---
title: "Heat exchanger types and LMTD derivation"
subject: "Heat Transfer"
chapter: 6
key: "semester_5/heat_transfer:classification-heat-exchangers-lmtd-derivation"
source: https://www.bctengineer.com/q/semester_5/heat_transfer/classification-heat-exchangers-lmtd-derivation
site: BCT Engineering
---

# Heat exchanger types and LMTD derivation

Heat Transfer · Chapter 6: Applications of Heat Transfer

- Asked in: year not given, 10 marks
- Video: [LMTD, counter vs parallel flow, temp. crossing, Shell & tube heat exchangers, examples, correction](https://www.youtube.com/watch?v=EEWWf120TKQ) — Concept Connect (Chintan Modi)
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/heat_transfer/classification-heat-exchangers-lmtd-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_5/heat_transfer/chapter-6

## Question

(a) Classify heat exchangers according to flow arrangement and construction, with one application of each. (b) Derive the expression for the logarithmic mean temperature difference (LMTD) for a counterflow heat exchanger, stating the assumptions.

## Short answer

Heat exchangers are classified by flow arrangement (parallel, counter, cross) and construction (double pipe, shell and tube, plate, etc.). Parallel flow gives uniform wall temperature (heaters), counterflow highest effectiveness (oil coolers). LMTD for counterflow: under steady state, constant U and c, negligible losses, from d(ΔT)/ΔT = –U(1/Ch – 1/Cc)dA, integration yields ΔT_lm = (ΔT1–ΔT2)/ln(ΔT1/ΔT2) and Q = UA ΔT_lm.

## Answer

#### (a) Classification
**By flow arrangement**
- **Parallel flow:** both fluids enter at the same end and flow in the same direction. The temperature difference is largest at inlet; the cold outlet can never exceed the hot outlet. It gives a fairly uniform wall temperature, so it is used in simple heaters and coolers where the wall temperature must be limited.
- **Counterflow:** fluids flow in opposite directions. Gives the highest mean temperature difference and effectiveness (oil coolers, regenerative systems).
- **Cross flow:** fluids flow at right angles (car radiator, air coolers), mixed or unmixed.

**By construction**
- **Double pipe** (pipe-in-pipe): simple, for small duties (small oil and chemical coolers).
- **Shell and tube:** a bundle of tubes in a shell with baffles; most common in power and process plants (condensers, boilers).
- **Compact (plate-fin):** large area per volume, for gases (aircraft, automobiles).
- **Plate type:** corrugated plates, easy to clean (food, dairy).
- **Direct contact** (cooling tower) and **regenerators** (stored heat) are other categories.

#### (b) LMTD for counterflow
Assumptions: steady state; constant $U$, $c_{ph}$ and $c_{pc}$; no heat loss to surroundings; no phase change; negligible axial conduction.

```text
 Th1 -->--------------------->-- Th2
 Tc2 --<---------------------<-- Tc1
      dA element at distance x
```
For an element of area $dA$: with $\Delta T = T_h - T_c$:

$$
dQ = -\dot m_hc_h\,dT_h = -\dot m_cc_c\,dT_c = U\,\Delta T\,dA
$$

Taking $x$ along the hot-fluid direction, $T_h$ and $T_c$ both fall as $x$ increases. Writing $C_h = \dot m_hc_h$ and $C_c = \dot m_cc_c$:

$$
d(\Delta T) = dT_h - dT_c = -dQ\left(\frac{1}{C_h}-\frac{1}{C_c}\right)
$$

Substituting $dQ = U\Delta T\,dA$:

$$
\frac{d(\Delta T)}{\Delta T} = -U\left(\frac{1}{C_h}-\frac{1}{C_c}\right)dA
$$

Integrate over the full area, from end 1 ($\Delta T_1 = T_{h1}-T_{c2}$) to end 2 ($\Delta T_2 = T_{h2}-T_{c1}$):

$$
\ln\frac{\Delta T_2}{\Delta T_1} = -UA\left(\frac{1}{C_h}-\frac{1}{C_c}\right)
$$

The total heat is $Q = C_h(T_{h1}-T_{h2}) = C_c(T_{c2}-T_{c1})$, so $\dfrac{1}{C_h}-\dfrac{1}{C_c} = \dfrac{(T_{h1}-T_{h2})-(T_{c2}-T_{c1})}{Q} = \dfrac{\Delta T_1-\Delta T_2}{Q}$. Therefore

$$
\ln\frac{\Delta T_2}{\Delta T_1} = -\frac{UA(\Delta T_1-\Delta T_2)}{Q}
\;\Rightarrow\;
Q = UA\,\frac{\Delta T_1-\Delta T_2}{\ln(\Delta T_1/\Delta T_2)} = UA\,\Delta T_{lm}
$$

$$
\boxed{\Delta T_{lm}=\frac{\Delta T_1-\Delta T_2}{\ln(\Delta T_1/\Delta T_2)}}
$$

The same expression holds for parallel flow, with $\Delta T_1$, $\Delta T_2$ taken at the inlet and outlet ends. If $\Delta T_1 = \Delta T_2$ then $\Delta T_{lm} = \Delta T_1$. For cross-flow and multi-pass exchangers, $Q = UAF\Delta T_{lm,cf}$ with a correction factor $F<1$.

## Check yourself

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

1. Which flow arrangement gives the highest mean temperature difference and effectiveness?
   A. Parallel flow
   B. Counterflow
   C. Cross flow
   D. All are equal
   Answer: B. In counterflow the cold fluid can leave hotter than the hot fluid's outlet, giving the best performance. (Key point: Flow arrangements)

2. Shell and tube exchangers are characterised by
   A. a bundle of tubes in a shell with baffles; common in condensers and boilers
   B. corrugated plates for dairy use
   C. stored heat in a matrix
   D. fluids mixing directly in a tower
   Answer: A. They are the most common type in power and process plants. (Key point: Shell and tube use)

3. Which is an assumption used in the LMTD derivation?
   A. U varies strongly along length
   B. Heat is lost to the surroundings
   C. Specific heats vary with temperature
   D. Steady state with constant U and specific heats, no heat loss and no phase change
   Answer: D. These make dQ = U dT dA integrable with constant coefficients. (Key point: LMTD assumptions)

4. The LMTD is given by
   A. (dT1 + dT2)/2
   B. (dT1 - dT2)/ln(dT1/dT2)
   C. sqrt(dT1 dT2)
   D. ln(dT1/dT2)/(dT1 - dT2)
   Answer: B. Integrating d(dT)/dT = -U(1/Ch - 1/Cc)dA and using Q = Ch(Th1-Th2) = Cc(Tc2-Tc1) gives Q = UA (dT1 - dT2)/ln(dT1/dT2). (Key point: LMTD result)

## Related in Heat Transfer

- [Effectiveness-NTU analysis of a heat exchanger](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Aeffectiveness-ntu-counterflow-oil-water) — Heat Transfer, chapter 6
- [Oil cooler area by LMTD method](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Aoil-cooler-lmtd-area-counter-parallel) — Heat Transfer, chapter 6
- [Heat loss from a cylinder in cross flow](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Across-flow-air-cylinder-heat-loss) — Heat Transfer, chapter 3
- [Plane wall with temperature dependent conductivity](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Avariable-thermal-conductivity-plane-wall) — Heat Transfer, chapter 2

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