Water/Air Tubular Heat Exchanger — Effectiveness, Duty, Outlet Temperatures: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, December 2018, Q8

Question 8: Water/Air Tubular Heat Exchanger — Effectiveness, Duty, Outlet Temperatures

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. Fifty parallel brass tubes ($d_i=2.3\text{ cm}$, $d_o=2.6\text{ cm}$, $L=6.7\text{ m}$ each) carry water ($\dot m_w=10\text{ kg/s}$, entering at 75°C) inside a tube-bundle exchanger; air ($\dot m_a=1.6\text{ kg/s}$, entering at 15°C) flows over the outside of the bundle. $\bar h_i=470$, $\bar h_o=210\text{ W/m}^2\cdot{}^{\circ}\text{C}$.

Given data
Quantity	Symbol	Value
Number / length of tubes	$N,L$	50 / 6.7 m
Tube ID / OD	$d_i,d_o$	2.3 cm / 2.6 cm
Inside / outside coefficients	$\bar h_i,\bar h_o$	470 / 210 W/m²·°C
Water flow / inlet temperature	$\dot m_w,T_{w,in}$	10 kg/s, 75°C
Air flow / inlet temperature	$\dot m_a,T_{a,in}$	1.6 kg/s, 15°C

Find. The exchanger effectiveness $\varepsilon$; the heat transfer rate $\dot Q$; the water and air outlet temperatures.

Fifty parallel brass tubes carry hot water; air crosses the tube bundle inside the surrounding casing and is heated.

Check: no brass thermal conductivity is given, so the (thin, high-conductivity) tube-wall conduction resistance is neglected next to the two convective resistances — standard for a metal tube this thin.

Approach. Build the overall $UA$ from the inside and outside convective resistances (areas based on $d_i$ and $d_o$ respectively, $N$ tubes in parallel); find $C_{min},C_{max},C_r=C_{min}/C_{max}$ and $NTU=UA/C_{min}$; since $C_r$ turns out to be very small here, the effectiveness–NTU relation collapses to the single-stream limit regardless of the exact flow arrangement (crossflow/counterflow all agree to within about 1% at this $C_r$), so $\varepsilon=1-e^{-NTU}$ is used directly.

Heat transfer areas and overall $UA$. $$A_i=\pi d_iLN=\pi\times0.023\times6.7\times50=24.21\text{ m}^2$$ $$A_o=\pi d_oLN=\pi\times0.026\times6.7\times50=27.36\text{ m}^2$$ $$\frac{1}{UA}=\frac{1}{\bar h_iA_i}+\frac{1}{\bar h_oA_o}=\frac{1}{470\times24.21}+\frac{1}{210\times27.36}$$ $$\boxed{UA=3818\text{ W/}{}^{\circ}\text{C}}$$
Capacity rates and $NTU$. $$C_w=\dot m_wc_{p,w}=10\times4186=41{,}860\text{ W/K},\qquad C_a=\dot m_ac_{p,a}=1.6\times1005=1608\text{ W/K}$$ $$C_{min}=C_a=1608\text{ W/K},\qquad C_r=\frac{C_{min}}{C_{max}}=\frac{1608}{41{,}860}=0.0384$$ $$NTU=\frac{UA}{C_{min}}=\frac{3818}{1608}$$ $$\boxed{NTU=2.374}$$
Effectiveness. With $C_r\approx0.038\ll1$, the water side is so much larger in capacity rate that it barely cools as it gives up heat — effectively a constant-temperature source, so every standard flow arrangement (counterflow, crossflow, either fluid mixed) converges to the same single-stream result to within about 1%: $$\varepsilon=1-e^{-NTU}=1-e^{-2.374}$$ $$\boxed{\varepsilon=0.907=90.7\%}$$ (An exact counterflow calculation at $C_r=0.0384$ gives $\varepsilon=0.902$ — a 0.5-point difference, confirming the arrangement is nearly irrelevant here.)
Heat transfer rate and outlet temperatures. $$\dot Q_{max}=C_{min}(T_{w,in}-T_{a,in})=1608\times(75-15)=96{,}480\text{ W}$$ $$\dot Q=\varepsilon\,\dot Q_{max}=0.907\times96{,}480$$ $$\boxed{\dot Q=87.5\text{ kW}}$$ $$T_{a,out}=T_{a,in}+\frac{\dot Q}{C_a}=15+\frac{87{,}500}{1608}$$ $$\boxed{T_{a,out}=69.4\,{}^{\circ}\text{C}}$$ $$T_{w,out}=T_{w,in}-\frac{\dot Q}{C_w}=75-\frac{87{,}500}{41{,}860}$$ $$\boxed{T_{w,out}=72.9\,{}^{\circ}\text{C}}$$

Question 8 — results
Quantity	Value
Overall $UA$	3818 W/°C
$NTU$ (based on $C_{min}$ = air side)	2.374
Effectiveness $\varepsilon$	90.7%
Heat transfer rate $\dot Q$	87.5 kW
Air outlet temperature	69.4°C
Water outlet temperature	72.9°C

Topic. Effectiveness–NTU analysis of a tube-bundle (tubular) heat exchanger with very unequal capacity rates.

Key relations. $UA$ from parallel convective resistances, $1/UA=1/(\bar h_iA_i)+1/(\bar h_oA_o)$; $NTU=UA/C_{min}$, $C_r=C_{min}/C_{max}$; single-stream limit $\varepsilon\to1-e^{-NTU}$ as $C_r\to0$.

Why this works. Water's capacity rate here (41.9 kW/K) is about 26× air's (1.6 kW/K), so $C_r\approx0.038$ is close enough to zero that the exchanger behaves almost exactly like air being heated against an effectively CONSTANT-temperature water bath — the same "infinite reservoir" idea used in Question 7, except here it is DERIVED from the given flow rates rather than stated outright, and the small residual $C_r$ is what makes the flow-arrangement choice (crossflow vs. counterflow) a minor, sub-1% correction rather than a first-order effect.

Common pitfall. Basing $UA$ on a single area (using $A_i$ for BOTH convective terms, or vice versa) instead of the correct area for each side; identifying $C_{min}$ as the water side by mistake (water has the larger $\dot m$, but air has the much smaller $\dot mc_p$ because its specific heat and flow rate are both far smaller).

Source. F. P. Incropera and D. P. DeWitt, Fundamentals of Heat and Mass Transfer, 7th ed., Ch. 11 (effectiveness–NTU method, overall heat transfer coefficient for tube banks).

Question 8 of 8: Water/Air Tubular Heat Exchanger — Effectiveness, Duty, Outlet Temperatures

Question 8: Water/Air Tubular Heat Exchanger — Effectiveness, Duty, Outlet Temperatures