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Sensible heating of air numerical

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%).

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=101.325p = 101.325 kPa; saturation pressure at 15 °C is 1.706 kPa and at 30 °C is 4.247 kPa.

Sensible heating means ω\omega is constant.

Inlet state

pv1=0.70×1.706=1.194 kPap_{v1} = 0.70 \times 1.706 = 1.194\ \text{kPa} ω1=0.622 1.194101.325−1.194=0.00742 kg/kg\omega_1 = 0.622\,\frac{1.194}{101.325 - 1.194} = 0.00742\ \text{kg/kg} v1=0.287×288.15101.325−1.194=0.8259 m3/kg dry airv_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+ω(2501+1.88 t)h = 1.005\,t + \omega(2501 + 1.88\,t):

h1=1.005×15+0.00742(2501+28.2)=33.84 kJ/kgh2=1.005×30+0.00742(2501+56.4)=49.13 kJ/kg\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

m˙a=V˙1v1=2000.8259=242.2 kg/min=4.04 kg/s\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=m˙a(h2−h1)=242.2×(49.13−33.84)60=61.7 kWQ = \dot m_a (h_2 - h_1) = \frac{242.2 \times (49.13 - 33.84)}{60} = 61.7\ \text{kW}

(c) Final relative humidity

ϕ2=pvps,30=1.1944.247=0.281\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 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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