Question

A perfectly insulated double pipe heat exchanger is operating at steady state. Saturated steam enters the inner pipe at 100 °C and leaves as saturated water at 100 °C. Cooling water enters the outer pipe at 75 °C and exits at 95 °C. The overall heat transfer coefficient is 1 kW m$^{-2}$ K$^{-1}$ and the heat transfer area is 1 m$^{2}$. The average specific heat capacity of water is 4.2 kJ kg$^{-1}$ K$^{-1}$. The required cooling water flow rate is \(\underline{\hspace{2cm}}\) kg s$^{-1}$ (rounded off to two decimal places).

Hint:

When steam condenses, always use latent heat in heat-transfer calculations since it dominates over sensible heat.

Answer 1

Correct Answer: 0.14 - 0.15

Step 1: Calculate the Log Mean Temperature Difference (LMTD)

For a double pipe heat exchanger with steam condensing at constant temperature (100°C):

Temperature differences:

• At inlet: $\Delta T_1 = T_{s} - T_{c,in} = 100 - 75 = 25°C$
• At outlet: $\Delta T_2 = T_{s} - T_{c,out} = 100 - 95 = 5°C$

LMTD: $$\Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1/\Delta T_2)}$$

$$\Delta T_{lm} = \frac{25 - 5}{\ln(25/5)} = \frac{20}{\ln(5)} = \frac{20}{1.609} = 12.43°C$$

Step 2: Calculate Heat Transfer Rate

Using the overall heat transfer equation:

$$Q = U \cdot A \cdot \Delta T_{lm}$$

$$Q = 1000 \times 1 \times 12.43 = 12,430 \text{ W} = 12.43 \text{ kW}$$

Step 3: Calculate Cooling Water Flow Rate

The heat gained by cooling water equals the heat transferred:

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

$$12,430 = \dot{m}_c \times 4200 \times (95 - 75)$$

$$12,430 = \dot{m}_c \times 4200 \times 20$$

$$12,430 = \dot{m}_c \times 84,000$$

$$\dot{m}_c = \frac{12,430}{84,000} = 0.148 \text{ kg/s}$$