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17-Phys-B6 Applied Thermodynamics and Heat Transfer · December 2017

Question 8 of 8: Heat Exchanger Design — Number of Parallel Tubes

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

Notes on this paper

Paper format. 98-Phys-B6 Applied Thermodynamics and Heat Transfer, National Examination December 2017 — a three-hour open-book examination; candidates are expected to bring both a thermodynamics text and a heat-transfer text to make use of the property tables. A complete examination is five questions — either three from Part A (Thermodynamics, Q1–Q4) and two from Part B (Heat Transfer, Q5–Q8), or two from Part A and three from Part B — every question carrying equal value; all eight are solved below as a complete study set. Candidates are invited to state any assumptions where a question is open to interpretation; this licence is used explicitly in Question 4 (the exam's own printed text is ambiguous about whether the piston displacement is 1.00 m³, read here as the intended value) and Question 6 (the external air stream is treated as an effectively infinite, constant-temperature reservoir since no air mass flow rate or duct is specified).

Reference texts. Y. A. Çengel and M. A. Boles, Thermodynamics: An Engineering Approach, 8th ed. (two-phase closed systems, flash chambers, steam turbines, vapour-compression refrigeration, reciprocating compressors); F. P. Incropera and D. P. DeWitt, Fundamentals of Heat and Mass Transfer, 7th ed. (composite cylindrical conduction, internal and external forced convection correlations, natural convection from a vertical plate, heat-exchanger LMTD analysis). Saturation and superheat property values below were computed (Bell et al., IAPWS-95 / REFPROP-quality equations of state for water, ammonia and R134a) and cross-checked against the printed appendix tables on pages 5–8 of the source exam, which they matched to 3–4 significant figures throughout.

Question 8: Heat Exchanger Design — Number of Parallel Tubes

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. A tube-bundle heat exchanger: steam condensing at a single saturation temperature on the outside of a bank of parallel tubes heats a chemical solution flowing inside them from $65\,{}^{\circ}\text{C}$ to $93\,{}^{\circ}\text{C}$; each tube's inside/outside convection coefficients and the tube-wall conductivity are given.

Given data
QuantitySymbolValue
Total heat duty$\dot Q$200 kW
Solution specific heat$c_p$3.26 kJ/kg·K
Solution inlet / outlet temperature$T_{in}/T_{out}$$65/93\,{}^{\circ}\text{C}$
Steam (condensing) pressure$P_{steam}$250 kPa
Tube outside / inside diameter$D_o/D_i$4.0 / 3.0 cm
Tube length (each)$L$3 m
Tube thermal conductivity$k_{tube}$111 W/m·K
Inside / outside convection coefficient$h_i/h_o$3400 / 7300 W/m²·K

Find. The number of parallel tubes $N$ needed to supply the 200 kW duty.

steam condensing at 250 kPa on tube bundle (casing)chemical solution in, 65°Cout, 93°CN parallel tubes, 4.0 cm OD × 3.0 cm ID × 3 m long, each carrying the flowN = ?
Tube-bundle exchanger: steam condenses at a single saturation temperature on the outside of $N$ parallel tubes; the chemical solution flows inside, one shared inlet/outlet header feeding all tubes equally.

Approach. Because the outer (steam) side is condensing, it stays at ONE saturation temperature along the whole tube length, so the usual four-temperature LMTD collapses to just the two terminal differences on the tube side. Build the overall $UA$ for a SINGLE tube from its three series resistances (inside convection, tube-wall conduction, outside convection, each on its own area), find the duty one tube can carry, then divide the total duty by that to get $N$ — rounded UP, since a fractional tube cannot be built.

  1. Saturation temperature and LMTD. At 250 kPa, $T_{sat}=127.41\,{}^{\circ}\text{C}$. With the solution rising from 65 to $93\,{}^{\circ}\text{C}$ against this constant temperature, $$\Delta T_1=T_{sat}-T_{in}=62.41\,{}^{\circ}\text{C}, \qquad \Delta T_2=T_{sat}-T_{out}=34.41\,{}^{\circ}\text{C}$$ $$LMTD=\frac{\Delta T_1-\Delta T_2}{\ln(\Delta T_1/\Delta T_2)}=\frac{62.41-34.41}{\ln(62.41/34.41)}$$ $$\boxed{LMTD=47.03\,{}^{\circ}\text{C}}$$
  2. Overall $UA$ for one tube. $A_i=2\pi r_iL=2\pi(0.015)(3)=0.2827\text{ m}^2$; $A_o=2\pi r_oL=2\pi(0.020)(3)=0.3770\text{ m}^2$. $$\begin{aligned} R_i&=\frac1{h_iA_i}=1.040\times10^{-3}\text{ K/W} \\ R_{wall}&=\frac{\ln(r_o/r_i)}{2\pi k_{tube}L}=\frac{\ln(4/3)}{2\pi\times111\times3}=1.375\times10^{-4}\text{ K/W} \end{aligned}$$ $$\begin{aligned} R_o&=\frac1{h_oA_o}=3.634\times10^{-4}\text{ K/W} \\ R_{tot}&=R_i+R_{wall}+R_o=1.541\times10^{-3}\text{ K/W} \end{aligned}$$ $$UA_{tube}=\frac1{R_{tot}}=648.9\text{ W/K}$$
  3. Duty per tube and number of tubes. $$\dot Q_{tube}=UA_{tube}\times LMTD=648.9\times47.03=30{,}518\text{ W}=30.52\text{ kW}$$ $$N=\frac{\dot Q}{\dot Q_{tube}}=\frac{200}{30.52}=6.55$$ A fractional tube cannot be built, and 6 tubes would only supply $6\times30.52=183.1\text{ kW}$ — short of the 200 kW duty — so round UP: $$\boxed{N=7\text{ tubes}}$$ (7 tubes supply $7\times30.52=213.6\text{ kW}$, comfortably meeting the duty.)
Question 8 — results
QuantityValue
Steam saturation temperature $T_{sat}$$127.41\,{}^{\circ}\text{C}$
LMTD$47.03\,{}^{\circ}\text{C}$
Overall conductance per tube $UA_{tube}$648.9 W/K
Duty per tube $\dot Q_{tube}$30.52 kW
Exact tube count6.55
Number of tubes (built)7
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