In a double-pipe heat exchanger (10 m), hot fluid flows in annulus and cold fluid in inner pipe.
Temperatures vary as:
$T_h(x) = 80 - 3x$,
$T_c(x) = 20 + 2x$,
where $T$ in °C and $x$ in m.
The logarithmic mean temperature difference (LMTD) is
$T_h(x) = 80 - 3x$,
$T_c(x) = 20 + 2x$,
where $T$ in °C and $x$ in m.
The logarithmic mean temperature difference (LMTD) is
Show Hint
- 24.6
- 27.9
- 30.0
- 50.0
The Correct Option is B
Solution and Explanation
$T_{h,in}=80$, $T_{c,in}=20$.
$\Delta T_1 = 80 - 20 = 60^\circ$C.
[4pt] At outlet $(x=10)$:
$T_{h,out} = 80 - 30 = 50$, $T_{c,out} = 20 + 20 = 40$.
$\Delta T_2 = 50 - 40 = 10^\circ$C.
[6pt] LMTD formula: \[ \Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)} = \frac{60 - 10}{\ln(60/10)} = \frac{50}{\ln 6} \approx 27.9. \]
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