7.4 Analyze sensible heating, cooling, and dehumidification | University of Alberta MEC E 340: Applied Thermodynamics

University of Alberta MEC E 340: Applied Thermodynamics

Psychrometry and Moist Air

MEC E 340 Applied Thermodynamics — Study topic 7.4

Consider moist air flowing through a heating or cooling device: state 1 is the inlet and state 2 is the outlet. Treat the device as a steady-flow control volume, with dry air and water vapour as the moist-air mixture; liquid water may also leave during dehumidification. Heat transfer into the control volume is positive. For ordinary air-conditioning calculations, balances are often written per kilogram of dry air, because the dry-air flow rate is unchanged through a device with no leakage. Sensible heating or cooling changes temperature without changing the air’s water-vapour content. Dehumidification removes some water vapour, so both the moisture balance and the energy carried by the condensate may matter. The examples state all property values they use; no chart or table values are assumed.

What you will learn

Distinguish sensible heating or cooling from dehumidification using dry-bulb temperature and humidity ratio.
Apply steady-flow mass and energy balances to moist-air processes on a dry-air basis.
Calculate heat transfer and condensate flow using only supplied properties and assumptions.
Check signs, units, and physical meaning of calculated results.

1. Set up the moist-air states

At each numbered state, identify the dry-bulb temperature,

T

, and humidity ratio,

w

. The humidity ratio is the mass of water vapour divided by the mass of dry air, commonly reported in kilograms of water vapour per kilogram of dry air. The dry-air mass flow rate is written as

\dot m_{da}

. When helpful, moist-air enthalpy

h

is given per kilogram of dry air.

A psychrometric chart or supplied property data can relate moist-air properties. Do not infer a dew point, relative humidity, saturation boundary, or enthalpy unless the problem supplies enough information or explicitly states an assumption. In this lesson, numerical properties are supplied directly.

w=\frac{m_v}{m_{da}}

Use state numbers consistently: 1 for inlet, 2 for outlet.
Keep the dry-air basis clear when using humidity ratios and enthalpies.
A change in temperature alone does not establish whether moisture condenses.

2. Apply mass and energy balances

For steady operation without dry-air leakage, dry-air mass flow is the same at inlet and outlet. If no water is added or removed, the humidity ratio is also unchanged: this is the defining moisture balance for sensible heating or cooling. A sensible heater raises

T

at constant

w

; a sensible cooler lowers

T

at constant

w

, provided no condensation occurs.

For dehumidification, let liquid condensate leave at rate

\dot m_c

. The water balance is the inlet vapour flow minus the outlet vapour flow. On a dry-air basis, that difference is

\dot m_{da}(w_1-w_2)

. The outlet humidity ratio must therefore be below the inlet value for positive condensate production.

For a steady device with negligible kinetic and potential energy changes and no shaft work, the first law says heat transfer equals the energy out minus the energy in, including the liquid condensate. A negative heat-transfer result under the chosen sign convention means heat is removed from the control volume. If condensate energy is negligible by an explicitly stated approximation, it may be omitted; otherwise include its supplied enthalpy.

\dot Q=\dot m_{da}(h_2-h_1)+\dot m_c h_l

Sensible process: $w_1=w_2$ and no condensate leaves.
Dehumidification: $w_2<w_1$ and liquid water leaves the control volume.
State the heat sign convention before interpreting the result.

3. Interpret the processes and check results

A temperature–humidity-ratio (

T

–

w

) sketch helps distinguish processes. Sensible heating or cooling is drawn at constant humidity ratio; dehumidification moves to a lower humidity ratio and may also lower temperature. These sketches are qualitative, not property charts: their coordinates do not provide numerical properties or show a saturation boundary.

After calculating, check that units agree. A mass flow rate in kilograms per second multiplied by an enthalpy difference in kilojoules per kilogram gives kilowatts. A positive sensible-heating load under the stated convention means heat enters the air. A cooling load should be reported clearly as heat removed, or as negative heat transfer into the control volume. For dehumidification, verify that condensate flow is nonnegative and that the outlet humidity ratio is lower.

