14.2 Psychrometrics & Air Conditioning Systems

14.2 Psychrometrics & Air Conditioning Systems

Psychrometrics is the branch of thermodynamic science concerned with the physical and thermal properties of moist air—a binary mixture of dry air and water vapor. In tropical climates like the Philippines, HVAC engineers must precisely evaluate psychrometric properties to calculate sensible and latent cooling loads, size cooling coils, and ensure indoor thermal comfort.

Psychrometric Parameters & Fundamental Governing Equations

At standard atmospheric pressure ($P = 101.325 \text{ kPa}$ at sea level), the thermodynamic state of moist air is uniquely fixed by any two independent intensive properties.

1. Dry-Bulb Temperature ($DBT, t$ in $^\circ\text{C}$): The true air temperature measured by a standard thermometer shielded from direct radiation.
2. Wet-Bulb Temperature ($WBT, t_{wb}$ in $^\circ\text{C}$): The adiabatic saturation temperature measured by a thermometer covered with a water-wetted wick exposed to rapid air stream motion ($V > 4 \text{ m/s}$).
3. Dew-Point Temperature ($DPT, t_{dp}$ in $^\circ\text{C}$): The saturation temperature corresponding to the partial pressure $P_v$ of water vapor. Cooling air at constant pressure below its DPT causes moisture condensation.
4. Partial Vapor Pressure ($P_v$ in $\text{kPa}$): The partial pressure exerted by water vapor molecules in moist air. Saturated vapor pressure at dry-bulb temperature $t$ is designated $P_{sat}(t)$ or $P_g$.
5. Relative Humidity ($RH, \phi$): The ratio of partial vapor pressure $P_v$ to saturation pressure $P_g$ at dry-bulb temperature $t$: $\phi = \frac{P_v}{P_g} \times 100\%$
6. Specific Humidity / Humidity Ratio ($\omega$ in $\text{kg}{w}/\text{kg}{da}$ or $\text{g}{w}/\text{kg}{da}$): The mass ratio of water vapor $m_w$ to dry air $m_a$: $\omega = \frac{m_w}{m_a} = 0.622 \frac{P_v}{P - P_v}$ where $0.622 = \frac{M_w}{M_a} = \frac{18.015}{28.97}$ is the ratio of molecular weights of water vapor and dry air.
7. Specific Enthalpy ($h$ in $\text{kJ/kg}_{da}$): Total thermal energy of moist air per kilogram of dry air: $h = c_{pa} t + \omega (h_{fg,0} + c_{pv} t) = 1.006 t + \omega (2501 + 1.805 t) \quad (\text{kJ/kg}_{da}$ where $c_{pa} = 1.006 \text{ kJ/kg}\cdot^\circ\text{C}$, $h_{fg,0} = 2501 \text{ kJ/kg}$ (latent heat of vaporization at $0^\circ\text{C}$), and $c_{pv} = 1.805 \text{ kJ/kg}\cdot^\circ\text{C}$.
8. Specific Volume ($v$ in $\text{m}^3/\text{kg}_{da}$): Volume of moist air mixture per unit mass of dry air: $v = \frac{R_a T}{P - P_v} = \frac{0.28704 (t + 273.15)}{P - P_v} \quad (\text{m}^3/\text{kg}_{da}$

Basic Psychrometric Processes

1. Sensible Heating: Horizontal movement to the right on psychrometric chart ($DBT \uparrow$, $\omega = \text{const}$). $\dot{Q}_s = \dot{m}_a c_{pa} (t_2 - t_1) = \dot{m}_a (1.006) (t_2 - t_1) \quad (\text{kW})$

2. Sensible Cooling: Horizontal movement to the left ($DBT \downarrow$, $\omega = \text{const}$). $\dot{Q}_s = \dot{m}_a c_{pa} (t_1 - t_2) \quad (\text{kW})$

3. Cooling & Dehumidification: Standard tropical air conditioning process. Air passes over a cooling coil maintained below its dew point ($t_{\text{surface}} < DPT$).

