n a concentric tube counter-flow heat exchange, hot oil enters at 102°C and leaves at 65°C. Cold water enters at 25°C and leaves at 42°C. The log mean temperature difference (LMTD) is ________ °C (Round off to one decimal place).

The correct answer is 49.3.

1. Understanding Counter-Flow Heat Exchangers
In a concentric tube counter-flow heat exchanger, the hot fluid and the cold fluid flow in opposite directions. To analyze the heat transfer process, we first identify the temperatures of both fluids at their respective inlet and outlet locations:
- Hot fluid (oil) inlet temperature (T_hi) = 102°C
- Hot fluid (oil) outlet temperature (T_he) = 65°C
- Cold fluid (water) inlet temperature (T_ci) = 25°C
- Cold fluid (water) outlet temperature (T_ce) = 42°C

2. Temperature Differences at Each End
For a counter-flow configuration, the temperature difference between the two fluids at the ends of the heat exchanger is defined as:
- At the inlet end of the hot fluid (where the cold water exits):
ΔT_i = T_hi - T_ce = 102°C - 42°C = 60°C
- At the exit end of the hot fluid (where the cold water enters):
ΔT_e = T_he - T_ci = 65°C - 25°C = 40°C

3. Calculating the Log Mean Temperature Difference (LMTD)
The Log Mean Temperature Difference (LMTD) is defined by the following formula:

$${\mathrm{LMTD}}_{}=\frac{\Delta T_i-\Delta T_e}{\ln \left(\frac{\Delta T_i}{\Delta T_e}\right)}$$

Substituting the values of ΔT_i and ΔT_e into the formula:

$$\mathrm{LMTD}=\frac{60-40}{\ln \left(\frac{60}{40}\right)}$$

$$\mathrm{LMTD}=\frac{20}{\ln \left(1.5\right)}$$

Evaluating the natural logarithm term:
ln(1.5) ≈ 0.405465

Now, perform the final division:
LMTD ≈ 20 / 0.405465 ≈ 49.326°C

Rounding off to one decimal place, we obtain:
LMTD = 49.3°C