3.3 Basic Psychrometric Processes: Sensible Heating/Cooling, Humidification & Dehumidification

3.3 Basic Psychrometric Processes: Sensible Heating/Cooling, Humidification & Dehumidification

Every HVAC air-conditioning cycle consists of a sequence of fundamental thermodynamic state changes. Understanding the exact path of each process on the psychrometric chart, along with its governing mass and energy balances, is vital for predicting equipment performance and passing the PE exam.

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1. Directional Vector Map of Psychrometric Processes

From any initial state point $(T_{db1}, W_1)$ on the psychrometric chart, air conditioning processes move in eight characteristic directions:

1. Sensible Heating (Due East $\to$): Dry-bulb temperature increases at constant humidity ratio ($W = \text{const}$, $T_{dp} = \text{const}$, $p_v = \text{const}$). Relative humidity decreases.
2. Sensible Cooling (Due West $\leftarrow$): Dry-bulb temperature decreases at constant humidity ratio ($W = \text{const}$, $T_{dp} = \text{const}$). Relative humidity increases until saturation ($100%\text{ RH}$) is reached.
3. Humidification with Heating (North-East $\nearrow$): Both temperature and moisture content increase (e.g., heated steam or hot water spray).
4. Cooling and Dehumidification (South-West $\swarrow$): Temperature and moisture content both decrease. Occurs when air passes over a cooling coil whose effective surface temperature is below the entering air dew point ($T_{coil} < T_{dp1}$).
5. Humidification with Cooling / Evaporative Cooling (North-West $\nwarrow$): Adiabatic saturation along lines of constant wet-bulb / constant enthalpy. Moisture evaporates into the air, extracting sensible heat and lowering $T_{db}$ while increasing $W$ and $\text{RH}$.
6. Chemical / Desiccant Dehumidification (South-East $\searrow$): Moisture is adsorbed by a solid or liquid desiccant. The latent heat of sorption is released into the airstream, causing $T_{db}$ to increase as $W$ decreases along a nearly constant enthalpy line.
7. Isothermal Humidification (Due North $\uparrow$): Pure dry steam injection at $212^\circ\text{F}$ adds moisture with minimal change in dry-bulb temperature.
8. Pure Dehumidification (Due South $\downarrow$): Moisture extraction at constant dry bulb (rarely achieved in practice without reheat).

2. Governing Equations for Basic Processes

1. Sensible Heating & Cooling Coils

When air flows across a hot water, steam, or electric heating coil (or a dry cooling coil operating above the entering air dew point $T_{coil} > T_{dp}$), no moisture is added or removed:

$W_1 = W_2 \quad \text{and} \quad T_{dp1} = T_{dp2}$

$\dot{q}_s = \dot{m}_{da} c_{p,ma} (T_2 - T_1) = 1.08 \times \text{CFM} \times (T_2 - T_1) \quad [\text{Btu/hr}]$

For electric resistance heating coils, thermal capacity is directly related to electrical power $P_{kW}$:

$P_{kW} = \frac{\dot{q}_s}{3412.14} = \frac{1.08 \times \text{CFM} \times \Delta T}{3412.14}$

2. Cooling and Dehumidification (Wet Cooling Coils)

When the effective surface temperature of a cooling coil ($T_s$) is below the dew point of the entering air ($T_s < T_{dp1}$), water vapor condenses on the coil fins:

$\text{Total Coil Load: } \dot{q}_{total} = \dot{m}_{da} (h_1 - h_2) = 4.5 \times \text{CFM} \times (h_1 - h_2) \quad [\text{Btu/hr}]$ $\text{Sensible Load: } \dot{q}_{sensible} = 1.08 \times \text{CFM} \times (T_1 - T_2) \quad [\text{Btu/hr}]$ $\text{Latent Load: } \dot{q}_{latent} = \dot{q}_{total} - \dot{q}_{sensible} = 4840 \times \text{CFM} \times (W_1 - W_2) \quad [\text{Btu/hr}]$

$\text{Sensible Heat Ratio (SHR): } \text{SHR} = \frac{\dot{q}_{sensible}}{\dot{q}_{total}} = \frac{\dot{q}_s}{\dot{q}_s + \dot{q}_l}$

$\text{Condensate Water Drainage Rate: } \dot{m}_{cond} = \dot{m}_{da} (W_1 - W_2) = 60 \times 0.075 \times \text{CFM} \times (W_1 - W_2) \quad [\text{lb}_m/\text{hr}]$ $\text{Condensate Volumetric Rate: } \text{GPH} = \frac{\dot{m}_{cond}}{8.33\text{ lb/gal}} \quad [\text{gallons per hour}]$

3. Direct Evaporative Cooling (Swamp Coolers)

In a direct evaporative cooler, unsaturated air passes through a wetted media where water evaporates without external heat addition (adiabatic process). The heat of vaporization is supplied entirely by the air's sensible heat:

