METBD 330: Thermodynamics

METBD 330: Thermodynamics

Gas-Vapor Mixtures and Air Conditioning

Air contains water vapor.

This is important for human comfort and for A/C system design.

Consider the "AIR" as a mixture of:

DRY AIR = Nitrogen, Oxygen, etc.
H₂O Vapor

Temperature Range being studied (typically): -10 to +50 ^oC

Treat the DRY AIR as an IDEAL GAS with constant specific heat

C_P = 1.005 kJ/kg-^oK (use this for h = C_PT or Dh = C_PDT)

ALSO, we can treat the water vapor as an IDEAL GAS: P v = R T

Together, the dry air and the water vapor make a "mixture" where

P = P_a + P_v

P is the total (actual) pressure
P_a is the partial pressure of the DRY AIR
P_v is the partial pressure of the H₂O VAPOR (also called the vapor pressure)

FOR THE WATER VAPOR, evaluate enthalpy (h) using either:

the saturated water table (A-4), where h_v = h_g(T) or,
by the formula, h_v = h_g(T) = 2501.3 + 1.82 T which gives h in kJ/kg when you use T in ^oC.

A note on subscripts: ‘a’ is for the dry air, ‘v’ is for the water vapor

SPECIFIC AND RELATIVE HUMIDITY

Specific humidity, w, (or absolute humidity) gives the measure of the water vapor in the air:

w = m_v/ m_a (kg of H₂O vapor/kg dry air)

using the ideal gas law: w = 0.622 ( P_v/ P_a)

or: w = 0.622 [ P_v/ ( P - P_v)]

Air can hold only a certain amount of water vapor. The limit is reached when P_v reaches the saturation pressure of water at the temperature of the mixture. The air is then saturated with moisture and is called SATURATED AIR.

Any moisture added to the saturated air will condense.

EXAMPLE:

Air at 25^oC, 100 kPa

From Table A-4: P_{SAT@25 C}= 3.169 kPa

IF: P_v = 0, we have dry air (no H₂O vapor)

0< P_v < 3.169 kPa, we have unsaturated air

P_v = 3.169 kPa, we have saturated air

Relative Humidity, f, is a measure of amount of moisture in the air relative to the maximum amount it can hold at that temperature.

f = m_v/ m_g (kg of H₂O vapor/kg of H₂O vapor in saturated air)

with P_g = P_SAT@T, f = P_v/ P_g

According to this definition: f will always be a number between 0 and 1 ( or we can write it as a percentage)

Absolute and relative humidity relations:

The enthalpy of the air is: h = h_a + w h_g (in kJ/kg of dry air)

EXAMPLE:

A 5m x 5m x 3m room contains air at 25^oC and 100 kPa at a relative humidity of 75%. Determine:

the partial pressure of the dry air,
the specific humidity of the air,
the enthalpy per unit mass of dry air, and
the mass of dry air and of water vapor in the room.

(a) P_a = P - P_v P_v = f P_g = f P_{SAT@25 C} = (0.75)(3.169 kPa) = 2.38 kPa

_{P_a = 100 kPa - 2.38 kPa = 97.62 kPa}

(b)

h = (1.005 kJ/kg-^oC)(25^oC) + (0.0152)(2547.2 kJ/kg)

h = 63.8 kJ/kg dry air

(d) Both the water vapor and dry air fill the room completely:

So, V_a = V_v = V_ROOM

Then, using the ideal gas law:

Likewise, the mass of water vapor could be found:

OR, by using m_v = wm_a = (0.0152)(85.61 kg) = 1.3 kg

THE DEW POINT TEMPERATURE

If air is cooled at constant pressure, condensation begins at the dew point temperature: T_dp =T _SAT@Pv

As air cools at constant pressure, the vapor pressure of the water, P_v remains constant. When the temperature drops to the saturation temperature for that pressure, condensation begins. Further cooling follows the saturated vapor line.

WET-BULB TEMPERATURE

Used to measure the temperature of saturated air
A thermometer with a wet wick covering the bulb - as the water evaporates it cools - after steady conditions are reached, the wet bulb temperature, T_WB, is recorded.
Along with the dry bulb temperature (no wick - normal reading), the specific and relative humidity (w and f) can be determined.