\dot m_c=\dot m_{da}(w_1-w_2)

A $T$ – $w$ diagram is schematic unless property coordinates are supplied.
Report whether a heat-load number means heat added or heat removed.
Check both the moisture balance and the energy balance.

Worked example

Sensible heating at constant humidity ratio

Moist air flows steadily through a heater at

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

. Its temperature rises from

18^\circ\mathrm{C}

at state 1 to

30^\circ\mathrm{C}

at state 2. The supplied model gives a constant specific heat of

1.02\ \mathrm{kJ/(kg_{da}\cdot K)}

. Assume no moisture is added or removed, no shaft work, and negligible kinetic and potential energy changes. Find the heat-transfer rate into the air.

Sensible heating on a schematic T–w plane

Moist air is the flowing working mixture; state 1 to state 2 is sensible heating at constant humidity ratio. Schematic, not to scale.

Define the process
The control volume contains the heater and its flowing moist air. The stated no-moisture-transfer assumption makes this a sensible-heating process, so the humidity ratio is unchanged. Use heat into the control volume as positive.
Use the sensible energy model
The supplied constant-specific-heat model relates the heat added per unit dry-air mass to the temperature rise. Multiply by the dry-air mass flow rate to obtain power.
$\dot Q=\dot m_{da}c_p(T_2-T_1)$
Substitute with units
The temperature difference is $12\ \mathrm{K}$ . The kilojoules per second convert directly to kilowatts.
$\dot Q=(0.80)(1.02)(30-18)=9.79\ \mathrm{kW}$

Answer: The heater supplies approximately

9.79\ \mathrm{kW}

to the moist air.

Check: The outlet is warmer, so positive heat into the air is physically consistent. The humidity ratio remains constant by assumption.

Worked example

Sensible cooling without condensation

A steady cooler handles moist air at

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

. The air temperature falls from

27^\circ\mathrm{C}

to

16^\circ\mathrm{C}

. The supplied model gives

c_p=1.02\ \mathrm{kJ/(kg_{da}\cdot K)}

. Assume no condensation, no shaft work, and negligible kinetic and potential energy changes. Find the heat removed from the air.

Sensible cooling on a schematic T–w plane

Moist air is the flowing working mixture; state 1 to state 2 is sensible cooling at constant humidity ratio. Schematic, not to scale.

Identify the moisture assumption
The problem explicitly rules out condensation. The humidity ratio is therefore constant, and the process is sensible cooling rather than dehumidification.
$w_2=w_1$
Calculate heat into the control volume
Use the same supplied sensible model as for heating. Because the outlet temperature is lower, the result for heat into the control volume will be negative.
$\dot Q=(1.20)(1.02)(16-27)=-13.46\ \mathrm{kW}$
State the cooling load
The negative sign means energy leaves the air. The magnitude of the heat removed is $13.46\ \mathrm{kW}$ .

Answer: The cooler removes approximately

13.46\ \mathrm{kW}

from the air.

Check: The temperature drop produces negative heat transfer into the control volume, consistent with cooling. Constant humidity ratio follows from the stated no-condensation assumption.

Worked example

Dehumidification with condensate energy included

A steady dehumidifier receives moist air at

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

. The inlet and outlet humidity ratios are

0.014

and

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

. Supplied moist-air enthalpies are

h_1=62

and

h_2=38\ \mathrm{kJ/kg_{da}}

; the leaving liquid water enthalpy is

h_l=42\ \mathrm{kJ/kg}

. Assume no shaft work and negligible kinetic and potential energy changes. Find the condensate flow and heat removed.

Dehumidification on a schematic T–w plane

Moist air flows from state 1 to state 2 while liquid condensate leaves. The state coordinates are qualitative; the plot is schematic, not to scale.

