved through the mixer, so the outlet dry-air flow equals the sum of inlet dry-air flows. Water vapour is also conserved if no liquid water enters, leaves, or forms. Its mass flow in a stream is the dry-air flow multiplied by that stream's humidity ratio.

Combining these balances gives the outlet humidity ratio as a dry-air-flow-weighted average. It must lie between the inlet humidity ratios when both inlet flows are positive and there is no condensation or other water source. This provides a useful first check.

\dot m_{da,3}=\sum_i\dot m_{da,i},\qquad \omega_3=\frac{\sum_i\dot m_{da,i}\omega_i}{\dot m_{da,3}}

Use dry-air flow, not total moist-air flow, as the weighting factor.
If liquid water forms or enters, include its mass and energy in the balances; do not use the no-liquid formula unchanged.

3. Apply the steady-flow energy balance

For a steady mixer with no shaft work and negligible kinetic- and potential-energy changes, the energy carried in by the streams plus heat transferred into the control volume equals the energy carried out. If the mixer is adiabatic, the heat-transfer term is zero.

With enthalpy expressed per kilogram of dry air, multiply each stream's enthalpy by its dry-air mass flow rate. The outlet enthalpy is another dry-air-flow-weighted average for an adiabatic mixer. If heat transfer is specified, include it with a clear sign convention: heat into the control volume is positive. These balances find the outlet humidity ratio and enthalpy; a psychrometric chart or supplied property relation may then be needed to find temperature or relative humidity. Do not infer those properties from enthalpy alone without the required data.

\sum_i\dot m_{da,i}h_i+\dot Q=\dot m_{da,3}h_3

A kilowatt is a kilojoule per second, so heat-transfer rate and enthalpy-flow terms can be combined consistently.
State any property relation or chart data used to convert the mixed enthalpy and humidity ratio into other moist-air properties.

4. Solve systematically and check the model

First list the known inlet flow rates, humidity ratios, and enthalpies, with units and bases. Next calculate the outlet dry-air flow, then apply the water-vapour balance, and finally apply the energy balance. This order keeps the water and energy calculations separate and makes the basis visible.

Check that the outlet humidity ratio lies within the inlet range when no liquid water or other water source is involved. For an adiabatic mixer, check that the outlet enthalpy lies between the inlet enthalpies. For a non-adiabatic mixer, account for the heat transfer before making that comparison. A failed check often signals a reversed heat sign, an incorrect flow weighting, or a mismatch between total-air and dry-air bases.

\omega_{\min}\leq\omega_3\leq\omega_{\max}

A mixing junction with multiple inlets is best represented by its branches and balances; a single-inlet device sketch would be misleading.
Only calculate temperature or relative humidity when the needed moist-air property data or relation is supplied.

Worked example

Adiabatic mixing of two streams

Two moist-air streams mix steadily in an insulated junction. Stream 1 has

\dot m_{da,1}=2.0\ \mathrm{kg_{da}/s}

,

\omega_1=0.010\ \mathrm{kg_v/kg_{da}}

, and

h_1=45\ \mathrm{kJ/kg_{da}}

. Stream 2 has

\dot m_{da,2}=1.0\ \mathrm{kg_{da}/s}

,

\omega_2=0.020\ \mathrm{kg_v/kg_{da}}

, and

h_2=70\ \mathrm{kJ/kg_{da}}

. Find the outlet dry-air flow, humidity ratio, and enthalpy. Assume no liquid water enters or forms, no shaft work, and negligible kinetic- and potential-energy changes. The inlet property values are supplied data.