$h_1 \approx h_2 \quad \text{and} \quad T_{wb1} = T_{wb2}$

$\text{Saturation Effectiveness (}\epsilon\text{): } \epsilon = \frac{T_{db,in} - T_{db,out}}{T_{db,in} - T_{wb,in}}$

$T_{db,out} = T_{db,in} - \epsilon \left( T_{db,in} - T_{wb,in} \right)$

$\text{Water Consumption Rate: } \dot{m}_{water} = \dot{m}_{da} (W_2 - W_1) = 4.5 \times \text{CFM} \times (W_2 - W_1) \quad [\text{lb/hr}]$

4. Steam Humidification vs. Water Spray Humidification

• Steam Humidification (Isothermal): Steam at $212^\circ\text{F}$ ($h_g = 1150.5\text{ Btu/lb}$) injected into an airstream causes negligible change in dry-bulb temperature ($\Delta T \approx +1^\circ\text{F}$ for typical additions): $\dot{m}_{steam} = \dot{m}_{da} (W_2 - W_1) = 4.5 \times \text{CFM} \times (W_2 - W_1) \quad [\text{lb/hr}]$
• Water Spray Humidification (Adiabatic): Atomized unheated liquid water spray absorbs latent heat from the airstream, reducing dry-bulb temperature while raising humidity ratio along the wet-bulb line.

3. Step-by-Step Worked Example: Cooling & Dehumidifying Coil Sizing

Problem Statement

An air-handling unit conditioning a lecture auditorium handles $8,000\text{ CFM}$ of air at standard sea-level pressure ($14.696\text{ psia}$). The air enters the chilled-water cooling coil at State 1 and leaves at State 2:

• Entering Air (State 1): $T_{db1} = 82.0^\circ\text{F}$, $T_{wb1} = 68.0^\circ\text{F}$ ($h_1 = 32.40\text{ Btu/lb}_{da}$, $W_1 = 0.0116\text{ lb}w/\text{lb}{da}$)
• Leaving Air (State 2): $T_{db2} = 54.0^\circ\text{F}$, $T_{wb2} = 52.5^\circ\text{F}$ ($h_2 = 21.75\text{ Btu/lb}_{da}$, $W_2 = 0.0081\text{ lb}w/\text{lb}{da}$)

Calculate:

1. Total cooling capacity in Btu/hr and tons of refrigeration
2. Sensible cooling capacity in Btu/hr
3. Latent cooling capacity in Btu/hr
4. Coil Sensible Heat Ratio (SHR)
5. Water condensation removal rate in pounds per hour (lb/hr) and gallons per hour (GPH)

Solution Steps

Step 1: Calculate Total Cooling Capacity $\dot{q}_{total} = 4.5 \times \text{CFM} \times (h_1 - h_2)$ $\dot{q}_{total} = 4.5 \times 8,000 \times (32.40 - 21.75) = 36,000 \times 10.65 = 383,400\text{ Btu/hr}$ $\text{Tons of Refrigeration} = \frac{383,400\text{ Btu/hr}}{12,000\text{ Btu/hr}\cdot\text{ton}} = 31.95\text{ tons}$

Step 2: Calculate Sensible Cooling Capacity $\dot{q}_{sensible} = 1.08 \times \text{CFM} \times (T_{db1} - T_{db2})$ $\dot{q}_{sensible} = 1.08 \times 8,000 \times (82.0 - 54.0) = 8,640 \times 28.0 = 241,920\text{ Btu/hr} = 20.16\text{ tons}$

Step 3: Calculate Latent Cooling Capacity $\dot{q}_{latent} = \dot{q}_{total} - \dot{q}_{sensible} = 383,400 - 241,920 = 141,480\text{ Btu/hr} = 11.79\text{ tons}$ (Verification via humidity ratio: $\dot{q}_l = 4840 \times 8000 \times (0.0116 - 0.0081) = 38,720,000 \times 0.0035 = 135,520\text{ Btu/hr}$, well within rounding of psychrometric chart lookups).

Step 4: Calculate Sensible Heat Ratio (SHR) $\text{SHR} = \frac{\dot{q}_{sensible}}{\dot{q}_{total}} = \frac{241,920}{383,400} = 0.631$

Step 5: Calculate Condensate Drainage Rate $\dot{m}_{cond} = 60 \times \rho \times \text{CFM} \times (W_1 - W_2) = 4.5 \times 8,000 \times (0.0116 - 0.0081)$ $\dot{m}_{cond} = 36,000 \times 0.0035 = 126.0\text{ lb}_m/\text{hr}$ $\text{Condensate Flow Rate (GPH)} = \frac{126.0\text{ lb/hr}}{8.33\text{ lb/gal}} = 15.13\text{ GPH}$ (A continuous 15.1 gallons per hour must be drained away from the coil pan).