Chapter SM 4: Heat and Mass Exchangers
SM 4.1 Introduction
In a number of HVAC components, water evaporates into or condenses out of an air stream. In an evaporative cooler, a warm and relatively dry air stream is cooled through evaporation from either a wet surface or water sprays. Such units are used for space cooling in the dry southern western areas of the country. In a humidifier, water or steam is sprayed into an air stream to increase the humidity level. Humidifiers are used to increase the humidity of dry and uncomfortable air that enters the building in very cold climates. Air washers are used to entrain particles and to clean the air flow through the spraying water into the air stream. These units are sometimes used instead of mechanical filters. Enthalpy exchangers are devices that exchange heat and moisture between two air streams, and often a desiccant is used as the exchange medium. In summer these units are used to remove the heat and moisture from incoming ventilation air and transfer it to exhaust air, which reduces the load on the cooling coil. In winter operation the exchanger is used to preheat and humidify the ventilation air. All of the devices described here fit specific niches in HVAC systems.
The basic thermodynamic, heat, and mass transfer principles for these devices will be presented in this chapter. The performance of these heat and mass exchangers can usually be expressed in terms of the parameters used for heat exchangers (Chapter 13) and cooling coils (Chapter 14).
SM 4.2 Evaporative Coolers
An evaporative cooler is shown schematically in Figure 4.1. A warm and relatively dry air stream, usually from the ambient, enters the device and passes over wetted pads or surfaces. Evaporation of the water cools and humidifies the air stream. The exchange surfaces are maintained wet by a circulating water flow. An alternative to using wetted surfaces is to spray water into the air, but it is more difficult to control the degree of saturation in this method. The evaporative cooler process is shown on a psychrometric diagram in Figure 4.1.
Figure 4.1 Schematic of evaporative cooler process
The air outlet state can be related to the inlet state through application of mass and energy balances. A steady-state mass balance on the water is
(4.1)
where and are the air and water flow rates, respectively, and win and wout are the inlet and outlet humidity ratios. Equation 4.1 relates the mass of water that evaporates to the humidity levels. The energy balance for the process incorporates the water mass flow rate from equation 4.1
(4.2)
where hin and hout are the enthalpies of the entering and leaving air /water vapor mixture, respectively, and hw is the enthalpy of the entering water stream. For evaporator cooler applications the last term in the energy balance is negligible (Section 4.4). The evaporative cooling process is then essentially a constant enthalpy process for the air stream. Evaporative cooling does not reduce the enthalpy of the air stream and thus would not reduce the load on a cooing coil if the unit were followed by a conventional air conditioner. Evaporative coolers are often used as the only source of cooling in very dry climates such as the Southwest. The air stream entering the space is then cooler and more comfortable and the outlet humidity is not so high that the air feels damp.
Because enthalpy and wet bulb lines are essentially parallel on a psychrometric chart, it is also accurate to think of the process as occurring along a constant wet bulb line. An effectiveness parameter is useful in characterizing the performance of an evaporative cooler. It is convenient to define a temperature effectiveness in terms of the approach of the outlet temperature to the wet bulb.
(4.3)
where Twb is the wet-bulb temperature of the inlet air stream. The effectiveness depends on the transfer coefficients, the surface area, and air flow rate (the Ntu) in the same manner as for sensible heat exchangers and cooling coils. Typical values of the effectiveness range between 60 to 80 %. Evaporative coolers are discussed in Section 5.7.
Evaporative coolers provide cooling at the expense of humidification of the air stream, but are limited by the wet-bulb temperature of the air. It is possible to provide another stage of cooling using an indirect evaporative cooler. In this device, an air stream is first cooled in an evaporative cooler and then used as the cold flow in a heat exchanger to cool another air stream. The cooled air stream can then be evaporatively cooled to provide a temperature lower than that with a single stage evaporative cooler. The arrangement is shown schematically and on psychrometric coordinates in Figure 4.2.
Figure 4.2 Schematic operation of an indirect evaporator cooler.
The desired output is the cooled outlet air stream, which is at the humidity ratio of the incoming ambient but at a lower temperature. The performance of the system can be determined in terms of the effectivenesses of the evaporative cooler and heat exchanger, eec and ehx, respectively. The cold outlet temperature of the evaporative cooler, Tc, is evaluated in terms of the effectiveness and wet bulb temperature using equation 4.3. The outlet temperature of the sensibly cooled stream, Tout, is then given in terms of the heat exchanger effectiveness and the temperature leaving the evaporative cooler, Tc. For equal capacitance rates for the heat exchanger, the outlet temperature can be expressed in terms of the inlet temperature, the wet bulb temperature, and the effectivenesses of evaporative cooler and heat exchanger as.
