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25-Nav-A1 Fundamentals of Naval Architecture · May-98-Mar-A1 2017

Question 8 of 8: Crossflow Finned-Tube Heat Exchanger

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

Notes on this paper

National Examinations, May 2017 — 98-Mar-A1 Applied Thermodynamics and Heat Transfer, 3 hours, open book (Part A: Thermodynamics, Part B: Heat Transfer; 5 of 8 questions required, all 8 answered below for full study coverage).

Reference texts: Cengel & Boles, Thermodynamics: An Engineering Approach; Sonntag, Borgnakke & Van Wylen, Fundamentals of Thermodynamics; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer.

It is solved as the thermodynamics/heat-transfer exam it actually is.

Question 8: Crossflow Finned-Tube Heat Exchanger (20 marks)

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.

Given data
QuantitySymbolValue
Air mass flow, in/out temperature$\dot m_a,T_{a,i}/T_{a,o}$0.80 kg/s, 30°C→7°C
Water mass flow, inlet temperature$\dot m_w,T_{w,i}$0.76 kg/s, 3°C
Overall coefficient$U$55 W/m²°C
Specific heats$c_{p,a},c_{p,w}$1005, 4180 J/kg·K
water 0.76 kg/s, 3°C in ↓air 0.80 kg/s, 30°C→7°Ccrossflow finned-tube exchanger, both fluids unmixed, U=55 W/m²°C
Crossflow finned-tube exchanger: water through the tubes, air across the finned bank.

Find. Required heat-exchanger area; then the % reduction in heat-transfer rate if $\dot m_w$ is halved.

Approach. Duty from the air-side energy balance fixes $\dot Q$ and hence effectiveness $\varepsilon$; solve the both-fluids-unmixed crossflow $\varepsilon$–NTU relation for NTU and thus $A$; repeat with the new $C_r$ at the same $UA$ to find the new duty.

  1. Heat duty and water outlet temperature. $$\dot Q=\dot m_ac_{p,a}(T_{a,i}-T_{a,o})=0.80(1005)(23)=18{,}490\text{ W}$$ $$T_{w,o}=T_{w,i}+\dfrac{\dot Q}{\dot m_wc_{p,w}}=3+\dfrac{18{,}490}{0.76(4180)}=8.82\,{}^{\circ}\text{C}$$
  2. Capacity rates and effectiveness. $C_a=\dot m_ac_{p,a}=804\text{ W/K}$, $C_w=\dot m_wc_{p,w}=3177\text{ W/K}$; since $C_a
  3. Solve for NTU (both fluids unmixed). Numerically solving $$\varepsilon=1-\exp\!\left\{\dfrac{1}{C_r}\text{NTU}^{0.22}\left[e^{-C_r\,\text{NTU}^{0.78}}-1\right]\right\}=0.852$$ gives $\text{NTU}\approx2.43$.
  4. Required area. $$\boxed{A=\dfrac{\text{NTU}\cdot C_{min}}{U}=\dfrac{2.43(804)}{55}=35.6\text{ m}^2}$$
  5. Water flow halved ($\dot m_w=0.38$ kg/s) — new duty at the same $UA$. $C_{min}$ is still $C_a=804\text{ W/K}$ (unchanged) and $UA$ is unchanged, so $\text{NTU}=UA/C_{min}=2.43$ stays the same, but $C_{max}$ drops to $C_{w,new}=1588\text{ W/K}$, so $C_{r,new}=804/1588=0.506$. Re-solving the same $\varepsilon$–NTU relation with the new $C_r$ gives $\varepsilon_{new}=0.783$, so $$\dot Q_{new}=\varepsilon_{new}C_{min}(T_{a,i}-T_{w,i})=0.783(804)(27)=17{,}010\text{ W}$$
  6. Percentage reduction. $$\boxed{\dfrac{\dot Q-\dot Q_{new}}{\dot Q}=\dfrac{18{,}490-17{,}010}{18{,}490}=8.0\%}$$
Question 8 — final results
QuantityValue
Heat duty$\dot Q = 18.49\text{ kW}$
Water outlet temperature$T_{w,o} = 8.82\,{}^{\circ}\text{C}$
Required heat-exchanger area$A = 35.6\text{ m}^2$
New duty, water flow halved$\dot Q_{new} = 17.01\text{ kW}$
Reduction in heat-transfer rate$8.0\%$
Check: modelled as a both-fluids-unmixed crossflow exchanger (the standard default for a finned-tube air/water HX when the flow-arrangement detail isn't given in the source figure) using the closed-form $\varepsilon$–NTU correlation rather than a graphical LMTD correction-factor chart.
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