Counter-Flow Tubular Water–Helium Heat Exchanger: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, December 2013, Q8

Question 8: Counter-Flow Tubular Water–Helium Heat Exchanger

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. Counter-flow tubular exchanger; hot water on the outer (casing) side at 36 kg/min, entering $80\,{}^{\circ}\text{C}$, leaving $50\,{}^{\circ}\text{C}$; helium on the tube side, entering $30\,{}^{\circ}\text{C}$, $C_p=5139$ J/kg·K; overall coefficient $U=115$ W/m²·K; surface area $A=22$ m².

Given data
Quantity	Symbol	Value
Hot-water flow rate	$\dot m_h$	36 kg/min (0.600 kg/s)
Hot-water inlet temperature	$T_{h,in}$	$80\,{}^{\circ}\text{C}$
Hot-water exit temperature	$T_{h,out}$	$50\,{}^{\circ}\text{C}$
Helium inlet temperature	$T_{c,in}$	$30\,{}^{\circ}\text{C}$
Helium specific heat	$C_{p,He}$	5139 J/kg·K
Overall heat-transfer coefficient	$U$	115 W/m²·K
Surface area	$A$	22 m²

Find. The helium mass flow rate $\dot m_{He}$ and its exit temperature $T_{c,out}$.

Counter-flow temperature profile: the two streams flow in opposite directions along the exchanger, so the helium's exit (highest) temperature sits at the same end as the hot water's inlet.

Approach. The hot-side energy balance alone already fixes the total duty $Q$ (all four hot-side quantities are given). With $Q$ known, the counter-flow LMTD equation $Q=UA\,\Delta T_{lm}$ has only the helium exit temperature as an unknown (the cold-side inlet is given), so it is solved implicitly for $T_{c,out}$; the helium energy balance then gives its capacity rate $\dot m_{He}C_{p,He}$ and hence $\dot m_{He}$.

Duty from the hot-water side. With $\dot m_h=36/60=0.600$ kg/s and $C_{p,water}\approx4.1875$ kJ/kg·K (evaluated at the water's mean temperature, $65\,{}^{\circ}\text{C}$), $$Q = \dot m_h C_{p,water}(T_{h,in}-T_{h,out}) = 0.600\times4.1875\times(80-50)$$ $$\boxed{Q = 75.37\text{ kW}}$$
Helium exit temperature from the LMTD equation. For counter flow, with $\Delta T_1=T_{h,in}-T_{c,out}$ and $\Delta T_2=T_{h,out}-T_{c,in}=50-30=20\,{}^{\circ}\text{C}$ fixed, $$Q = UA\,\Delta T_{lm} = UA\,\frac{\Delta T_1-\Delta T_2}{\ln(\Delta T_1/\Delta T_2)}$$ With $UA=115\times22=2530$ W/K $=2.530$ kW/K and $Q=75.37$ kW fixed, this is one equation in the single unknown $T_{c,out}$ (through $\Delta T_1$); solving numerically, $$\boxed{T_{c,out} = 37.65\,{}^{\circ}\text{C}}$$ (check: $\Delta T_1=80-37.65=42.35$, $\Delta T_{lm}=(42.35-20)/\ln(42.35/20)=29.79\,{}^{\circ}\text{C}$, $UA\,\Delta T_{lm}=2.530\times29.79=75.4$ kW — matches $Q$.)
Helium capacity rate and mass flow rate. From the helium-side energy balance, $$\dot m_{He}C_{p,He} = \frac{Q}{T_{c,out}-T_{c,in}} = \frac{75.37}{37.65-30} = \frac{75.37}{7.65} = 9.86\text{ kW/K}$$ $$\boxed{\dot m_{He} = \frac{9.86\times10^3}{5139} = 1.918\text{ kg/s} = 115.1\text{ kg/min}}$$

Question 8 — results
Quantity	Value
Heat-exchanger duty $Q$	75.37 kW
Helium exit temperature $T_{c,out}$	$37.65\,{}^{\circ}\text{C}$
Helium capacity rate	9.86 kW/K
Helium mass flow rate $\dot m_{He}$	1.918 kg/s (115.1 kg/min)

Topic. Counter-flow heat-exchanger sizing via the log-mean-temperature-difference (LMTD) method.

Key relations. Energy balance $Q=\dot m C_p\Delta T$ on each stream; $Q=UA\,\Delta T_{lm}$, $\Delta T_{lm}=(\Delta T_1-\Delta T_2)/\ln(\Delta T_1/\Delta T_2)$ for counter flow, with $\Delta T_1$, $\Delta T_2$ the temperature differences at the two exchanger ends.

Why this works. The hot stream alone already supplies enough information (both its terminal temperatures and its own capacity rate) to pin down the total duty $Q$ without knowing anything about the cold stream; that duty then acts as a known target that the LMTD equation must reproduce, leaving only ONE genuine unknown ($T_{c,out}$, since the LMTD's other end-difference $\Delta T_2$ is already fixed by the given data) to solve for numerically.

Common pitfall. Using the co-flow (parallel-flow) LMTD pairing ($\Delta T_1=T_{h,in}-T_{c,in}$, $\Delta T_2=T_{h,out}-T_{c,out}$) instead of the counter-flow pairing used here, which for a counter-flow exchanger compares each hot-stream terminus against the cold-stream terminus at the SAME physical end, not the same "inlet/outlet" label.

Source. F. P. Incropera and D. P. DeWitt, Fundamentals of Heat and Mass Transfer, 7th ed., Ch. 11 (heat exchangers, LMTD method).

Question 8 of 8: Counter-Flow Tubular Water–Helium Heat Exchanger

Question 8: Counter-Flow Tubular Water–Helium Heat Exchanger