Question

A shell-and-tube heat exchanger is used for cooling crude oil from $400~\text{K}$ to $360~\text{K}$. Crude oil flows through the tube at $3650~\text{kg/h}$. Water enters the shell side at $310~\text{K}$ and has a flow rate of $1600~\text{kg/h}$. Assume $c_{p,\text{oil}}=2.5~\text{kJ/(kg.K)}$, $c_{p,\text{w}}=4.187~\text{kJ/(kg.K)}$, overall $U=300~\text{W/(m}^2\!\cdot\!\text{K)}$, and countercurrent flow. The required heat–transfer area is ............... m$^2$ (rounded to one decimal place).

Hint:

For variable outlet on the cold side, use energy balance to get $T_{c,\text{out}}$, then compute countercurrent LMTD and area from $Q=UA\Delta T_{\text{lm}}$.

Answer 1

Step 1: Heat duty from hot side.
$\dot m_h = 3650/3600 = 1.0139~\text{kg/s}$, \; $\Delta T_h = 400 - 360 = 40~\text{K}$.
\[ Q = \dot m_h c_{p,h}\Delta T_h = 1.0139 \times 2500 \times 40 = 1.014 \times 10^5~\text{W}. \] Step 2: Cold outlet temperature.
$\dot m_c = 1600/3600 = 0.4444~\text{kg/s}$; \; $\Delta T_c = \dfrac{Q}{\dot m_c c_{p,c}} = \dfrac{1.014 \times 10^5}{0.4444 \times 4187} = 54.5~\text{K}$.
$T_{c,\text{out}} = 310 + 54.5 = 364.5~\text{K}$.
Step 3: LMTD for countercurrent flow.
$\Delta T_1 = T_{h,\text{in}} - T_{c,\text{out}} = 400 - 364.5 = 35.5~\text{K}$,
$\Delta T_2 = T_{h,\text{out}} - T_{c,\text{in}} = 360 - 310 = 50~\text{K}$.
\[ \Delta T_{\text{lm}} = \frac{\Delta T_2 - \Delta T_1}{\ln(\Delta T_2/\Delta T_1)} = \frac{50 - 35.5}{\ln(50/35.5)} = 42.35~\text{K}. \] Step 4: Area from $Q = UA \Delta T_{\text{lm}}$.
\[ A = \frac{Q}{U \,\Delta T_{\text{lm}}} = \frac{1.014 \times 10^5}{300 \times 42.35} = 7.98~\text{m}^2 \approx 8.0~\text{m}^2. \]