22-Mec-A1 Applied Thermodynamics and Heat Transfer · December 2018
Question 8 of 8: Shell-and-tube water-to-air heat exchanger
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Examination 16-Mec-A1, 3 hours, open book. Eight questions of equal value: Part A — Thermodynamics (Q1–Q4) and Part B — Heat Transfer (Q5–Q8). A complete paper is any five questions (three from one part and two from the other).
Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — ideal-gas mixtures, the air-standard Otto cycle, wet-region steam properties, the throttling calorimeter, the steady-flow energy equation, the regenerative gas-turbine (Brayton) cycle and vapour-compression refrigeration; Çengel & Ghajar, Heat and Mass Transfer (6th ed.) and Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — radial conduction through composite cylinders, conduction with internal heat generation, internal-flow convection with a constant surrounding-fluid temperature, and the effectiveness–NTU method for shell-and-tube exchangers. Steam properties are IAPWS-consistent (equivalent to the steam tables); ammonia properties are read from the saturated- and superheated-ammonia tables appended to the examination; air and combustion gases are treated as ideal gases with constant specific heats ($\gamma=1.4$, $R=0.287\ \text{kJ/kg}\cdot\text{K}$, $c_p=1.005\ \text{kJ/kg}\cdot\text{K}$).
Given. 50 parallel brass tubes, $d_i=2.3$ cm, $d_o=2.6$ cm, $L=6.7$ m each. Water (in tubes) $\dot m_w=10$ kg/s at 75 °C; air (in shell) $\dot m_a=1.6$ kg/s at 15 °C. $\bar h_i=470$ (water side), $\bar h_o=210\ \text{W/m}^2\text{°C}$ (air side). $c_{p,w}=4180$, $c_{p,a}=1007\ \text{J/kg°C}$; brass wall neglected.
Find. the effectiveness, the heat-transfer rate, and both outlet temperatures.
Figure 8 — Shell-and-tube arrangement: water through 50 parallel tubes, air over the bundle in the shell. Air (the smaller heat-capacity stream) is $C_\text{min}$, so it undergoes the larger temperature swing.
Approach. Compute the two capacity rates to identify $C_\text{min}$; build $UA$ from the inside and outside surface conductances; get NTU and the one-shell-pass effectiveness; then the duty and both outlets.
Capacity rates.
$$C_w=\dot m_w c_{p,w}=10(4180)=41{,}800,\qquad C_a=\dot m_a c_{p,a}=1.6(1007)=1611\ \text{W/°C}$$
Air is $C_\text{min}=1611$ W/°C; $C_r=C_\text{min}/C_\text{max}=1611/41800=0.0385$.
Surface areas and $UA$. $A_i=50\pi d_iL=50\pi(0.023)(6.7)=24.2\ \text{m}^2$; $A_o=50\pi d_oL=27.4\ \text{m}^2$. Neglecting the brass wall,
$$UA=\left(\frac{1}{\bar h_iA_i}+\frac{1}{\bar h_oA_o}\right)^{-1}=\left(\frac{1}{470(24.2)}+\frac{1}{210(27.4)}\right)^{-1}=3818\ \text{W/°C}$$
NTU and effectiveness (one shell pass). $\text{NTU}=UA/C_\text{min}=3818/1611=2.37$. With $\sqrt{1+C_r^2}=1.0007$,
$$\varepsilon=\frac{2}{(1+C_r)+\sqrt{1+C_r^{2}}\;\dfrac{1+e^{-\text{NTU}\sqrt{1+C_r^{2}}}}{1-e^{-\text{NTU}\sqrt{1+C_r^{2}}}}}=0.891$$
$\varepsilon=0.89$
(Because $C_r\approx0$, this is essentially $\varepsilon=1-e^{-\text{NTU}}=0.906$; the shell-pass form trims it slightly to 0.891.)
Heat-transfer rate and outlet temperatures.
$$\dot Q=\varepsilon\,C_\text{min}(T_{w,i}-T_{a,i})=0.891(1611)(75-15)=8.61\times10^{4}\ \text{W}=86.1\ \text{kW}$$
$$T_{a,o}=15+\frac{\dot Q}{C_a}=15+\frac{86{,}100}{1611}=68.4\ ^\circ\text{C},\qquad T_{w,o}=75-\frac{\dot Q}{C_w}=75-\frac{86{,}100}{41{,}800}=72.9\ ^\circ\text{C}$$
$T_{a,o}=68.4$ °C, $\;T_{w,o}=72.9$ °C
