Heat Exchanger Area Calculation

The problem requires calculating the heat exchange area ($A$) for a counter-flow heat exchanger. We are given flow rates, inlet/outlet temperatures for the cold fluid, inlet temperature for the hot fluid, specific heat, and the overall heat transfer coefficient.

1. Calculate Heat Transfer Rate ($Q$)

First, calculate the rate of heat absorbed by the cold water. This represents the total heat transfer ($Q$) in the exchanger.

Given:

• Cold water flow rate, $m_c = 0.1 \text{ kg/s}$
• Cold water inlet temperature, $T_{c,in} = 20^\circ C$
• Cold water outlet temperature, $T_{c,out} = 70^\circ C$
• Specific heat of water, $c_p = 4200 \text{ J/(kg K)}$

The heat transfer rate is calculated using:

$ Q = m_c \cdot c_p \cdot (T_{c,out} - T_{c,in}) $

Substituting the values:

$ Q = (0.1 \text{ kg/s}) \cdot (4200 \text{ J/(kg K)}) \cdot (70^\circ C - 20^\circ C) $

$ Q = 0.1 \cdot 4200 \cdot 50 $

$ Q = 21000 \text{ W} $

2. Calculate Hot Water Outlet Temperature ($T_{h,out}$)

Assuming the heat lost by the hot fluid equals the heat gained by the cold fluid, we can find the hot water's outlet temperature.

Given:

• Hot water flow rate, $m_h = 0.1 \text{ kg/s}$
• Hot water inlet temperature, $T_{h,in} = 90^\circ C$

The heat transfer rate equation for the hot fluid is:

$ Q = m_h \cdot c_p \cdot (T_{h,in} - T_{h,out}) $

Substituting the known values:

$ 21000 \text{ W} = (0.1 \text{ kg/s}) \cdot (4200 \text{ J/(kg K)}) \cdot (90^\circ C - T_{h,out}) $

$ 21000 = 420 \cdot (90 - T_{h,out}) $

Solve for $T_{h,out}$:

$ 90 - T_{h,out} = \frac{21000}{420} = 50 $

$ T_{h,out} = 90^\circ C - 50^\circ C = 40^\circ C $

3. Calculate Log Mean Temperature Difference (LMTD)

For a counter-flow heat exchanger, the LMTD is calculated using the temperature differences at the two ends.

Temperature differences:

• End 1: $\Delta T_1 = T_{h,in} - T_{c,out} = 90^\circ C - 70^\circ C = 20^\circ C$
• End 2: $\Delta T_2 = T_{h,out} - T_{c,in} = 40^\circ C - 20^\circ C = 20^\circ C$

The LMTD formula is:

$ \text{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)} $

Since $\Delta T_1 = \Delta T_2 = 20^\circ C$, the LMTD is equal to this temperature difference.

$ \text{LMTD} = 20^\circ C \quad (\text{or } 20 \text{ K}) $

4. Calculate Heat Exchange Area ($A$)

Use the fundamental heat exchanger design equation:

$ Q = U \cdot A \cdot \text{LMTD} $

Given:

• Overall heat transfer coefficient, $U = 2000 \text{ W/(m}^2 \text{ K)}$

Rearrange to solve for Area ($A$):

$ A = \frac{Q}{U \cdot \text{LMTD}} $

Substitute the calculated values:

$ A = \frac{21000 \text{ W}}{(2000 \text{ W/(m}^2 \text{ K)}) \cdot (20 \text{ K})} $

$ A = \frac{21000}{40000} $

$ A = 0.525 \text{ m}^2 $

The required heat exchange area is 0.525 m$^2$.