   • Apparatus Dew Point ($ADP$): Effective surface saturation temperature of the cooling coil.
   • Bypass Factor ($BF$): Fraction of air that passes through the coil untransformed due to imperfect contact: $BF = \frac{t_2 - t_{ADP}}{t_1 - t_{ADP}} = \frac{\omega_2 - \omega_{ADP}}{\omega_1 - \omega_{ADP}} = \frac{h_2 - h_{ADP}}{h_1 - h_{ADP}}$
   • Contact Factor ($CF$): Fraction of air coming into perfect contact with coil surface ($CF = 1 - BF$).

4. Adiabatic Humidification (Evaporative Cooling): Air passes through a water spray recirculated at wet-bulb temperature ($h \approx \text{const}$, $WBT = \text{const}$, $DBT \downarrow$, $\omega \uparrow$).

Adiabatic Mixing of Two Air Streams

In central HVAC systems, outdoor fresh air (stream 1, mass flow $\dot{m}{a1}$) mixes adiabatically with return air from the building space (stream 2, mass flow $\dot{m}{a2}$) to form mixed supply air (stream 3, mass flow $\dot{m}{a3} = \dot{m}{a1} + \dot{m}_{a2}$).

Mass and energy conservation yield: $\text{Dry Air Mass Balance:} \quad \dot{m}_{a3} = \dot{m}_{a1} + \dot{m}_{a2}$ $\text{Moisture Balance:} \quad \dot{m}_{a1} \omega_1 + \dot{m}_{a2} \omega_2 = \dot{m}_{a3} \omega_3$ $\text{Enthalpy Balance:} \quad \dot{m}_{a1} h_1 + \dot{m}_{a2} h_2 = \dot{m}_{a3} h_3$

On the psychrometric chart, state 3 lies on the straight line connecting state 1 and state 2, positioned such that: $h_3 = \frac{\dot{m}_{a1} h_1 + \dot{m}_{a2} h_2}{\dot{m}_{a1} + \dot{m}_{a2}}, \quad \omega_3 = \frac{\dot{m}_{a1} \omega_1 + \dot{m}_{a2} \omega_2}{\dot{m}_{a1} + \dot{m}_{a2}}, \quad t_3 \approx \frac{\dot{m}_{a1} t_1 + \dot{m}_{a2} t_2}{\dot{m}_{a1} + \dot{m}_{a2}}$

Air Conditioning Load Calculations & SHF

Cooling load calculations account for heat gain entering the space:

• Room Sensible Heat ($RSH$): Conduction through walls/roof/windows ($Q = U A \Delta T$), solar radiation through glass ($Q = A \cdot \text{SHGC} \cdot I$), internal lighting, equipment motors, occupant sensible heat.
• Room Latent Heat ($RLH$): Moisture gain from human respiration and perspiration, cooking/steam equipment, water vapor infiltration.
• Total Room Heat ($TRH$): $TRH = RSH + RLH$.

Sensible Heat Factor (SHF)

The ratio of sensible heat to total heat load determines the slope of the room process line on the psychrometric chart:

$\text{RSHF} = \frac{RSH}{RSH + RLH} = \frac{RSH}{TRH}$

The Grand Sensible Heat Factor ($GSHF$) includes the outdoor ventilation air load imposed directly on the cooling coil:

$GSHF = \frac{TSH}{TSH + TLH} = \frac{RSH + OASH}{TRH + OALH}$

ASHRAE Standard 62.1 specifies minimum outdoor air rates to maintain indoor air quality (IAQ): $V_{ot} = R_p P_{z} + R_a A_{z}$ where $R_p$ is outdoor airflow rate per person (L/s$\cdot$person or cfm/person), $P_z$ is zone population, $R_a$ is outdoor airflow rate per unit area (L/s$\cdot\text{m}^2$), and $A_z$ is zone floor area.

