3.5 Sensible Heat Ratio (SHR), Apparatus Dew Point (ADP) & Coil Bypass Factor

3.5 Sensible Heat Ratio (SHR), Apparatus Dew Point (ADP) & Coil Bypass Factor

Designing a cooling coil requires satisfying both the sensible cooling load (temperature reduction) and the latent cooling load (moisture removal) simultaneously. The Sensible Heat Ratio (SHR), Apparatus Dew Point (ADP), and Coil Bypass Factor (BF) are the engineering parameters that link space cooling loads to physical coil construction and required airflow.

1. Sensible Heat Ratio (SHR) Definitions

Sensible Heat Ratio (also called Sensible Heat Factor, SHF) is the ratio of sensible heat to total heat:

$\text{SHR} = \frac{\dot{q}_{sensible}}{\dot{q}_{total}} = \frac{\dot{q}_s}{\dot{q}_s + \dot{q}_l}$

1. Room Sensible Heat Ratio (RSHR)

Governs the slope of the Room Condition Line (RCL) on the psychrometric chart connecting the room design state ($R$) and the supply air state ($S$):

$\text{RSHR} = \frac{q_{s,room}}{q_{s,room} + q_{l,room}}$

Where $q_{s,room}$ and $q_{l,room}$ include internal loads (people, lighting, equipment, envelope conduction, solar gain) but exclude outdoor ventilation air loads.

2. Grand Sensible Heat Ratio (GSHR)

Governs the slope of the Grand Condition Line (GCL) connecting the mixed air state ($M$) to the apparatus dew point ($ADP$):

$\text{GSHR} = \frac{q_{s,total}}{q_{t,total}} = \frac{q_{s,room} + q_{s,vent}}{q_{t,room} + q_{t,vent}}$

Where:

• $q_{s,vent} = 1.08 \times \text{CFM}{oa} \times (T{oa} - T_{room})$
• $q_{l,vent} = 4840 \times \text{CFM}{oa} \times (W{oa} - W_{room})$

3. Effective Sensible Heat Ratio (ESHR)

Used when coil bypass factor $BF > 0$. It accounts for the fraction of outdoor ventilation air that bypasses the coil fins and enters the conditioned space directly:

$\text{ESHR} = \frac{q_{s,room} + (1 - BF) q_{s,vent}}{q_{t,room} + (1 - BF) q_{t,vent}}$

2. Apparatus Dew Point (ADP) & Coil Bypass Factor (BF)

Apparatus Dew Point (ADP)

The Apparatus Dew Point is the effective average surface temperature of the cooling coil fins and tubes. Geometrically on the psychrometric chart, the $ADP$ is the point where the condition line intersects the $100%\text{ RH}$ saturation curve ($T_{db} = T_{wb} = T_{dp} = T_{adp}$).

Coil Bypass Factor ($BF$)

In a physical cooling coil, not all air molecules make direct thermal contact with the cold fin surfaces. A fraction of the air ($BF$) passes through between fins unaffected, while the remaining fraction ($1 - BF$) comes into complete thermal equilibrium at the $ADP$.

$\text{Bypass Factor: } BF = \frac{T_{leaving} - T_{adp}}{T_{entering} - T_{adp}} = \frac{W_{leaving} - W_{adp}}{W_{entering} - W_{adp}} = \frac{h_{leaving} - h_{adp}}{h_{entering} - h_{adp}}$

$\text{Contact Factor: } CF = 1 - BF = \frac{T_{entering} - T_{leaving}}{T_{entering} - T_{adp}}$

Rearranging to find leaving coil air temperature ($T_{leaving}$):

$T_{leaving} = T_{adp} + BF \left( T_{entering} - T_{adp} \right) = (1 - BF) T_{adp} + BF \cdot T_{entering}$

Parameters Governing Coil Bypass Factor ($BF$)

3. Supply Airflow (CFM) Calculation with Coil Bypass

To satisfy the room sensible cooling load ($q_{s,room}$), the supply airflow must be sized based on the temperature difference between the room design temperature ($T_{room}$) and the actual supply air temperature ($T_{supply}$):

$\text{CFM}_{sa} = \frac{q_{s,room}}{1.08 \times (T_{room} - T_{supply})}$

Expressing $T_{supply}$ (leaving coil) in terms of $ADP$ and $BF$: $T_{room} - T_{supply} = T_{room} - \left[ T_{adp} + BF(T_{room} - T_{adp}) \right] = (1 - BF)(T_{room} - T_{adp})$

$\text{CFM}_{sa} = \frac{q_{s,room}}{1.08 \times (1 - BF) \times (T_{room} - T_{adp})}$

4. Step-by-Step Worked Example: Full SHR, ADP, and Airflow Sizing

Problem Statement

A conditioned space has a design sensible cooling load of $q_{s,room} = 194,400\text{ Btu/hr}$ and a latent cooling load of $q_{l,room} = 64,800\text{ Btu/hr}$.

