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23-Chem-A3 Heat and Mass Transfer · December 2017

Question 3 of 6: Part A: Number of Tubes in a Steam-Heated Water Heater

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

Notes on this paper

National Exams — December 2017 — 16-Chem-A3 Heat and Mass Transfer. Three-hour, open-book exam (one textbook of the candidate’s choice; any non-communicating calculator). Format: two parts — Part A — Heat Transfer (Q1–Q3) and Part B — Mass Transfer (Q4–Q6). All problems are worth 25 points; the rubric requires at least two problems attempted from each part and marks only the first two per part, so a complete paper is four questions (100 points). All six problems are solved below for completeness. Property data are quoted as stated in each problem; where a value applies to a film condition it is used at the stated film temperature.

Reference texts: F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — free/forced convection correlations, conduction shape factors, the heat–mass-transfer analogy; J. P. Holman, Heat Transfer (McGraw-Hill) — natural-convection constants $C,m$ and cylinder conduction; R. E. Treybal, Mass-Transfer Operations (McGraw-Hill) — wetted-wall $k_G$, gas absorption and the Chilton–Colburn analogy; R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (Wiley); supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).

Question 3 — Part A: Number of Tubes in a Steam-Heated Water Heater (25 points)

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. Water is heated 20 → 45 °C inside brass tubes ($D_o=0.025$ m, $D_i=0.0225$ m, $L=4$ m, $k_w=111.65$ W/m·K) at velocity $v=61.2/60=1.02$ m/s; steam condenses at 110 °C outside with $h_o=4650$ W/m²K, and the total condensation rate is 1.25 kg/s ($\lambda=2230$ kJ/kg).

DatumValue
Tube $D_o$ / $D_i$ / $L$25 mm / 22.5 mm / 4 m
Water velocity in tube1.02 m/s
Water $\rho,\ \nu,\ c_p$995.7 kg/m³, $6.59\times10^{-7}$ m²/s, 4174 J/kg·K
Water $k$0.617 W/m·K
Steam: $T_\text{sat}$, $h_o$, $\lambda$110 °C, 4650 W/m²K, 2230 kJ/kg
Steam condensed (total)1.25 kg/s

Find. The number of parallel brass tubes $N$ needed to carry the water and transfer the required duty.

One brass tube (of the bundle) — steam condensing outside, water insidewater 20 °C45 °Cwater, v = 1.02 m/ssteam at 110 °C (h₦ = 4650 W/m²K)steam 110 °Cwater 20→45 °CLMTD = 76.8 °C
Figure 3 — One representative tube of the bundle: water is heated from 20 to 45 °C inside while steam condenses at a constant 110 °C outside. The condensing-steam temperature is flat, giving the log-mean $\Delta T$ shown.

Approach. Build the overall coefficient $U_o$ (water-side $h_i$ from Dittus–Boelter, tube-wall resistance, given steam-side $h_o$), take the total duty from the steam-condensation load, form the LMTD for a condensing stream, and size the area $A_o=Q/(U_o\,\text{LMTD})$; the tube count follows as $A_o$ divided by the area of one tube.

  1. Water-side coefficient by Dittus–Boelter. With $Re=vD_i/\nu = (1.02)(0.0225)/6.59\times10^{-7}=3.48\times10^{4}$ (turbulent) and $Pr=\nu\rho c_p/k = 4.44$, using $Nu=0.023\,Re^{0.8}Pr^{0.4}$ (fluid being heated): $$Nu = 0.023(3.48\times10^{4})^{0.8}(4.44)^{0.4}=180,\qquad h_i=\frac{Nu\,k}{D_i}=\frac{(180)(0.617)}{0.0225}=4920\ \text{W/m}^2\text{K}.$$
  2. Overall coefficient on the outside area. Combine the three resistances (outside film, brass wall, inside film referred to $A_o$): $$\frac{1}{U_o}= \frac{1}{h_o}+\frac{D_o\ln(D_o/D_i)}{2k_w}+\frac{D_o}{D_i\,h_i} = 2.15\times10^{-4}+1.18\times10^{-5}+2.26\times10^{-4},$$ $$U_o = \boxed{2210\ \text{W/m}^2\text{K}}.$$ The brass wall contributes almost nothing; the two water/steam films dominate about equally.
  3. Log-mean temperature difference. The steam side is isothermal at 110 °C, so with $\Delta T_1=110-20=90$ K and $\Delta T_2=110-45=65$ K, $$\text{LMTD}=\frac{\Delta T_1-\Delta T_2}{\ln(\Delta T_1/\Delta T_2)}=\frac{90-65}{\ln(90/65)}=76.8\ \text{K}.$$
  4. Duty from the steam load. The stated condensation rate sets the total heat delivered: $$Q = \dot m_s\lambda = (1.25)(2230\times10^{3}) = 2.79\times10^{6}\ \text{W} = 2788\ \text{kW}.$$
  5. Required area and tube count. The outside area needed and the area of one 4-m tube are $$A_o=\frac{Q}{U_o\,\text{LMTD}}=\frac{2.788\times10^{6}}{(2210)(76.8)}=16.4\ \text{m}^2,\qquad a_\text{tube}=\pi D_o L = \pi(0.025)(4)=0.314\ \text{m}^2,$$ $$N=\frac{A_o}{a_\text{tube}}=\frac{16.4}{0.314}=52.3\;\Rightarrow\;\boxed{N \approx 53\ \text{brass tubes}}.$$ Rounding the 52.3 up to the next whole tube gives 53 tubes to transfer the condensing-steam load.
QuantityResult
Water-side $h_i$≈ 4920 W/m²K
Overall $U_o$≈ 2210 W/m²K
LMTD76.8 K
Duty (steam load)≈ 2788 kW
Required area / area per tube16.4 m² / 0.314 m²
Number of tubes required≈ 53
Check
The printed data are not self-consistent: at 1720 kg/hr the water duty would be only $\dot m c_p\Delta T\approx50$ kW, about 56× smaller than the stated 1.25 kg/s steam-condensation load (2788 kW). An energy balance requires the two to match, which they do if the water rate is read as ≈ 1720 kg/min (duty ≈ 2.9 MW) — a units slip in the source. Following the exam’s “state your assumptions” instruction, the governing duty is taken from the explicit steam load, the water velocity is used only for $h_i$, and the tube count follows from the heat-transfer area: $N\approx53$. (A single 4-m tube would suffice only for the literal 50-kW reading; the velocity/throughput constraint would round the count up slightly further, to the order of 55–70, so 53–56 tubes is the design range.)