(4.4)
Commercially manufactured indirect evaporative coolers usually combine the evaporation and heat transfer effects into one unit. Water is brought in and flows down along sheets of plastic or metal that also form passages between which the air flows. This allows the heat transfer and evaporation to occur simultaneously. Example 4.2 illustrates the cooling that can be obtained through an indirect evaporative cooler.
"Example 4.1 An indirect evaporative cooling system combines the evaporative cooler of example 4.1 (80 % effectiveness) with a heat exchanger of 70 % effectiveness. The air flow rate through both units is 2500 cfm and the ambient is 90 F and 5 % relative humidity. Determine the outlet state and the cooling potential."
"Problem specifications"
T_a = 90 “F” “Ambient temperature”
RH_a = 0.05 “Ambient RH”
p_atm = 14.7 “psia” “Ambient pressure”
V_dot = 2500 “cfm” “Volume flow rate”
eff_EC = 0.8 “Evaporative cooler eff.”
eff_HX = 0.7 “Heat exchanger eff”
"Determine the properties of the incoming air stream and the mass flow rate of the air streams"
T_wb_a = WetBulb(AirH2O,t=T_a, p=p_atm, r=RH_a) “F” “Ambient wet-bulb”
w_a = HumRat(AirH2O,t=T_a, p=p_atm, r=RH_a) “lbm/lbm” “Ambient humidity”
rho_a = density(AirH2O,t=T_a, p=p_atm, r=RH_a) “lbm/ft3” “Ambient density”
cp_a = specheat(AirH2O,t=T_a, p=p_atm, r=RH_a) “Btu/lbm-F” “Air specific heat”
m_dot = rho_a*V_dot*Convert(1/min,1/hr) “lbm/hr” “Mass flow rate”
"For the first evaporative cooler, determine the outlet temperature. The outlet temperature will be the sink for cooling the second air stream from the ambient."
eff_EC = (T_a -T_1)/(T_a - T_wb_a) “Evaporative cooler eff”
"The humidity ratio of the coolant stream is"
w_2 = humrat(AirH2O,T=T_1, p=p_atm, B=T_wb_a) “lbm/lbm” “Outlet humidity”
"For the heat exchanger, determine the outlet temperatures of the supply stream (T_3) and the stream that was used to cool the exhaust stream (T_2). The capacitance rates of the supply and coolant stream are equal. The effectiveness based on the supply stream is"
eff_HX = (T_a -T_3)/(T_a - T_1) “Heat exchanger eff”
"The outlet humidity of the supply stream (w_3) equals that of the ambient since the supply stream is only cooled sensibly."
w_3 =w_a “lbm/lbm” “Humidity ratio”
"The outlet temperature of the coolant stream is determined. Because the supply and coolant stream capacitance ratios are equal the energy balance reduces to"
(T_2- T_1) = (T_a - T_3) “F” “Energy balance”
w_2 = w_1 “lbm/lbm” “Humidity ratio”
"The supply stream is then evaporatively cooled in an evaporative cooler. The effectiveness is the same as for the first evaporative cooler. The wet-bulb temperature that of the supply stream at state 3."
T_wb_3 = wetbulb(AirH2O,T=T_3, p=p_atm, w=w_3) “F” “Supply wet-bulb”
eff_EC = (T_3 -T_supply)/(T_3 - T_wb_3) “Evaporative cooler eff”
"The humidity of the supply state is"
w_supply = humrat(AirH2O,T=T_supply, p=p_atm, B=T_wb_3) “lbm/lbm” “Supply humidity”
RH_supply = RelHum(AirH2O,T=T_supply,P=p_atm,w=w_supply) “Supply RH”
"The sensible cooling that could be achieved for a building with a temperature of 78 F with the supply flow is"
T_z = 78 “F” “Zone temperature”
Cool = m_dot*cp_a*(T_z - T_supply) “Btu/hr” “Cooling potential”
Tons = Cool*convert(Btu/hr,tons) “tons” “Cooling potential”
Results and Discussion
The ambient has a wet bulb temperature of 55.4 F, and is cooled to 62.3 F in the first evaporative cooler. The air stream with this temperature is then used to sensibly cool the supply stream from the ambient from 90 F to 70.6 F. The wet-bulb temperature of this stream is reduced to 47.0 F.
In the second evaporative cooler the supply stream is brought to a condition of 51.8 F and 70 % RH. The supply stream at this temperature could provide a cooling potential of 68,380 Btu/hr, or 5.7 tons. This is 70 % more cooling potential than that provided by the single stage evaporative cooler.
If the cooling potential is greater than desired, the effectiveness of the second evaporative cooler can be decreased by reducing the water flow rate. This would increase the supply temperature and reduce the relative humidity, and possible make for more comfortable conditions.
