Question

A shell and tube heat exchanger is used as a steam condenser. Coolant water enters the tube at 300 K at a rate of 100 kg/s. The overall heat transfer coefficient is 1500 W/m².K, and total heat transfer area is 400 m². Steam condenses at a saturation temperature of 350 K. Assume that the specific heat of coolant water is 4000 J/kg.K. The temperature of the coolant water coming out of the condenser is \(\underline{\hspace{1cm}}\) K (round off to the nearest integer).

Hint:

The temperature of the coolant exiting a condenser is determined using energy balance and the heat transfer equation for the system.

Answer 1

Correct Answer: 337 - 341

Step 1: Calculate the Maximum Possible Heat Transfer

The maximum heat transfer occurs when the outlet water temperature approaches the steam temperature:

$$Q_{max} = \dot{m} \times c_p \times (T_{steam} - T_{in})$$

$$Q_{max} = 100 \times 4000 \times (350 - 300)$$

$$Q_{max} = 100 \times 4000 \times 50 = 20,000,000 \text{ W} = 20 \text{ MW}$$

Step 2: Calculate the Log Mean Temperature Difference (LMTD)

For a condenser, the steam temperature remains constant at $T_{steam} = 350$ K.

The LMTD is:

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

where:

• $\Delta T_1 = T_{steam} - T_{in} = 350 - 300 = 50$ K
• $\Delta T_2 = T_{steam} - T_{out}$

Step 3: Calculate Actual Heat Transfer

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

Also, for the coolant water:

$$Q = \dot{m} \times c_p \times (T_{out} - T_{in})$$

Step 4: Solve for Outlet Temperature

From the heat balance:

$$Q = 1500 \times 400 \times \text{LMTD} = 100 \times 4000 \times (T_{out} - 300)$$

$$600,000 \times \text{LMTD} = 400,000 \times (T_{out} - 300)$$

Now, $\Delta T_2 = 350 - T_{out}$

$$\text{LMTD} = \frac{50 - (350 - T_{out})}{\ln\left(\frac{50}{350 - T_{out}}\right)} = \frac{T_{out} - 300}{\ln\left(\frac{50}{350 - T_{out}}\right)}$$

Substituting into the equation:

$$600,000 \times \frac{T_{out} - 300}{\ln\left(\frac{50}{350 - T_{out}}\right)} = 400,000 \times (T_{out} - 300)$$

$$\frac{600,000}{\ln\left(\frac{50}{350 - T_{out}}\right)} = 400,000$$

$$\ln\left(\frac{50}{350 - T_{out}}\right) = \frac{600,000}{400,000} = 1.5$$

$$\frac{50}{350 - T_{out}} = e^{1.5} = 4.4817$$

$$350 - T_{out} = \frac{50}{4.4817} = 11.16$$

$$T_{out} = 350 - 11.16 = 338.84 \text{ K}$$

Answer: The temperature of the coolant water coming out of the condenser is 339 K.