Hot and cold fluids enter a parallel flow double tube heat exchanger at 100°C and 15°C, respectively. The heat capacity rates of hot and cold fluids are Ch = 2000 W/K and Cc = 1200 W/K, respectively. If the outlet temperature of the cold fluid is 45°C, the log mean temperature difference (LMTD) of the heat exchanger is ________ K (round off to two decimal places).

The correct answer is 57.71.

Step 1: Identify the given data
Type of heat exchanger: Parallel flow double tube heat exchanger
Inlet temperature of hot fluid, $T_{\mathrm{hi}}=100\mathrm{°C}$
Inlet temperature of cold fluid, $T_{\mathrm{ci}}=15\mathrm{°C}$
Outlet temperature of cold fluid, $T_{\mathrm{co}}=45\mathrm{°C}$
Heat capacity rate of hot fluid, $C_h=2000\mathrm{W/K}$
Heat capacity rate of cold fluid, $C_c=1200\mathrm{W/K}$

Step 2: Determine the outlet temperature of the hot fluid ($T_{\mathrm{ho}}$)
Using the conservation of energy principle, the heat lost by the hot fluid must equal the heat gained by the cold fluid:
$Q_{\mathrm{lost}}=Q_{\mathrm{gained}}$
$C_h(T_{\mathrm{hi}}-T_{\mathrm{ho}})=C_c(T_{\mathrm{co}}-T_{\mathrm{ci}})$

Substitute the given values into the energy equation:
$2000(100-T_{\mathrm{ho}})=1200(45-15)$
$2000(100-T_{\mathrm{ho}})=1200\left(30\right)$
$2000(100-T_{\mathrm{ho}})=36000$
$100-T_{\mathrm{ho}}=\frac{36000}{2000}$
$100-T_{\mathrm{ho}}=18$
$T_{\mathrm{ho}}=100-18=82\mathrm{°C}$

Step 3: Calculate the temperature differences at the inlet and outlet
For a parallel flow heat exchanger, the fluids enter at the same end (Section 1) and exit at the same end (Section 2).
Temperature difference at the inlet (Section 1):
$\Delta T_1=T_{\mathrm{hi}}-T_{\mathrm{ci}}=100-15=85K$
Temperature difference at the outlet (Section 2):
$\Delta T_2=T_{\mathrm{ho}}-T_{\mathrm{co}}=82-45=37K$

Step 4: Calculate the Log Mean Temperature Difference (LMTD)
The formula for the Log Mean Temperature Difference is:
$\mathrm{LMTD}=\frac{\Delta T_1-\Delta T_2}{\ln \left(\frac{\Delta T_1}{\Delta T_2}\right)}$

Substitute the calculated values into the formula:
$\mathrm{LMTD}=\frac{85-37}{\ln \left(\frac{85}{37}\right)}$
$\mathrm{LMTD}=\frac{48}{\ln \left(2.2973\right)}$
$\mathrm{LMTD}\approx \frac{48}{0.8317}\approx 57.71K$

Thus, the log mean temperature difference (LMTD) of the heat exchanger is 57.71 K (or 57.71 °C, since temperature differences are equivalent in Kelvins and Celsius).