NivaarExam PrepOfficial exam papers ↗

22-Mec-A1 Applied Thermodynamics and Heat Transfer · December 2014

Question 6 of 8: Water Warmed Flowing Through a Duct in Warm Air

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Open-book, 3-hour paper. Part A (Thermodynamics, Q1–Q4) and Part B (Heat Transfer, Q5–Q8); the rubric grades any five (three from one part, two from the other), all of equal value. All eight questions are solved in full. Freon-12 property values are read from the saturated and superheated tables printed in the exam appendix (pages 4–5). Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed.); Çengel & Ghajar, Heat and Mass Transfer (6th ed.); Incropera et al., Fundamentals of Heat and Mass Transfer (8th ed.).

Question 6: Water Warmed Flowing Through a Duct in Warm Air (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Square duct $0.20\times0.20\ \text{m}$; water in $20\ ^\circ\text{C}$ at $V=0.05\ \text{m/s}$ ($\rho=998$, $c_p=4182$); air $T_\infty=50\ ^\circ\text{C}$; outer $h=150\ \text{W/m}^2\text{}\cdot\text{K}$; length $L=1\ \text{m}$.

Find. Heat gained over 1 m and the water exit temperature.

water 5 cm/s20°C≈20.4°Cambient air 50°C, hₒ=150 W/m²Ksquare duct 20 cm × 20 cm, length L = 1 m
Figure 7 — Warm air (50 °C) heats water flowing through the square duct; the outer film $h_o=150$ controls the transfer.

Approach. Take the outer film as the controlling resistance ($U\approx h_o$, water-side resistance negligible), find the wetted area and water capacity rate, and use the single-stream exponential approach toward the constant air temperature.

  1. Mass flow. $\dot m=\rho V A_\text{cross}=998(0.05)(0.20^2)=\boxed{1.996\ \text{kg/s}}$; $\dot mc_p=8347\ \text{W/K}$.
  2. Surface area. Perimeter $P=4(0.20)=0.80\ \text{m}$; $A_s=PL=0.80\ \text{m}^2$; $UA_s=150(0.80)=120\ \text{W/K}$.
  3. Exit temperature. $\dfrac{T_\infty-T_o}{T_\infty-T_i}=e^{-UA_s/\dot mc_p}=e^{-0.01438}=0.9857$; $T_o=50-30(0.9857)=\boxed{20.4\ ^\circ\text{C}}$.
  4. Heat gained. $\dot Q=\dot mc_p(T_o-T_i)=8347(0.428)=\boxed{3.57\ \text{kW}}$.

With nearly 2 kg/s of water passing through only 1 m of duct, the residence time is tiny, so the water temperature barely moves — it rises just 0.4 °C and picks up about 3.6 kW. Over a much longer run the water would asymptote toward the 50 °C air, but here the huge thermal mass of the stream dominates.

Check: The internal water-side film is taken as negligible against the given outer $h_o=150$; because the water coefficient is on the order of $10^3$–$10^4\ \text{W/m}^2\text{K}$, this changes $U$ (and $\dot Q$) by only a few percent.
QuantityResult
Water mass flow≈ 1.996 kg/s
Wetted area (1 m)0.80 m²
Water exit temperature≈ 20.4 °C
Heat gained over 1 m≈ 3.57 kW