Procedure:

Compute the specific humidity of the saturated (wet bulb) air.

and P₂ is the total (actual) pressure

Compute the specific humidity of the air around the dry bulb

Compute the relative humidity of the air around the dry bulb

Compute the enthalpy of the air per unit mass of dry air

h₁ = h_a1 + w₁h_v1 = C_PT_DB + w₁ h_g@T-DB

EXAMPLE (Air at 1 atm. = 101.325 kPa):

Given: dry- and wet-bulb temperatures: T_DB = 25^oC, T_WB = 15^oC

Find: specific humidity (w), relative humidity (f), and enthalpy (h)

at 25^oC, P_SAT = 3.169 kPa, h_g = 2547.2 kJ/kg
at 15^oC, P_SAT = 1.705 kPa, h_f = 62.99 kJ/kg, h_fg = 2465.9 kJ/kg

Procedure:

Compute the specific humidity of the saturated (wet bulb) air.

Compute the specific humidity of the air around the dry bulb

Compute the relative humidity of the air around the dry bulb

Compute the enthalpy of the air per unit mass of dry air

h₁ = h_a1 + w₁h_v1 = C_PT_DB + w₁ h_g@T-DB

= (1.005 kJ/kg-^oC)(25^oC) + (0.00653)(2547.2 kJ/kg)

= 41.8 kJ/kg dry air

THE PSYCHROMETRIC CHART

Tabulates properties at a specific air pressure
Different charts for different pressures
Used extensively in air conditioning design

EXAMPLE:

Given the dry- and wet-bulb temperatures:

T_DB = 25^oC
T_WB = 15^oC
(Air at 1 atm. pressure)

Find: the specific humidity (w), relative humidity (f), and enthalpy (h) from the psychrometric chart.

Read to the right margin, w= 6.5 g of H₂O/kg of dry air

= 0.0065 kg of H₂O/kg of dry air

Read from the curves of relative humidity, f = 33 %

Read from the upper left enthalpy scale, h = 42 kJ/kg of dry air

Read from the lines of specific volume, v = 0.853 m³/kg of dry air

Given any two of these properties, the others can be read from the chart.

SPECIFICATIONS FOR HUMAN COMFORT

AIR CONDITIONING

can mean cooling or heating !
(also includes humidification or de-humidification)

Air conditioning processes are shown on the psychrometric chart:

Most air conditioning systems can be treated as steady-flow processes.

For the dry air:

For the water vapor:

For the energy balance:

SIMPLE HEATING OR COOLING

(w = constant, f decreases)

Ignore H₂O vapor and use:

EXAMPLE:

Outdoor air at 10^oC, 1 atm. and 30% relative humidity flows into an air conditioning system at a rate of 45 m³/minute. The air is to be heated to 22^oC. Determine the rate of heat supply required (in kW).

From the chart:

At the inlet, for a dry-bulb temperature of 10^oC and f_i = 30%

w_i = 2.3 g H₂O/kg dry air = 0.0023 kg H₂O/kg dry air

h_i = 16 kJ/kg dry air and v_i = 0.805 m³/ kg dry air

At the exit:

Knowing w_e= w_i and T_e = 22^oC, you follow the horizontal line of constant specific humidity until to reach a dry-bulb temperature of 22^oC and read, h_e = 28 kJ/kg dry air

then,

HEATING WITH HUMIDIFICATION

(w increases from added H₂O, f goes up or down, depending on T)

Methods: Steam Injection (adds heat too) or water spray (cools the air)

EXAMPLE:

The same outdoor air as in the previous example (flowing into an air conditioning system at 45 m³/min.), after being heated to 22^oC, is then humidified by an injection of hot steam to a final state of 25^oC and 60% relative humidity. Find the mass flow rate of steam required.

The conservation of mass through this humidifying section is:

From the chart, at T_DP = 25^oC and f = 60%,

w_out= 12 g H₂O/kg dry air= 0.012 kg H₂O/kg dry air

then,

COOLING WITH DE-HUMIDIFICATION

Condense out some H₂O

w decreases, f reaches 100%

EXAMPLE:

Air enters a window A/C system at 1 atm., 30^oC, 80 % relative humidity at a rate of 10 m³/min. The air leaves the A/C system as saturated air at 14^oC. Also, part of the moisture which condensed during the process is removed at 14^oC. Determine the rates of heat and moisture removal from the air.

The mass flow rate of dry air remains constant during the entire process.

The amount of moisture in the air decreases due to dehumidification (w₃< w₁)

First, the air cools without condensation (w₂= w₁)

Then, continues to cool, with condensation and f₃= f₂

For the dry air mass:

For the water mass:

The energy balance:

From the psychrometric chart:

h₁ = 85.4 kJ/kg dry air

w₁ = 0.0216 kg H₂O/kg dry air

v₁ = 0.889 m³/kg dry air

h₃ = 39.3 kJ/kg dry air

w₃ = 0.010 kg H₂O/kg dry air

From the saturated water table (Table A-4): h_w = h_{f@14 C} = 58.8 kJ/kg

Then, at the inlet:

For the liquid water:

Energy balance: