---
title: "Sensible heating of air numerical"
subject: "Applied Thermodynamics"
chapter: 4
key: "semester_3/applied_thermodynamics:sensible-heating-load-numerical"
source: https://www.bctengineer.com/q/semester_3/applied_thermodynamics/sensible-heating-load-numerical
site: BCT Engineering
---

# Sensible heating of air numerical

Applied Thermodynamics · Chapter 4: Air-Conditioning

- Asked in: year not given, 6 marks
- Video: [Air Conditioning Processes Part 1 Sensible Cooling and Sensible Heating](https://www.youtube.com/watch?v=fX7co_kHbFY) — Anastacio G. Pantaleon, Jr.
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_thermodynamics/sensible-heating-load-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_thermodynamics/chapter-4

## Question

Air at 15 °C and 70% relative humidity flows at 200 m³/min into a heating coil and leaves at 30 °C. Find (a) the mass flow rate of dry air, (b) the heat transfer rate in the coil, and (c) the relative humidity of the air leaving the coil. Take $p = 101.325$ kPa; saturation pressure at 15 °C is 1.706 kPa and at 30 °C is 4.247 kPa.

## Short answer

With constant ω = 0.00742 kg/kg, v₁ = 0.8259 m³/kg gives ṁ_a = 200/0.8259 = 242 kg/min. Enthalpies: h₁ = 33.84 kJ/kg, h₂ = 49.13 kJ/kg. Heat transfer Q = (242/60)×(49.13−33.84) = 61.7 kW. Leaving RH = p_v/p_s,₃₀ = 1.194/4.247 = 0.281 (28%).

## Answer

Sensible heating means $\omega$ is constant.

#### Inlet state
$$
p_{v1} = 0.70 \times 1.706 = 1.194\ \text{kPa}
$$

$$
\omega_1 = 0.622\,\frac{1.194}{101.325 - 1.194} = 0.00742\ \text{kg/kg}
$$

$$
v_1 = \frac{0.287 \times 288.15}{101.325 - 1.194} = 0.8259\ \text{m}^3/\text{kg dry air}
$$

Enthalpies, $h = 1.005\,t + \omega(2501 + 1.88\,t)$:

$$
\begin{aligned}
h_1 &= 1.005 \times 15 + 0.00742(2501 + 28.2) = 33.84\ \text{kJ/kg}\\
h_2 &= 1.005 \times 30 + 0.00742(2501 + 56.4) = 49.13\ \text{kJ/kg}
\end{aligned}
$$

#### (a) Mass flow of dry air
$$
\dot m_a = \frac{\dot V_1}{v_1} = \frac{200}{0.8259} = 242.2\ \text{kg/min} = 4.04\ \text{kg/s}
$$

#### (b) Heat transfer
$$
Q = \dot m_a (h_2 - h_1) = \frac{242.2 \times (49.13 - 33.84)}{60} = 61.7\ \text{kW}
$$

#### (c) Final relative humidity
$$
\phi_2 = \frac{p_v}{p_{s,30}} = \frac{1.194}{4.247} = 0.281
$$

**Answer:** (a) 242 kg/min dry air; (b) about 61.7 kW; (c) 28 %.

## Check yourself

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

1. What is the key property used in sensible heating?
   A. RH stays constant
   B. Specific humidity omega stays constant, so omega2 = omega1
   C. DBT stays constant
   D. Enthalpy stays constant
   Answer: B. No moisture is added or removed, so omega is constant; this lets you evaluate h2 and RH2 at the same omega. (Key point: Constant omega)

2. How is the dry air mass flow found from 200 m^3/min?
   A. Multiply by v1 (0.8259)
   B. Divide by v1 = 0.8259 m^3/kg, giving 242 kg/min
   C. Multiply by the density of dry air at 30 °C
   D. Divide by omega
   Answer: B. Mass flow = volume flow / specific volume, using v = R_a T/(p - p_v) at the inlet. (Key point: Dry air mass flow)

3. The coil heat rate is given by:
   A. m_a (omega2 - omega1)
   B. m_a (h2 - h1) = 242.2 x 15.29 / 60 = 61.7 kW
   C. m_a x (t2 - t1) with no specific heat
   D. m_a h1
   Answer: B. Q equals dry air mass flow times the enthalpy rise (33.83 to 49.12 kJ/kg), about 61.7 kW. (Key point: Heat transfer rate)

4. Why does RH fall from 70% to about 28%?
   A. Vapour was removed
   B. p_v stays 1.194 kPa while p_s rises from 1.706 to 4.247 kPa at 30 °C
   C. p_v rises to 4.247 kPa
   D. Total pressure dropped
   Answer: B. The vapour pressure is unchanged, but warmer air has a much higher saturation pressure: 1.194/4.247 = 0.281. (Key point: Final relative humidity)

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