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

Question 7 of 8: Refrigeration Load on an Indoor Ice Rink

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 7: Refrigeration Load on an Indoor Ice Rink (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. $A_\text{ice}=35\times20=700\ \text{m}^2$; $T_\text{ice}=-8\ ^\circ\text{C}$, $T_\text{air}=10\ ^\circ\text{C}$, $T_\text{wall}=16\ ^\circ\text{C}$; $\varepsilon_\text{ice}=0.97$, $\varepsilon_\text{wall}=0.8$; $h=50\ \text{W/m}^2\text{}\cdot\text{K}$; enclosure area (walls + ceiling) $A_\text{surr}=6300\ \text{m}^2$.

Find. The heat-removal rate at the pad.

ice surface, 700 m², −8°Cbuilding enclosure — walls 16°C, air 10°Cconv (air→ice)rad (walls→ice)
Figure 8 — Two parallel gains into the ice: convection from the 10 °C air and net radiation from the 16 °C enclosure. Their sum is the pad refrigeration load.

Approach. The pad must remove whatever the ice surface gains at steady state — convection from the air plus net radiation from the enclosure walls — so evaluate each and add.

  1. Convection from air. $\dot Q_\text{conv}=hA_\text{ice}(T_\text{air}-T_\text{ice})=50(700)(10-(-8))=\boxed{630\ \text{kW}}$.
  2. Radiation resistances. Two-surface enclosure: $\dfrac{1-\varepsilon_\text{ice}}{\varepsilon_\text{ice}A_\text{ice}}+\dfrac{1}{A_\text{ice}}+\dfrac{1-\varepsilon_\text{wall}}{\varepsilon_\text{wall}A_\text{surr}}=1.512\times10^{-3}\ \text{m}^{-2}$.
  3. Net radiation. With $T_\text{wall}=289.15\ \text{K}$, $T_\text{ice}=265.15\ \text{K}$: $\dot Q_\text{rad}=\dfrac{\sigma(T_\text{wall}^4-T_\text{ice}^4)}{\sum R}=\dfrac{5.67\times10^{-8}(2.048\times10^{9})}{1.512\times10^{-3}}=\boxed{76.8\ \text{kW}}$.
  4. Total pad load. $\dot Q=\dot Q_\text{conv}+\dot Q_\text{rad}=630+76.8=\boxed{\approx707\ \text{kW}}$.

Convection dominates because the 18 °C air-to-ice gap acts over the full 700 m² at a healthy coefficient, delivering 630 kW; radiation from the warmer walls adds another 77 kW even though the surface-to-surface temperature difference is small, because the high ice emissivity and large view factor keep the space resistance low. The refrigeration plant on the pad must reject about 0.7 MW just to hold the ice — a reminder that indoor rinks are energy-intensive.

PathHeat gain
Convection (air → ice)≈ 630 kW
Radiation (walls → ice)≈ 76.8 kW
Total pad refrigeration load≈ 707 kW