Apply the water balance
The dry-air flow is unchanged. The decrease in vapour carried per kilogram of dry air leaves as liquid condensate.
$\dot m_c=(1.00)(0.014-0.009)=0.0050\ \mathrm{kg/s}$
Apply the energy balance
Heat into the control volume equals outlet energy minus inlet energy. Include both outlet moist air and the liquid condensate, using the supplied enthalpy values on their stated mass bases.
$\dot Q=(1.00)(38)+(0.0050)(42)-(1.00)(62)=-23.79\ \mathrm{kW}$
Interpret the sign
The negative result means heat leaves the control volume. The magnitude of heat removed is $23.79\ \mathrm{kW}$ .

Answer: The condensate flow rate is

0.0050\ \mathrm{kg/s}

, and the device removes

23.79\ \mathrm{kW}

of heat.

Check: The outlet humidity ratio is lower, giving a positive condensate rate. The outlet air and condensate together carry less energy than the inlet, so heat must be removed under the stated assumptions.

Common mistakes and how to avoid them

Assuming every cooling process removes moisture.

Correction: Cooling is sensible only when moisture content stays constant. Condensation requires evidence or an explicit assumption that water vapour is removed.

Using total moist-air mass flow with properties stated per kilogram of dry air.

Correction: Match each flow basis to the property basis. Here, humidity ratio and moist-air enthalpy are per unit dry-air mass.

Ignoring condensate in a dehumidifier energy balance without stating an approximation.

Correction: Include the liquid-water enthalpy when it is supplied and the problem does not authorize neglecting it.

Calling a negative heat-transfer result a negative cooling load.

Correction: With heat into the control volume positive, negative heat transfer means heat removal. Report the cooling-load magnitude and direction clearly.

Lesson summary

Sensible heating or cooling changes dry-bulb temperature while humidity ratio remains constant.
For dehumidification, condensate flow follows from the reduction in humidity ratio on a dry-air basis.
Use a steady-flow energy balance with consistent enthalpy bases and include condensate energy when specified.
Check mass balance, heat-transfer sign, units, and the physical direction of each process.

Check your understanding

Question 1

A cooler operates with no condensation or moisture addition. Which quantity remains constant between inlet and outlet?

Dry-bulb temperature
Humidity ratio
Moist-air enthalpy
Heat transfer

Show answer and explanation

Humidity ratio

With no water added or removed, the water-vapour mass per unit dry-air mass is unchanged. Temperature and enthalpy can fall during cooling.

Question 2

Air flows at

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

and its humidity ratio falls from

0.012

to

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

. What is the condensate flow rate?

$0.002\ \mathrm{kg/s}$
$0.004\ \mathrm{kg/s}$
$0.006\ \mathrm{kg/s}$
$0.010\ \mathrm{kg/s}$

Show answer and explanation

0.002\ \mathrm{kg/s}

The humidity-ratio decrease is

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

. Multiplying by

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

gives

0.002\ \mathrm{kg/s}

.

Question 3

Using heat into a control volume as positive, what does a negative steady heat-transfer rate indicate for a cooling device?

Heat enters the device
The dry-air mass balance is violated
Heat leaves the device
The outlet humidity ratio must be zero

Show answer and explanation

Heat leaves the device

A negative value under this convention means energy is transferred out as heat. It does not by itself determine the outlet humidity ratio.

Key terms

Dry-bulb temperature: The ordinary air temperature used to describe the thermal state of moist air.
Humidity ratio: Water-vapour mass divided by dry-air mass in a moist-air mixture.
Sensible heating or cooling: A temperature change in which the humidity ratio remains constant and no moisture is added or removed.
Dehumidification: Removal of water vapour from moist air, commonly represented here by liquid condensate leaving the device.
Dry-air basis: A calculation basis that expresses mixture properties or water content per unit mass of dry air.

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7.4 · Analyze sensible heating, cooling, and dehumidification

University of Alberta MEC E 340: Applied Thermodynamics

MEC E 340 Applied Thermodynamics — Study topic 7.4

What you will learn

1. Set up the moist-air states

2. Apply mass and energy balances

3. Interpret the processes and check results

Sensible heating at constant humidity ratio

Sensible cooling without condensation

Dehumidification with condensate energy included

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