Heat Exchanger Area – Parallel vs Counterflow: 25-Nav-B2 Marine Engineering and Vibrations, May 2017, Q4

Question 4: Heat Exchanger Area – Parallel vs Counterflow (20 marks)

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 6.93 kg/s cooled 65.6→39.4°C by cold water 6.30 kg/s entering at 10.0°C, through a 0.0254 m O.D. tube with $U=568$ W/m²·K based on the outer area.

Given data
Symbol	Value
ṁ_h	6.93 kg/s
T_h,in, T_h,out	65.6°C, 39.4°C
ṁ_c	6.30 kg/s
T_c,in	10.0°C
U (outer area)	568 W/m²·K
D_o	0.0254 m

Find. The required outer-tube heat transfer area for (a) a parallel-flow and (b) a counterflow tubular exchanger.

Parallel-flow versus counterflow tubular arrangements – both must reject the same duty Q, but the temperature approach at the outlet differs sharply between the two.

Approach. Find the duty Q and the cold-water outlet temperature from an energy balance, then compute the log-mean temperature difference (LMTD) for each arrangement and size the area from $Q=UA\,\Delta T_{lm}$.

Duty and cold-outlet temperature. $Q=\dot m_h c_p(T_{h,in}-T_{h,out})=6.93(4186)(65.6-39.4)=760.0\ \text{kW}$. Then $T_{c,out}=T_{c,in}+Q/(\dot m_c c_p)=10.0+\dfrac{760{,}035}{6.30(4186)}=\boxed{38.82^{\circ}\text{C}}$.
Parallel-flow LMTD. Both streams enter together: $\Delta T_1=T_{h,in}-T_{c,in}=55.6^{\circ}\text{C}$, $\Delta T_2=T_{h,out}-T_{c,out}=39.4-38.82=0.58^{\circ}\text{C}$. $$\Delta T_{lm,par}=\frac{\Delta T_1-\Delta T_2}{\ln(\Delta T_1/\Delta T_2)}=\frac{55.6-0.58}{\ln(55.6/0.58)}=12.06^{\circ}\text{C}$$ Area: $$A_{par}=\frac{Q}{U\,\Delta T_{lm,par}}=\frac{760{,}035}{568(12.06)}=\boxed{111.0\ \text{m}^2}$$
Counterflow LMTD. Hot inlet pairs with cold outlet: $\Delta T_1=T_{h,in}-T_{c,out}=65.6-38.82=26.78^{\circ}\text{C}$, $\Delta T_2=T_{h,out}-T_{c,in}=39.4-10.0=29.4^{\circ}\text{C}$. $$\Delta T_{lm,ctr}=\frac{29.4-26.78}{\ln(29.4/26.78)}=28.07^{\circ}\text{C}$$ Area: $$A_{ctr}=\frac{760{,}035}{568(28.07)}=\boxed{47.7\ \text{m}^2}$$

The parallel-flow arrangement needs more than twice the area of the counterflow arrangement for the identical duty. The reason is visible in the temperature-difference profile: because both streams start together in parallel flow, the driving temperature difference collapses to only 0.58°C by the outlet, starving the tail end of the exchanger of driving force, whereas counterflow keeps a healthy 27–29°C difference along the whole tube.

Final Results
Quantity	Value
Duty Q	760.0 kW
Cold outlet T_c,out	38.82°C
Parallel-flow LMTD	12.06°C
Parallel-flow area A	111.0 m²
Counterflow LMTD	28.07°C
Counterflow area A	47.7 m²

Topic: LMTD sizing of a tubular heat exchanger, parallel vs counterflow.

Key relations: Energy balance $Q=\dot mc_p\Delta T$; $\Delta T_{lm}=(\Delta T_1-\Delta T_2)/\ln(\Delta T_1/\Delta T_2)$; sizing $Q=UA\,\Delta T_{lm}$.

Why this works: The LMTD is the correctly-averaged driving temperature difference along an exchanger whose local $\Delta T$ decays exponentially with area; taking the log-mean rather than the arithmetic mean matters most exactly when the two end-differences are very unequal, as they are here for parallel flow.

Common pitfall: Pairing the wrong end-temperatures – parallel flow pairs (hot-in, cold-in) and (hot-out, cold-out); counterflow pairs (hot-in, cold-out) and (hot-out, cold-in). Swapping them silently produces the other arrangement's answer.

Source: Incropera, Fundamentals of Heat and Mass Transfer, ch. 11 (heat exchangers, LMTD method); Cengel, Heat and Mass Transfer, ch. 11.

Question 4 of 7: Heat Exchanger Area – Parallel vs Counterflow

Question 4: Heat Exchanger Area – Parallel vs Counterflow (20 marks)