Set the control-volume assumptions
Use a steady control volume around the mixing junction. It has two inlets and one outlet. The junction is adiabatic, has no shaft work, and has negligible kinetic- and potential-energy changes.
Balance dry air
The outlet carries all of the incoming dry air.
$\dot m_{da,3}=2.0+1.0=3.0\ \mathrm{kg_{da}/s}$
Balance water vapour
Multiply each humidity ratio by its dry-air flow to obtain the water-vapour flow, then divide their sum by the outlet dry-air flow.
$\omega_3=\frac{2.0(0.010)+1.0(0.020)}{3.0}=0.0133\ \mathrm{kg_v/kg_{da}}$
Balance energy
With no heat transfer or work, the incoming enthalpy flow equals the outgoing enthalpy flow. Enthalpies are on a dry-air basis.
$h_3=\frac{2.0(45)+1.0(70)}{3.0}=53.3\ \mathrm{kJ/kg_{da}}$

Answer: The outlet dry-air flow is

3.0\ \mathrm{kg_{da}/s}

, its humidity ratio is

0.0133\ \mathrm{kg_v/kg_{da}}

, and its enthalpy is

53.3\ \mathrm{kJ/kg_{da}}

.

Check: The outlet humidity ratio is between 0.010 and 0.020, and the outlet enthalpy is between 45 and 70 kJ/kg dry air, as expected for adiabatic mixing without liquid water.

Worked example

Unequal flow rates and a dry-air basis

Two streams mix in a steady, insulated junction. Stream 1 has

\dot m_{da,1}=1.5\ \mathrm{kg_{da}/s}

,

\omega_1=0.008\ \mathrm{kg_v/kg_{da}}

, and

h_1=35\ \mathrm{kJ/kg_{da}}

. Stream 2 has

\dot m_{da,2}=2.5\ \mathrm{kg_{da}/s}

,

\omega_2=0.016\ \mathrm{kg_v/kg_{da}}

, and

h_2=58\ \mathrm{kJ/kg_{da}}

. Find the outlet humidity ratio and enthalpy. Assume no liquid water enters or forms, no shaft work, and negligible kinetic- and potential-energy changes. Treat the stated properties as supplied data.

Find the total dry-air flow
Add the inlet dry-air flows to obtain the outlet flow.
$\dot m_{da,3}=1.5+2.5=4.0\ \mathrm{kg_{da}/s}$
Find the mixed humidity ratio
Weight each inlet humidity ratio by its own dry-air flow. The larger-flow stream contributes more to the outlet value.
$\omega_3=\frac{1.5(0.008)+2.5(0.016)}{4.0}=0.0130\ \mathrm{kg_v/kg_{da}}$
Find the mixed enthalpy
Apply the adiabatic energy balance using enthalpy per kilogram of dry air.
$h_3=\frac{1.5(35)+2.5(58)}{4.0}=49.375\ \mathrm{kJ/kg_{da}}$

Answer: The outlet humidity ratio is

0.0130\ \mathrm{kg_v/kg_{da}}

and the outlet enthalpy is

49.375\ \mathrm{kJ/kg_{da}}

.

Check: The humidity ratio lies between 0.008 and 0.016, and the enthalpy lies between 35 and 58 kJ/kg dry air. The results reflect the greater dry-air flow from stream 2.

Worked example

Mixing with heat supplied

A steady mixer receives stream 1 at

\dot m_{da,1}=1.0\ \mathrm{kg_{da}/s}

,

\omega_1=0.006\ \mathrm{kg_v/kg_{da}}

, and

h_1=30\ \mathrm{kJ/kg_{da}}

, and stream 2 at

\dot m_{da,2}=1.5\ \mathrm{kg_{da}/s}

,

\omega_2=0.014\ \mathrm{kg_v/kg_{da}}

, and

h_2=50\ \mathrm{kJ/kg_{da}}

. Heat is supplied to the control volume at

\dot Q=0.50\ \mathrm{kW}

. Find the outlet humidity ratio and enthalpy. Assume no liquid water enters or forms, no shaft work, and negligible kinetic- and potential-energy changes. The stream properties and heat-transfer rate are supplied.

Balance dry air and water vapour
Heat transfer does not add dry air or water vapour. Apply the same mass balances as for an adiabatic mixer.
$\dot m_{da,3}=1.0+1.5=2.5\ \mathrm{kg_{da}/s},\qquad \omega_3=\frac{1.0(0.006)+1.5(0.014)}{2.5}=0.0108\ \mathrm{kg_v/kg_{da}}$
Include heat in the energy balance
Heat into the control volume is positive. Since one kilowatt equals one kilojoule per second, the supplied heat rate can be added to the inlet enthalpy-flow rates.
$h_3=\frac{1.0(30)+1.5(50)+0.50}{2.5}=42.2\ \mathrm{kJ/kg_{da}}$

Answer: The outlet humidity ratio is

0.0108\ \mathrm{kg_v/kg_{da}}

and the outlet enthalpy is

42.2\ \mathrm{kJ/kg_{da}}

.

