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

Question 8 of 8: Maximum Oil Flow in a Tube-in-Sh​ell Cooler

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 8: Maximum Oil Flow in a Tube-in-Sh​ell Cooler (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. Water $\dot m_w=230\ \text{kg/hr}=0.0639\ \text{kg/s}$, in $35\ ^\circ\text{C}$, exit $\le99\ ^\circ\text{C}$ ($c_p=4180$); oil in $120\ ^\circ\text{C}$, $c_p=2100\ \text{J/kg}\cdot\text{K}$; $A=1.4\ \text{m}^2$, $U=280\ \text{W/m}^2\text{}\cdot\text{K}$.

Find. Maximum oil mass flowrate that can be cooled.

tube-in-sh​ell (counterflow)A=1.4 m², U=280oil 120°C≈113.5°Cwater 35°Cwater ≤99°C
Figure 9 — The water (small capacity rate) is $C_\text{min}$; capping its exit at 99 °C caps the duty, which sets the largest oil flow the unit can cool.

Approach. The 99 °C water-exit cap fixes the maximum duty. The water is the minimum-capacity stream, so pin the duty at the 99 °C limit, find the effectiveness and NTU, then solve the counterflow relation for the oil capacity rate and hence its flow.

  1. Water capacity and limiting duty. $C_w=\dot m_wc_p=0.0639(4180)=267\ \text{W/K}$; at the 99 °C cap $\dot Q=C_w(99-35)=\boxed{17.1\ \text{kW}}$.
  2. Effectiveness and NTU. Water is $C_\text{min}$, so $\dot Q_\text{max}=C_w(120-35)=22.7\ \text{kW}$; $\varepsilon=\dfrac{17.1}{22.7}=0.753$; $\text{NTU}=\dfrac{UA}{C_w}=\dfrac{392}{267}=1.47$.
  3. Solve for the capacity ratio. Counterflow $\varepsilon=\dfrac{1-e^{-\text{NTU}(1-C_r)}}{1-C_re^{-\text{NTU}(1-C_r)}}$ is met at $C_r=C_w/C_o=0.102$, so $C_o=267/0.102=2619\ \text{W/K}$.
  4. Oil flow. $\dot m_o=\dfrac{C_o}{c_{p,\text{oil}}}=\dfrac{2619}{2100}=\boxed{1.25\ \text{kg/s}}$ (≈ 4490 kg/hr); oil leaves at $120-\dot Q/C_o=\boxed{113.5\ ^\circ\text{C}}$.

Raising the oil flow beyond 1.25 kg/s would force the water outlet above the 99 °C limit (a larger oil capacity rate pushes the effectiveness higher, so the water absorbs more and boils toward saturation), so this is the ceiling. At the limit the oil is barely cooled — only 6.5 °C — because the tiny 230 kg/hr water stream simply cannot carry away more heat.

Check: A counterflow arrangement is assumed. A 1-sh​ell/2-tube pass unit would need an $F$-correction (~0.9), lowering the effective duty and the maximum oil flow by roughly 10 %; the water-cap logic is unchanged.
QuantityResult
Limiting duty (water at 99 °C)≈ 17.1 kW
Effectiveness / NTU0.753 / 1.47
Capacity ratio $C_r$≈ 0.102
Maximum oil flow≈ 1.25 kg/s (≈ 4490 kg/hr)
Oil exit temperature≈ 113.5 °C
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