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 ______________ K (round off to the nearest integer).

The correct answer is: 338.84 (or 339 if rounded to the nearest integer).

1. Identifying Given Data from the Image and Problem Statement:
From the provided image showing the step-by-step parameters, we have the following values:
- Coolant water inlet temperature:
$T_{\mathrm{ci}}=300\text{K}$
- Mass flow rate of coolant:
${\dot{m}}_c=100\text{kg/s}$
- Overall heat transfer coefficient:
$U=1500\text{W/(m}{2}^{\mathrm{}}·\text{K)}$
- Total heat transfer area:
$A=400{\text{m}}^2$
- Specific heat of coolant water:
$c_{\mathrm{pc}}=4000\text{J/(kg}·\text{K)}$
- Condensing steam saturation temperature (hot fluid inlet and outlet temperature):
$T_{\mathrm{sat}}=T_{\mathrm{hi}}=T_{\mathrm{he}}=350\text{K}$

2. Determination of Heat Capacity Rates:
For a condenser, the steam undergoes a phase change (condensation) at constant temperature. This means its effective specific heat capacity is infinitely large, so:
$C_{\mathrm{ph}}\to \infty =C_{\max }$
Therefore, the minimum heat capacity rate corresponds to the coolant water:
$C_{\min }={\dot{m}}_c\times c_{\mathrm{pc}}=100\times 4000=\mathrm{400,000}\text{W/K}$

3. Calculation of Number of Transfer Units (NTU):
The NTU is defined as:
$\mathrm{NTU}=\frac{UA}{C_{\min }}$
Substituting the given values:
$\mathrm{NTU}=\frac{1500\times 400}{\mathrm{400,000}}=1.5$

4. Calculating Heat Exchanger Effectiveness (ε):
For a heat exchanger where one fluid undergoes condensation (meaning the capacity ratio $R=\frac{C_{\min }}{C_{\max }}=0$), the effectiveness formula simplifies to:
$\epsilon =1-e^{-\mathrm{NTU}}$
Substituting NTU = 1.5:
$\epsilon =1-e^{-1.5}\approx 1-0.22313=0.77687$

5. Determining the Coolant Water Outlet Temperature (T_ce):
The effectiveness is also defined by temperature changes as:
$\epsilon =\frac{T_{\mathrm{ce}}-T_{\mathrm{ci}}}{T_{\mathrm{hi}}-T_{\mathrm{ci}}}$
Substituting our known values to solve for T_ce:
$1-e^{-1.5}=\frac{T_{\mathrm{ce}}-300}{350-300}$
$T_{\mathrm{ce}}-300=50\times (1-e^{-1.5})$
$T_{\mathrm{ce}}=300+50\times 0.77687$
$T_{\mathrm{ce}}=300+38.8434=338.8434\text{K}$
Rounding off to two decimal places gives 338.84 K.