ChemicalHeat TransfereasyFE ExamCourseworkPE / Advanced

Required Heat Exchanger Area

A heat exchanger must transfer $Q = 500\ \text{kW}$ .

The overall heat transfer coefficient is $U = 800\ \text{W/m}^2\text{K}$ and the log mean temperature difference is $40\ \text{K}$ .

Find the required heat transfer area.

Given

Q	500 kW	Heat duty
U	800 W/m^2*K	Overall heat transfer coefficient
LMTD	40 K	Log mean temperature difference

Hint 1

The exchanger design equation relates duty, coefficient, area, and temperature driving force.

Hint 2

A = Q/(U·ΔT_lm).

Hint 3

Check units: U is in W/m²K, so Q must be in W.

Worked solution — try the problem first

The design equation for an exchanger:

$Q = UA\,\Delta T_{lm} \quad\Longrightarrow\quad A = \frac{Q}{U\Delta T_{lm}}$

Convert the duty to watts to match $U$ :

$Q = 500\ \text{kW} = 500{,}000\ \text{W}$

$A = \frac{500{,}000}{800 \times 40} = \frac{500{,}000}{32{,}000} = 15.6\ \text{m}^2$

$\boxed{A = 15.6\ \text{m}^2}$

In practice you would add margin for fouling. Over time, deposits add thermal resistance and reduce the effective $U$ , so exchangers are commonly oversized by 10-25%, or specified with a fouling factor built into $U$ .

Note the LMTD is given here. In a real design you would compute it from the four terminal temperatures, and apply a correction factor $F$ for shell-and-tube geometries that are neither pure counterflow nor pure parallel flow: $Q = UAF\Delta T_{lm}$ .

Concepts:Heat exchanger designOverall heat transfer coefficientLMTD

Sign in to submit

Grading needs an account so your progress, attempts, and daily quota can be tracked. The problem statement and worked solution stay open to everyone.

Create an account