Examples 4.1 and 4.2 illustrate how evaporative coolers can be employed to provide air conditioning. The cooling cost is "free" in that only fan power is required. Fan power for an evaporative cooler is generally small relative to the cooling benefit. Indirect evaporative coolers have higher fan power requirements due to the pressure drop in the heat exchanger, and two stages of evaporative cooling are the largest number of stages that is economically feasible.
4.3 Spray Dehumidifiers
Spray dehumidifiers are occasionally used to both cool and dehumidify air. A water flow is cooled by a chiller and then introduced into the spray dehumidifier. The operation is similar to that for a cooling tower, with the difference that the entering water stream is cold. The spray dehumidifier differs from the evaporative cooler in that the water flow rate is high and only a small amount of water evaporates. Inside the dehumidifier, the air becomes saturated and cooled. A schematic of a spray dehumidifier and the psychrometric representation of the air and water processes are shown in Figure 4.3.
Figure 4.3 Schematic operation of a spray dehumidifier
An effectiveness is defined in terms of enthalpies. The coldest and driest possible exit state for the air stream is to leave in equilibrium with the water inlet state, which corresponds to saturation at the temperature of the incoming water. The effectiveness is then defined as the actual enthalpy change from inlet to outlet divided by the enthalpy difference between the air inlet and saturated air at the water inlet temperature.
(4.5)
For the conditions illustrated in Figure 4.3, the air inlet enthalpy is greater than that of saturated air at the water inlet temperature. Energy will then be transferred from the air to the water stream, the air will be cooled, and the humidity level will drop. The water condensed out of the air will leave with the water flow. In this manner a cold water spray will actually dehumidify the air. Example 4.3 illustrates the performance of a spray dehumidifier.
"Example 4.2 An air flow of 5000 L/s and at 30 C and 75 % RH enters a spray dehumidifier with an effectiveness of 0.8 and an inlet water temperature of 10 C. Determine the outlet state of the air and the water flow rate required for a 2 C temperature rise of the water through the spray dehumidifier."
"Problem Specifications"
p_atm = 101.3 "kPa" "Atmospheric pressure"
T_in= 30 "C" "Air inlet temperature"
RH_in = 0.75 "Relative humidity"
V_dot = 5000 "L/s" "Air volume flow rate"
eff = 0.8 "Dehum. effectiveness"
T_w_in= 10 "C" "Water inlet temp."
T_w_out = 12 "C" "Water outlet temp"
"Properties of the air flow at inlet conditions"
m_dot_a = V_dot*rho_in*convert(L,m3) "kg/s" "Water mass flow rate"
rho_in = density(AirH2O, T = T_in, p=p_atm, R=RH_in) "kg/m3""Air density"
w_in = HumRat(AirH2O, T = T_in, p=p_atm, R=RH_in) "kg/kg""Humidity ratio"
h_in = Enthalpy(AirH2O, T = T_in, p=p_atm, R=RH_in) "kJ/kg""Air enthalpy"
"Properties of the air flow at outlet conditions"
RH_out = 1 "Relative humidity"
w_out = HumRat(AirH2O, T = T_out , p=p_atm, R=RH_out ) "kg/kg""Humidity ratio"
h_out = Enthalpy(AirH2O, T = T_out , p=p_atm, R=RH_out ) "kJ/kg""Air enthalpy"
"Properties of the water flow"
h_w_in= Enthalpy(Water,T=T_w_in,P=p_atm) "kJ/kg" "Water enthalpy"
h_w_out = Enthalpy(Water,T=T_w_out,P=p_atm) "kJ/kg" "Water enthalpy"
"Spray dehumidifier effectiveness, equation 4.5"
eff = (h_in- h_out)/(h_in- h_w_sat_in) "Defn of effectiveness"
h_w_sat_in= Enthalpy(AirH2O, T = T_w_in, p=p_atm, R=1) "kJ/kg""Sat air enth. at T_w_in"
"Mass balance on the dehumidifier to determine the rate at which moisture is condensed out of the air stream"
m_dot_cond = m_dot_a*(w_in- w_out) "kg/s" "Condensation flow rate"
"Energy balance on the dehumidifier to determine the water flow rate that will produce a 2 C rise in the water temperature. The energy of the condensed water is included in the energy balance and its temperature is taken as that of the inlet water."
m_dot_a*(h_in- h_out) + m_dot_w*h_w_in- m_dot_w*h_w_out -m_dot_cond*h_w_out = 0 "kW"
Results and Discussion:
Solving the set of equations yields the amount of moisture condensed out of the air as 0.057 kg/s, an inlet water flow rate of 27.94 kg/s, and an outlet of 28.00 kg/s. The condensate flow rate is relatively small compared to the total water flow rate, but there still is some water addition to the flow stream. Ultimately this condensate flow will need to be drained off.