25-Nav-A1 Fundamentals of Naval Architecture · May-98-Mar-A1 2016

Question 8 of 8: Cross-Flow Water/Air Tube-Bank Heat Exchanger

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

Notes on this paper

National Examinations, May 2016 — 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: Cross-Flow Water/Air Tube-Bank Heat Exchanger

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. 40 tubes of 1 cm diameter, each spanning the 1 m width of a 1 m × 1 m duct (so each tube's length is 1 m), water inside and hot air in cross-flow outside.

Given data
Quantity	Symbol	Value
Tube count, diameter, length	$N,\,D,\,L$	40, 1 cm, 1 m
Water: $c_p$, inlet $T$, velocity	$c_{p,w},\,T_{w,i},\,V_w$	4180 J/kg·K, 18°C, 3 m/s
Air: $c_p$, inlet $T$, $P$, velocity	$c_{p,a},\,T_{a,i},\,P,\,V_a$	1010 J/kg·K, 130°C, 105 kPa, 12 m/s
Overall coefficient	$U$	80 W/m²·K

Find. Water and air outlet temperatures, and the total rate of heat transfer.

Approach. Get both mass flow rates from the given velocities and flow areas, form $C_{min},C_{max}$, compute $NTU=UA_s/C_{min}$ with $A_s=N\pi DL$, apply the crossflow (both fluids unmixed) effectiveness correlation, then back out $Q$ and both outlet temperatures.

Mass flow rates. Water fills the 40 tube bores: $\dot m_w=\rho_wV_w\!\left(N\dfrac{\pi}{4}D^2\right)=998(3)(40)(7.854\times10^{-5})\approx9.41\text{ kg/s}$. Air density at 130°C, 105 kPa (ideal gas): $\rho_a=P/(RT)=105{,}000/(287\times403.15)\approx0.908\text{ kg/m}^3$; the 40 tubes block a frontal area $N\,D\,L=0.40\text{ m}^2$ of the 1 m² duct, leaving a flow area of $0.60\text{ m}^2$: $$\dot m_a=\rho_aV_a(1-ND L)=0.908(12)(0.60)\approx6.53\text{ kg/s}$$
Heat capacity rates and NTU. $C_w=\dot m_wc_{p,w}=39{,}320\text{ W/K}$, $C_a=\dot m_ac_{p,a}=6600\text{ W/K}$ (air is the minimum-capacity fluid), $C_r=C_{min}/C_{max}=0.168$. Surface area $A_s=N\pi DL=1.257\text{ m}^2$, so $$NTU=\frac{UA_s}{C_{min}}=\frac{80(1.257)}{6600}\approx0.01523$$
Effectiveness and heat rate. For crossflow with both fluids unmixed at this very small NTU, $\varepsilon\approx NTU\approx0.01507$ (the exchanger is lightly loaded — only 40 small tubes across a 1 m² duct): $$Q=\varepsilon C_{min}(T_{a,i}-T_{w,i})=0.01507(6600)(130-18)$$ $$\boxed{Q\approx11.14\text{ kW}}$$
Outlet temperatures. $$T_{w,o}=T_{w,i}+\frac{Q}{C_w}=18+\frac{11{,}139}{39{,}320}\approx\boxed{18.28^\circ\text{C}}$$ $$T_{a,o}=T_{a,i}-\frac{Q}{C_a}=130-\frac{11{,}139}{6600}\approx\boxed{128.3^\circ\text{C}}$$

Question 8 — final results
Quantity	Value
Water mass flow rate	$\dot m_w\approx9.41\text{ kg/s}$
Air mass flow rate	$\dot m_a\approx6.53\text{ kg/s}$
NTU	$\approx0.0152$
Rate of heat transfer	$Q\approx11.14\text{ kW}$
Water outlet temperature	$\approx18.28^\circ\text{C}$
Air outlet temperature	$\approx128.3^\circ\text{C}$

Topic: Effectiveness–NTU method for a lightly-loaded cross-flow tube-bank heat exchanger.

Key relations: $NTU=UA_s/C_{min}$; crossflow (both fluids unmixed) effectiveness correlation; $Q=\varepsilon C_{min}\Delta T_{max}$.

Why this works: With only 40 slender tubes spanning a comparatively large 1 m² duct and a modest $U=80\text{ W/m}^2\text{K}$, the exchanger surface area is small relative to both fluids' heat capacity rates, so $NTU\ll1$ and the effectiveness reduces to essentially $NTU$ itself — a useful check for any lightly-loaded exchanger, since $\varepsilon\to NTU$ as $NTU\to0$ regardless of flow arrangement or $C_r$.

Common pitfall: forgetting that the 40 tubes physically block part of the duct's cross-section, which reduces the air-side flow area (and hence changes $\dot m_a$) below the full 1 m²; using the wrong tube length (the tubes span the duct's 1 m width, not some separate unstated flow-direction length).

Source: Cengel & Ghajar, Heat and Mass Transfer, Ch.11/13 (effectiveness–NTU method for heat exchangers); Incropera & DeWitt, Ch.11.

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