---
title: "Cooling and dehumidifying coil load"
subject: "Applied Thermodynamics"
chapter: 4
key: "semester_3/applied_thermodynamics:cooling-dehumidifying-coil-numerical"
source: https://www.bctengineer.com/q/semester_3/applied_thermodynamics/cooling-dehumidifying-coil-numerical
site: BCT Engineering
---

# Cooling and dehumidifying coil load

Applied Thermodynamics · Chapter 4: Air-Conditioning

- Asked in: year not given, 8 marks
- Video: [Cooling with dehumidification](https://www.youtube.com/watch?v=JwEW9yRXxw4) — tmarbach
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_thermodynamics/cooling-dehumidifying-coil-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_thermodynamics/chapter-4

## Question

Air at 30 °C and 60% RH flows at 120 m³/min through a cooling coil with an apparatus dew point of 8 °C and a bypass factor of 0.25. Determine (a) the condition of air leaving the coil, (b) the total, sensible and latent cooling loads in kW and TR, (c) the sensible heat factor, and (d) the rate of condensate removal. Take $p = 101.325$ kPa, 1 TR = 3.517 kW.

## Short answer

With apparatus dew point 8 °C and bypass factor 0.25, leaving air is at 13.5 °C, 0.00900 kg/kg, ~93% RH. Total cooling load 79.2 kW (22.5 TR), sensible 38.8 kW, latent 40.4 kW; SHF 0.49; condensate 57.6 kg/h.

## Answer

#### Entering air state 1 (30 °C, 60 %)
$p_s = 4.247$ kPa, $p_v = 0.6 \times 4.247 = 2.548$ kPa.

$$
\omega_1 = 0.01605\ \text{kg/kg},\quad h_1 = 71.19\ \text{kJ/kg},\quad v_1 = 0.8808\ \text{m}^3/\text{kg}
$$

$$
\dot m_a = \frac{120}{0.8808 \times 60} = 2.271\ \text{kg/s}
$$

#### ADP state A (8 °C, saturated)
$p_s(8\ ^\circ\text{C}) = 1.072$ kPa, so $\omega_A = 0.00665$ kg/kg and $h_A = 24.77$ kJ/kg.

#### (a) Leaving state 2
$$
\text{BPF} = \frac{t_2 - t_A}{t_1 - t_A} = \frac{\omega_2 - \omega_A}{\omega_1 - \omega_A} = 0.25
$$

$$
\begin{aligned}
t_2 &= 8 + 0.25(30 - 8) = 13.5\ ^\circ\text{C}\\
\omega_2 &= 0.00665 + 0.25(0.01605 - 0.00665) = 0.00900\ \text{kg/kg}\\
h_2 &= 1.005 \times 13.5 + 0.00900(2501 + 1.88 \times 13.5) = 36.30\ \text{kJ/kg}
\end{aligned}
$$

Relative humidity of the leaving air: $p_{v2} = 1.445$ kPa and $p_s(13.5\ ^\circ\text{C}) \approx 1.548$ kPa, so $\phi_2 \approx 0.93$ (about 93 %).

#### (b) Loads
Intermediate state $h'$ at ($t_2$, $\omega_1$): $h' = 1.005 \times 13.5 + 0.01605(2501 + 25.4) = 54.11$ kJ/kg.

$$
\begin{aligned}
Q_{total} &= \dot m_a(h_1 - h_2) = 2.271 \times (71.19 - 36.30) = 79.2\ \text{kW} = 22.5\ \text{TR}\\
Q_{sensible} &= \dot m_a(h_1 - h') = 2.271 \times (71.19 - 54.11) = 38.8\ \text{kW}\\
Q_{latent} &= \dot m_a(h' - h_2) = 2.271 \times (54.11 - 36.30) = 40.4\ \text{kW}
\end{aligned}
$$

(The enthalpy carried away by the condensate, under 1 kW, is small and neglected.)

#### (c) Sensible heat factor
$$
\text{SHF} = \frac{Q_s}{Q_{total}} = \frac{38.8}{79.2} = 0.49
$$

#### (d) Condensate
$$
\dot m_w = \dot m_a(\omega_1 - \omega_2) = 2.271 \times (0.01605 - 0.00900) = 0.0160\ \text{kg/s} = 57.6\ \text{kg/h}
$$

**Answer:** (a) 13.5 °C, $\omega$ = 0.00900 kg/kg, RH about 93 %; (b) total 79.2 kW (22.5 TR), sensible 38.8 kW, latent 40.4 kW; (c) SHF = 0.49; (d) 57.6 kg/h.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the leaving state found from ADP and BPF?
   A. t2 = tA + BPF (t1 - tA) = 13.5 °C, and similarly for omega
   B. t2 = tA
   C. t2 = t1 - BPF x tA
   D. t2 = t1 x (1 - BPF)
   Answer: A. BPF = (t2 - tA)/(t1 - tA), so t2 = 8 + 0.25 x 22 = 13.5 °C; omega2 is found the same way. (Key point: Leaving air state)

2. The sensible load is computed using the intermediate state h' at:
   A. (t1, omega2)
   B. (t2, omega1)
   C. (tA, omega_A)
   D. (t2, omega_A)
   Answer: B. Cooling at constant omega1 down to t2 is the sensible part: Q_s = m_a(h1 - h'); the rest, h' to h2, is latent. (Key point: Sensible vs latent split)

3. Which is correct?
   A. Q_total = m_a(h1 - h2) = 79.2 kW and SHF = 38.8/79.2 = 0.49
   B. Q_total = m_a(h1 - hA) and SHF = 2
   C. SHF = latent / total = 0.51
   D. Q_total is the sensible load only
   Answer: A. Total load uses h1 - h2 (2.271 x 34.86 = 79.2 kW); SHF is sensible over total. (Key point: Total load and SHF)

4. The condensate rate is obtained from:
   A. m_a (h1 - h2)
   B. m_a (omega1 - omega2) = 0.016 kg/s (57.6 kg/h)
   C. m_a (t1 - t2)
   D. m_a omega_A
   Answer: B. Water removed equals the drop in specific humidity times dry air flow. (Key point: Condensate removal)

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