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Step-by-Step Worked Sample Problem

Problem Statement: In a Manila commercial building central air conditioning system, outdoor ventilation air ($\dot{m}{a1} = 1.0 \text{ kg/s}$) at $35.0^\circ\text{C}$ DBT and $27.0^\circ\text{C}$ WBT ($h_1 = 85.0 \text{ kJ/kg}$, $\omega_1 = 0.0190 \text{ kg/kg}$) mixes adiabatically with return room air ($\dot{m}{a2} = 4.0 \text{ kg/s}$) at $24.0^\circ\text{C}$ DBT and $50%$ RH ($h_2 = 48.0 \text{ kJ/kg}$, $\omega_2 = 0.0093 \text{ kg/kg}$).

The resulting air mixture (state 3) passes through a cooling and dehumidifying coil having an Apparatus Dew Point ($ADP$) of $9.0^\circ\text{C}$ ($h_{ADP} = 27.0 \text{ kJ/kg}$, $\omega_{ADP} = 0.0072 \text{ kg/kg}$) and a Bypass Factor ($BF$) of $0.15$.

Calculate:

1. Total dry air mass flow rate ($\dot{m}_{a3}$) and enthalpy ($h_3$), humidity ratio ($\omega_3$), and dry-bulb temperature ($t_3$) of the mixed air entering the cooling coil.
2. Dry-bulb temperature ($t_4$) and enthalpy ($h_4$) of air leaving the cooling coil.
3. Total cooling capacity of the cooling coil in kW and Tons of Refrigeration (TR).

Solution Procedure:

Step 1: Calculate mixed air state 3 properties. $\dot{m}_{a3} = \dot{m}_{a1} + \dot{m}_{a2} = 1.0 + 4.0 = 5.0 \text{ kg/s}$

$\text{Enthalpy } h_3 = \frac{\dot{m}_{a1} h_1 + \dot{m}_{a2} h_2}{\dot{m}_{a3}} = \frac{1.0(85.0) + 4.0(48.0)}{5.0} = \frac{85.0 + 192.0}{5.0} = 55.4 \text{ kJ/kg}$

$\text{Humidity Ratio } \omega_3 = \frac{1.0(0.0190) + 4.0(0.0093)}{5.0} = \frac{0.0190 + 0.0372}{5.0} = 0.01124 \text{ kg/kg}$

$\text{Dry-Bulb Temp } t_3 \approx \frac{1.0(35.0) + 4.0(24.0)}{5.0} = \frac{35.0 + 96.0}{5.0} = 26.2^\circ\text{C}$

Step 2: Calculate coil leaving air state 4 properties using Bypass Factor ($BF = 0.15$). $BF = \frac{t_4 - t_{ADP}}{t_3 - t_{ADP}} \implies t_4 = t_{ADP} + BF(t_3 - t_{ADP})$ $t_4 = 9.0 + 0.15 \times (26.2 - 9.0) = 9.0 + 0.15 \times 17.2 = 9.0 + 2.58 = 11.58^\circ\text{C}$

$BF = \frac{h_4 - h_{ADP}}{h_3 - h_{ADP}} \implies h_4 = h_{ADP} + BF(h_3 - h_{ADP})$ $h_4 = 27.0 + 0.15 \times (55.4 - 27.0) = 27.0 + 0.15 \times 28.4 = 27.0 + 4.26 = 31.26 \text{ kJ/kg}$

Step 3: Calculate total cooling coil capacity ($\dot{Q}_{\text{coil}}$). $\dot{Q}_{\text{coil}} = \dot{m}_{a3} (h_3 - h_4) = 5.0 \text{ kg/s} \times (55.4 - 31.26) \text{ kJ/kg} = 5.0 \times 24.14 = 120.70 \text{ kW}$

Step 4: Convert cooling capacity to Tons of Refrigeration (TR). $\text{Capacity in TR} = \frac{120.70 \text{ kW}}{3.517 \text{ kW/TR}} = 34.32 \text{ TR}$