---
title: "Oil cooler area by LMTD method"
subject: "Heat Transfer"
chapter: 6
key: "semester_5/heat_transfer:oil-cooler-lmtd-area-counter-parallel"
source: https://www.bctengineer.com/q/semester_5/heat_transfer/oil-cooler-lmtd-area-counter-parallel
site: BCT Engineering
---

# Oil cooler area by LMTD method

Heat Transfer · Chapter 6: Applications of Heat Transfer

- Asked in: year not given, 8 marks
- Video: [Problem - LMTD Method of heat exchanger analysis | Heat Exchangers | HMT | Mod 4 | KTU | S6 MECH |](https://www.youtube.com/watch?v=mK2QDYjCjzQ) — Mohith Dadu
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/heat_transfer/oil-cooler-lmtd-area-counter-parallel
- Whole chapter: https://www.bctengineer.com/md/semester_5/heat_transfer/chapter-6

## Question

In a counterflow oil cooler, hot oil (cp = 2100 J/kg K) flows at 2 kg/s and is cooled from 100 °C to 60 °C by water (cp = 4180 J/kg K) flowing at 1.2 kg/s and entering at 20 °C. The overall heat transfer coefficient is 400 W/m^2 K. Determine (a) the heat duty, (b) the water outlet temperature, (c) the LMTD and the surface area required, and (d) the area if the exchanger were parallel flow with the same terminal temperatures.

## Short answer

Heat duty: Q = 2×2100×(100-60) = 168 kW. Water outlet: T_c2 = 20 + 168000/(1.2×4180) = 53.5°C. Counterflow LMTD: ΔT_lm = (46.5-40)/ln(46.5/40) = 43.2 K, A = 168000/(400×43.2) = 9.73 m². Parallel flow LMTD = (80-6.5)/ln(80/6.5) = 29.3 K, A = 168000/(400×29.3) = 14.3 m².

## Answer

#### (a) Heat duty

$$
Q = \dot m_hc_h(T_{h1}-T_{h2}) = 2\times2100\times(100-60) = 168\,000\ \text{W} = 168\ \text{kW}
$$

#### (b) Water outlet

$$
T_{c2} = T_{c1}+\frac{Q}{\dot m_cc_c} = 20 + \frac{168\,000}{1.2\times4180} = 20 + 33.49 = 53.5\ ^\circ\text{C}
$$

#### (c) Counterflow LMTD and area
```text
 Oil   100 ----------> 60
 Water  53.5 <---------- 20
```
$\Delta T_1 = T_{h1}-T_{c2} = 100 - 53.5 = 46.5$ K, $\Delta T_2 = T_{h2}-T_{c1} = 60 - 20 = 40$ K.

$$
\Delta T_{lm} = \frac{46.5-40}{\ln(46.5/40)} = \frac{6.51}{0.1505} = 43.2\ \text{K}
$$

$$
A = \frac{Q}{U\,\Delta T_{lm}} = \frac{168\,000}{400\times43.17} = 9.73\ \text{m}^2
$$

#### (d) Parallel flow
Both fluids enter at the same end: $\Delta T_1 = 100 - 20 = 80$ K, $\Delta T_2 = 60 - 53.5 = 6.5$ K.

$$
\Delta T_{lm,p} = \frac{80-6.5}{\ln(80/6.5)} = \frac{73.5}{2.510} = 29.3\ \text{K}
$$

$$
A_p = \frac{168\,000}{400\times29.29} = 14.3\ \text{m}^2
$$

Parallel flow needs about 47 percent more area for the same duty, showing why counterflow is preferred.

**Answer:** $Q = 168$ kW; $T_{c2} = 53.5$ °C; $\Delta T_{lm} = 43.2$ K and $A = 9.73$ m$^2$ (counterflow); $A = 14.3$ m$^2$ (parallel flow).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the heat duty found, and what is it?
   A. Q = m_c c_c x 20 = 100 kW
   B. Q = m_h c_h (Th1 - Th2) = 2 x 2100 x 40 = 168 kW
   C. Q = U A dT = 400 kW
   D. Q = m_h c_h x Th1 = 420 kW
   Answer: B. The oil side has all temperatures known, so Q = 2 x 2100 x (100 - 60) = 168 000 W. (Key point: Heat duty)

2. The water outlet temperature is
   A. 20 + 168000/(1.2 x 4180) = 53.5 C
   B. 100 C
   C. 60 C
   D. 20 + 168000/(2 x 2100) = 60 C
   Answer: A. The heat gained by water equals Q, so Tc2 = 20 + 33.5 = 53.5 C. (Key point: Water outlet temperature)

3. In counterflow, the end temperature differences are
   A. 100 - 20 = 80 K and 60 - 53.5 = 6.5 K
   B. 100 - 53.5 = 46.5 K and 60 - 20 = 40 K
   C. 100 - 60 = 40 K and 53.5 - 20 = 33.5 K
   D. 46.5 K and 6.5 K
   Answer: B. Counterflow pairs the hot inlet with the cold outlet and the hot outlet with the cold inlet, giving an LMTD of 43.2 K and A = 9.73 m^2. (Key point: Counterflow terminal differences)

4. For the same duty, parallel flow needs about 14.3 m^2 against 9.73 m^2 for counterflow because
   A. U is lower in parallel flow
   B. Q is smaller in parallel flow
   C. its LMTD is smaller (29.3 K)
   D. its LMTD is larger
   Answer: C. Parallel-flow terminal differences (80 and 6.5 K) give a lower LMTD, so more area is required. (Key point: Parallel versus counter)

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