Parallel Flow Heat Exchanger LMTD Calculation

This solution details the steps required to calculate the Log Mean Temperature Difference (LMTD) for a heat exchanger operating in a parallel flow configuration. LMTD is essential for determining the heat transfer rate and evaluating the performance of heat exchangers.

Heat Exchanger Parameters Provided

The problem provides the following data for the hot and cold water streams:

Energy Balance for Cold Water Outlet Temperature

To calculate the LMTD, we first need to find the outlet temperature of the cold water stream ($ T_{c,out} $). We can do this by applying the principle of energy balance, assuming the heat lost by the hot fluid equals the heat gained by the cold fluid, and that the specific heat capacities of both fluids are equal.

The energy balance equation is:

$$ \dot{m}_h c_{p,h} (T_{h,in} - T_{h,out}) = \dot{m}_c c_{p,c} (T_{c,out} - T_{c,in}) $$

Assuming \( c_{p,h} = c_{p,c} \), the equation simplifies to:

$$ \dot{m}_h (T_{h,in} - T_{h,out}) = \dot{m}_c (T_{c,out} - T_{c,in}) $$

Substitute the given values:

$$ 0.2 \text{ Kg/sec} \times (75^\circ C - 45^\circ C) = 0.5 \text{ Kg/sec} \times (T_{c,out} - 20^\circ C) $$

Calculate the heat transfer rate on the hot side:

$$ 0.2 \times 30^\circ C = 0.5 \times (T_{c,out} - 20^\circ C) $$

$$ 6 \text{ (Kg \cdot °C)/sec} = 0.5 \text{ Kg/sec} \times (T_{c,out} - 20^\circ C) $$

Now, solve for the cold water outlet temperature ($ T_{c,out} $):

$$ T_{c,out} - 20^\circ C = \frac{6 \text{ (Kg \cdot °C)/sec}}{0.5 \text{ Kg/sec}} $$

$$ T_{c,out} - 20^\circ C = 12^\circ C $$

$$ T_{c,out} = 12^\circ C + 20^\circ C $$

$$ T_{c,out} = 32^\circ C $$

The calculated outlet temperature for the cold water stream is 32°C.

Calculating LMTD for Parallel Flow Configuration

The Log Mean Temperature Difference (LMTD) is calculated using the following formula for a parallel flow heat exchanger:

$$ LMTD = \frac{\Delta T_1 - \Delta T_2}{\ln\left(\frac{\Delta T_1}{\Delta T_2}\right)} $$

Where:

• $ \Delta T_1 $ is the temperature difference between the hot and cold fluids at one end (usually the inlet end).
• $ \Delta T_2 $ is the temperature difference between the hot and cold fluids at the other end (usually the outlet end).

Let's calculate these temperature differences:

• Hot fluid inlet temperature ($ T_{h,in} $): 75°C
• Cold fluid inlet temperature ($ T_{c,in} $): 20°C
• Hot fluid outlet temperature ($ T_{h,out} $): 45°C
• Cold fluid outlet temperature ($ T_{c,out} $): 32°C

Calculate $ \Delta T_1 $:

$$ \Delta T_1 = T_{h,in} - T_{c,in} = 75^\circ C - 20^\circ C = 55^\circ C $$

Calculate $ \Delta T_2 $:

$$ \Delta T_2 = T_{h,out} - T_{c,out} = 45^\circ C - 32^\circ C = 13^\circ C $$

Now, substitute these values into the LMTD formula:

$$ LMTD = \frac{55^\circ C - 13^\circ C}{\ln\left(\frac{55^\circ C}{13^\circ C}\right)} $$

Simplify the numerator:

$$ LMTD = \frac{42^\circ C}{\ln\left(\frac{55}{13}\right)} $$

Calculate the ratio and its natural logarithm:

$$ \frac{55}{13} \approx 4.230769 $$

$$ \ln(4.230769) \approx 1.44225 $$

Finally, calculate the LMTD:

$$ LMTD = \frac{42^\circ C}{1.44225} $$

$$ LMTD \approx 29.12^\circ C $$

The Log Mean Temperature Difference for this parallel flow heat exchanger is approximately 29.12°C.