NivaarExam PrepOfficial exam papers ↗

22-Mec-A1 Applied Thermodynamics and Heat Transfer · Undated paper

Question 6 of 8: Ice-storage cooling tank

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

Notes on this paper

Paper format: National Examination 16-Mec-A1 Applied Thermodynamics and Heat Transfer — 3 hours, open book. Two parts: Part A (Thermodynamics, Q1–Q4) and Part B (Heat Transfer, Q5–Q8). The rubric asks for five questions (three from one part, two from the other), each of equal value; all eight are solved.

Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed.); Çengel & Ghajar, Heat and Mass Transfer (6th ed.); Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (8th ed.); Moran & Shapiro, Fundamentals of Engineering Thermodynamics. Property data: saturated ammonia and R-134a tables printed with the paper.

Question 6 — Ice-storage cooling tank (Part B, 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.

QuantitySymbolValue
Air flow$\dot m_a$0.5 kg/s
Air cooled from / to$T_\text{in},T_\text{out}$24 → 14 °C
Air specific heat$c_{p,a}$1.005 kJ/kg·K
Latent heat of fusion$h_{sf}$334 kJ/kg
Operating period$t$8 h
Initial ice fraction—80 % by mass

Find. (a) cooling rate $\dot Q$; (b) daily ice melt $m_\text{ice}$; (c) tank volume $V$.

Approach. Size the sensible cooling duty on the air stream, integrate it over the day into a latent-heat demand, then convert the required ice mass into a stored-water volume through the 80 % ice fraction.

  1. Cooling duty on the air stream. $$\dot Q=\dot m_a\,c_{p,a}\,(T_\text{in}-T_\text{out})=0.5(1.005)(24-14)=\boxed{5.03\ \text{kW}}$$
  2. Daily energy and ice melted. Over 8 h $=28\,800$ s the energy absorbed is $E=\dot Q\,t=5.025(28\,800)=1.447\times10^{5}$ kJ, so the melted mass is $$m_\text{ice}=\frac{E}{h_{sf}}=\frac{144{,}720}{334}=\boxed{433\ \text{kg}}$$
  3. Minimum tank volume. The tank must contain at least this much ice; at 80 % ice by mass the stored-water mass is $m_w=m_\text{ice}/0.80=542$ kg, giving $$V=\frac{m_w}{\rho_w}=\frac{541.6}{1000}=\boxed{0.542\ \text{m}^3}$$
QuantityResult
Cooling duty $\dot Q$5.03 kW
Daily cooling energy144.7 MJ
Ice melted per day433 kg
Minimum tank volume0.542 m³
Check — heavily reconstructed problem.
Question 6 is only partly specified: page 4 gives only disconnected fragments ("heat pump ... large tank of water ... energy source in winter", "air ... 0.01 kg/s", "$T_e=-14$ °C", "80 % ice ... $h_{sf}=334$ kJ/kg"). The governing idea — a phase-change (latent-heat) thermal store sized from an integrated load — is unambiguous, so a coherent, self-consistent ice-storage problem is posed here with a representative air flow (the fragmentary 0.01 kg/s gives a physically trivial ~11-litre tank). The method (sensible load → latent demand → stored-water volume) is what is graded.