Counterflow heat-exchanger area for a higher outlet temperature: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2018, Q8

Question 8 — Counterflow heat-exchanger area for a higher outlet temperature (Part B, 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. Counterflow exchanger. Cold fluid: $c_{p,c}=800\ \text{J/kg}\cdot\text{°C}$, $\dot m_c=2.4$ kg/s, in at 300 °C. Hot fluid: $c_{p,h}=960\ \text{J/kg}\cdot\text{°C}$, $\dot m_h=2.0$ kg/s, in at 1000 °C.

Quantity	Value
Cold capacity rate, $C_c=\dot m_c c_{p,c}$	2.4×800 = 1920 W/°C
Hot capacity rate, $C_h=\dot m_h c_{p,h}$	2.0×960 = 1920 W/°C
Cold inlet / hot inlet	300 °C / 1000 °C
Case 1 cold outlet	700 °C
Case 2 cold outlet (target)	800 °C

Find. the fractional increase in surface area (with $U$, flows and inlet temperatures unchanged) to raise the cold-fluid outlet from 700 °C to 800 °C.

Figure 8 — Counterflow temperature profiles. With equal capacity rates ($C_r=1$) the temperature difference is uniform along the exchanger, so the LMTD equals that constant gap: 300 °C in the base case, 200 °C for the more demanding 800 °C target.

Approach. The capacity rates are equal ($C_r=1$), so for counterflow the terminal temperature differences are equal and the LMTD is just that constant gap. Compute $UA=\dot Q/\text{LMTD}$ for each case; with $U$ fixed the area ratio equals the $UA$ ratio.

Case 1 duty and hot outlet. $$\dot Q_1=C_c(700-300)=1920(400)=768\ \text{kW},\qquad T_{h,o}=1000-\frac{\dot Q_1}{C_h}=1000-400=600\ ^\circ\text{C}$$
Case 1 LMTD and $UA$. Terminal differences: hot-in/cold-out $=1000-700=300$; hot-out/cold-in $=600-300=300$. Equal, so $\text{LMTD}_1=300\ ^\circ\text{C}$: $$UA_1=\frac{\dot Q_1}{\text{LMTD}_1}=\frac{768\,000}{300}=2560\ \text{W/}^\circ\text{C}$$
Case 2 duty and hot outlet. $$\dot Q_2=C_c(800-300)=1920(500)=960\ \text{kW},\qquad T_{h,o}=1000-\frac{960\,000}{1920}=500\ ^\circ\text{C}$$
Case 2 LMTD and $UA$. Terminal differences: $1000-800=200$ and $500-300=200$, so $\text{LMTD}_2=200\ ^\circ\text{C}$: $$UA_2=\frac{960\,000}{200}=4800\ \text{W/}^\circ\text{C}$$
Area increase (U constant). $$\frac{A_2}{A_1}=\frac{UA_2}{UA_1}=\frac{4800}{2560}=1.875\;\Rightarrow\;\frac{\Delta A}{A_1}=0.875$$ The surface area must increase by about 87.5 % (to 1.875× the original).

Quantity	Case 1 (700 °C)	Case 2 (800 °C)
Duty $\dot Q$	768 kW	960 kW
Hot outlet	600 °C	500 °C
LMTD	300 °C	200 °C
$UA$	2560 W/°C	4800 W/°C
Required area increase ≈ 87.5 % (A₂ = 1.875 A₁)

Concept & Theory

Topic. LMTD sizing of a counterflow heat exchanger and area scaling for a more demanding duty.

Key relations. $\dot Q=C\,\Delta T$ for each stream; LMTD $=(\Delta T_1-\Delta T_2)/\ln(\Delta T_1/\Delta T_2)$, reducing to the common value when $\Delta T_1=\Delta T_2$; $\dot Q=U A\,\text{LMTD}$; with $U$ fixed, $A_2/A_1=UA_2/UA_1$.

Why this works. Equal capacity rates make the hot and cold lines parallel on the $T$–position plot, so the temperature difference is constant and the log-mean is simply that gap — no logarithm needed. Pushing the cold outlet higher both raises the duty and shrinks the driving $\Delta T$, so the area must grow on two counts; taking the ratio cancels $U$ and leaves a clean comparison.

Common pitfall. Forgetting that raising the cold-outlet target lowers the LMTD (a smaller driving force needs more area, on top of the larger heat load), or trying to evaluate the $0/0$ log-mean directly when the two terminal differences are equal instead of using the constant-$\Delta T$ limit.

Source. Çengel & Ghajar, Heat and Mass Transfer §11 (heat exchangers; the LMTD method, counterflow arrangement).

Question 8 of 8: Counterflow heat-exchanger area for a higher outlet temperature

Question 8 — Counterflow heat-exchanger area for a higher outlet temperature (Part B, equal value)