• Indoor Room Design Condition: $T_{room} = 75.0^\circ\text{F}$ DB, $50%\text{ RH}$ ($W_{room} = 0.00925\text{ lb/lb}$)
• Outdoor Ventilation Air: $2,000\text{ CFM}$ at $95.0^\circ\text{F}$ DB, $78.0^\circ\text{F}$ WB ($W_{oa} = 0.0168\text{ lb/lb}$)
• Selected Chilled Water Coil: Bypass Factor $BF = 0.10$

Calculate:

1. Room Sensible Heat Ratio (RSHR)
2. Grand Sensible Heat Ratio (GSHR)
3. Apparatus Dew Point ($ADP$) temperature from psychrometric alignment (RCL intersection with saturation curve yields $T_{adp} = 51.0^\circ\text{F}$)
4. Leaving coil air temperature ($T_{leaving}$)
5. Required supply airflow rate ($\text{CFM}_{sa}$)

Solution Steps

Step 1: Calculate Room Sensible Heat Ratio (RSHR) $\text{RSHR} = \frac{q_{s,room}}{q_{s,room} + q_{l,room}} = \frac{194,400}{194,400 + 64,800} = \frac{194,400}{259,200} = 0.750$

Step 2: Calculate Ventilation Loads and GSHR $q_{s,vent} = 1.08 \times \text{CFM}_{oa} \times (T_{oa} - T_{room}) = 1.08 \times 2,000 \times (95.0 - 75.0) = 2,160 \times 20.0 = 43,200\text{ Btu/hr}$ $q_{l,vent} = 4840 \times \text{CFM}_{oa} \times (W_{oa} - W_{room}) = 4840 \times 2,000 \times (0.0168 - 0.00925) = 9,680,000 \times 0.00755 = 73,084\text{ Btu/hr}$ $q_{s,total} = q_{s,room} + q_{s,vent} = 194,400 + 43,200 = 237,600\text{ Btu/hr}$ $q_{t,total} = (q_{s,room} + q_{l,room}) + (q_{s,vent} + q_{l,vent}) = 259,200 + 116,284 = 375,484\text{ Btu/hr}$ $\text{GSHR} = \frac{q_{s,total}}{q_{t,total}} = \frac{237,600}{375,484} = 0.633$

Step 3: Determine Apparatus Dew Point ($ADP$) Plotting the RSHR line ($0.75$) through the room state ($75.0^\circ\text{F}$ DB, $50%\text{ RH}$) to the saturation curve yields: $T_{adp} = 51.0^\circ\text{F}$

Step 4: Calculate Leaving Coil Air Temperature ($T_{leaving}$) $T_{leaving} = T_{adp} + BF \times (T_{room} - T_{adp}) = 51.0^\circ\text{F} + 0.10 \times (75.0^\circ\text{F} - 51.0^\circ\text{F})$ $T_{leaving} = 51.0 + 0.10(24.0) = 51.0 + 2.4 = 53.4^\circ\text{F}$

Step 5: Calculate Required Supply Airflow ($\text{CFM}_{sa}$) $\Delta T = T_{room} - T_{leaving} = 75.0^\circ\text{F} - 53.4^\circ\text{F} = 21.6^\circ\text{F}$ $\text{CFM}_{sa} = \frac{q_{s,room}}{1.08 \times \Delta T} = \frac{194,400}{1.08 \times 21.6} = \frac{194,400}{23.328} = 8,333\text{ CFM}$ (Verification via direct formula: $\text{CFM} = \frac{194,400}{1.08 \times (1 - 0.10) \times (75 - 51)} = \frac{194,400}{1.08 \times 0.90 \times 24} = \frac{194,400}{23.328} = 8,333\text{ CFM}$).