Counterflow Tube-in-Shell Heat Exchanger (Water → Helium): 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2013, Q8

Question 8: Counterflow Tube-in-Shell Heat Exchanger (Water → Helium) (equal value)

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. Hot water (shell side): $\dot m_w=36$ kg/min $=0.6$ kg/s, $c_{p,w}=4180\ \text{J/kg}\cdot^\circ\text{C}$, in 80 °C, out 50 °C. Helium (tube side): $c_{p,\text{He}}=5139\ \text{J/kg}\cdot^\circ\text{C}$, in 30 °C. Counterflow; $U=115\ \text{W/m}^2\cdot^\circ\text{C}$, $A=22\ \text{m}^2$.

Find. The helium mass flow rate $\dot m_\text{He}$ and its outlet temperature.

Figure 7 — Counterflow temperature profile, plotted against position with the water flowing left→right and the helium right→left. Water cools 80→50 °C; the helium therefore enters at the right-hand end at 30 °C and leaves at the left-hand end at 37.8 °C, so its line rises towards the left. End differences ΔT₁=42.2 °C (left) and ΔT₂=20 °C (right) set the log-mean.

Approach. The water side is fully known, so its duty fixes $\dot Q$; $\dot Q=UA\,\Delta T_\text{lm}$ then gives the required log-mean, which (with the known cold-inlet end) pins the helium outlet; a final energy balance on the helium gives its flow.

Heat duty from the water side.$$\dot Q=\dot m_w c_{p,w}(80-50)=0.6(4180)(30)=\boxed{75.24\ \text{kW}}.$$
Required log-mean temperature difference. From $\dot Q=UA\,\Delta T_\text{lm}$,$$\Delta T_\text{lm}=\frac{\dot Q}{UA}=\frac{75\,240}{115(22)}=29.7\ ^\circ\text{C}.$$
Helium outlet temperature. For counterflow, $\Delta T_2=T_{w,\text{out}}-T_{\text{He,in}}=50-30=20\ ^\circ\text{C}$ and $\Delta T_1=80-T_{\text{He,out}}$. Solving $\Delta T_\text{lm}=(\Delta T_1-\Delta T_2)/\ln(\Delta T_1/\Delta T_2)=29.7$ gives $\Delta T_1=42.2\ ^\circ\text{C}$, hence$$T_{\text{He,out}}=80-42.2=\boxed{37.8\ ^\circ\text{C}}.$$
Helium mass flow. An energy balance on the helium ($\dot Q$ into it):$$\dot m_\text{He}=\frac{\dot Q}{c_{p,\text{He}}(T_{\text{He,out}}-T_{\text{He,in}})}=\frac{75\,240}{5139(37.8-30)}=\boxed{1.88\ \text{kg/s}}.$$

Check

The energy balance closes: helium receives $1.88(5139)(7.8)\approx75.2$ kW, matching the water duty. Water $c_p$ is taken at the 65 °C mean (4180 J/kg·°C); using 4186 shifts $\dot Q$ and $\dot m_\text{He}$ by <0.2 %.

Quantity	Result
Heat duty	˙Q ≈ 75.24 kW
Log-mean ΔT (counterflow)	≈ 29.7 °C
Helium outlet temperature	≈ 37.8 °C
Helium mass flow rate	≈ 1.88 kg/s

Concept & Theory

Topic. Counterflow heat exchanger sizing by the LMTD method.

Key relations. Duty $\dot Q=\dot mc_p\Delta T$; rate $\dot Q=UA\,\Delta T_\text{lm}$; counterflow $\Delta T_\text{lm}=(\Delta T_1-\Delta T_2)/\ln(\Delta T_1/\Delta T_2)$.

Why this works. One stream (water) is fully specified, so its enthalpy drop fixes the transferred heat without knowing the other flow. The rate equation then converts $U$, $A$ and the duty into the log-mean driving difference; with one end difference known, the other — and thus the helium outlet — follows, and a final balance sizes the helium stream.

Common pitfall. Using an arithmetic-mean $\Delta T$ instead of the log-mean, or pairing the end differences incorrectly for counterflow (each end matches the streams physically present there). A single-shell–pass correction factor is unnecessary here because the exchanger is true counterflow (F = 1).

Source. Incropera et al., Fundamentals of Heat and Mass Transfer §11.3–11.4 (LMTD method, counterflow); Çengel & Ghajar, Heat and Mass Transfer §11.

Question 8 of 8: Counterflow Tube-in-S​hell Heat Exchanger (Water → Helium)

Question 8: Counterflow Tube-in-S​hell Heat Exchanger (Water → Helium) (equal value)

Question 8 of 8: Counterflow Tube-in-Shell Heat Exchanger (Water → Helium)

Question 8: Counterflow Tube-in-Shell Heat Exchanger (Water → Helium) (equal value)