Check: The humidity ratio is between the inlet values. The adiabatic weighted enthalpy would be 42.0 kJ/kg dry air; the supplied heat raises it by 0.20 kJ/kg dry air, consistent with heat entering.

Common mistakes and how to avoid them

Weighting inlet properties by total moist-air flow when the supplied values are on a dry-air basis.

Correction: Use dry-air mass flow consistently with humidity ratio and enthalpy per kilogram of dry air.

Leaving heat transfer out of the energy balance because the streams are mixing.

Correction: Include any stated heat transfer, with heat into the control volume positive under the convention used here.

Assuming a mixed-stream temperature or relative humidity can be found from the mass balances alone.

Correction: Use the mixed enthalpy and humidity ratio with supplied property data or a stated psychrometric relation to determine additional properties.

Using the no-condensation water balance when liquid water enters or forms.

Correction: State the liquid-water assumption first. If liquid water is part of the process, include its mass and energy in the appropriate balances.

Lesson summary

Model the mixer as a steady open control volume and number the inlet and outlet states.
Conserve dry air and water vapour separately; for no liquid-water transfer, humidity ratio is weighted by dry-air flow.
Apply the steady-flow energy balance on the same dry-air basis, including heat transfer when present.
Check units, signs, and whether the outlet values are physically consistent with the inlet streams and assumptions.

Check your understanding

Question 1

Two adiabatic inlet streams have equal dry-air flow rates and humidity ratios 0.008 and 0.014 kg water vapour/kg dry air. With no liquid water, what is the outlet humidity ratio?

0.008
0.011
0.014
0.022

Show answer and explanation

0.011

Equal dry-air flows give equal weighting, so the outlet value is the arithmetic mean: 0.011 kg water vapour/kg dry air.

Question 2

For an adiabatic mixer with no shaft work and negligible kinetic- and potential-energy changes, which balance determines outlet enthalpy?

The sum of inlet enthalpy flow rates equals the outlet enthalpy flow rate.
The outlet enthalpy equals the sum of inlet enthalpies, regardless of flow rates.
The outlet enthalpy must equal the enthalpy of the larger-flow inlet.
The outlet enthalpy is found from the water-vapour balance alone.

Show answer and explanation

The sum of inlet enthalpy flow rates equals the outlet enthalpy flow rate.

The steady-flow energy balance equates total incoming and outgoing enthalpy rates under these assumptions; each specific enthalpy is multiplied by its dry-air flow.

Question 3

A mixer receives 0.40 kW of heat. With heat into the control volume defined as positive, how does this term enter the energy balance?

As $\dot Q=-0.40\ \mathrm{kW}$
As $\dot Q=0$ because mixing is occurring
As $\dot Q=+0.40\ \mathrm{kW}$
As a term in the dry-air mass balance

Show answer and explanation

As

\dot Q=+0.40\ \mathrm{kW}

The stated sign convention makes heat entering the control volume positive, so the energy balance includes

+0.40\ \mathrm{kW}

.

Key terms

Control volume: A selected region in space through which mass and energy may flow.
Dry-air mass flow rate: The rate at which the dry-air part of a moist-air stream passes a location, reported here in kg dry air/s.
Humidity ratio: Water-vapour mass divided by dry-air mass in a moist-air sample.
Moist-air enthalpy: An energy property of the moist-air mixture; in these examples it is reported per kilogram of dry air.

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7.5 · Balance mass and energy for mixing moist-air streams

University of Alberta MEC E 340: Applied Thermodynamics

MEC E 340 study topic 7.5

What you will learn

1. Choose the basis and define the states

2. Apply the mass balances

3. Apply the steady-flow energy balance

4. Solve systematically and check the model

Adiabatic mixing of two streams

Unequal flow rates and a dry-air basis

Mixing with heat supplied

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through MEC E 340

About this lesson and its review