Consider the flow of saturated steam at \(270.1 \mathrm{kPa}\) that flows through the shell side of a shell-and-tube heat exchanger while the water flows through 4 tubes of diameter \(1.25 \mathrm{~cm}\) at a rate of \(0.25 \mathrm{~kg} / \mathrm{s}\) through each tube. The water enters the tubes of heat exchanger at \(20^{\circ} \mathrm{C}\) and exits at \(60^{\circ} \mathrm{C}\). Due to the heat exchange with the cold fluid, steam is condensed on the tubes external surface. The convection heat transfer coefficient on the steam side is \(1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the fouling resistance for the steam and water may be taken as \(0.00015\) and \(0.0001 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), respectively. Using the NTU method, determine \((a)\) effectiveness of the heat exchanger, \((b)\) length of the tube, and \((c)\) rate of steam condensation.

First, we need to determine the heat capacity rate of the water. The heat capacity rate of the water (\(\dot{C}_{w}\)) can be evaluated using the water flow rate (\(\dot{m}_{w}\)) and the specific heat (\(c_{p,w}\)) of the water at constant pressure: $$\dot{C}_{w}=\dot{m}_{w} c_{p,w}$$ The water flow rate through each tube is given as \(0.25 kg/s\). Since there are 4 tubes, the total water flow rate is: $$\dot{m}_{w}=0.25 kg/s \times 4 = 1 kg/s$$ The specific heat of the water can be approximated as \(4.18 kJ/kg \cdot K = 4180 W/kg \cdot K\). Now we can calculate the heat capacity rate of the water: $$\dot{C}_{w}= 1 kg/s \times 4180 W/kg \cdot K = 4180 W/K$$ Next, we need to determine the heat capacity rate of the steam (\(\dot{C}_{s}\)). We can use the enthalpy of saturated steam at the given pressure (\(270.1 kPa\)) and assuming a saturated liquid in the outlet. Use steam tables to get enthalpy values for saturated steam \(h_{v}\) and saturated liquid \(h_{f}\): $$h_{v}=2706.7 kJ/kg$$ $$h_{f}= 698.10 kJ/kg$$ The heat capacity rate of the steam can be calculated as: $$\dot{C}_{s}=\dot{m}_{s} \times (h_{v} - h_{f})$$ where \(\dot{m}_{s}\) is the steam mass flow rate. At this point, we still have an unknown variable, \(\dot{m}_{s}\), which we will determine in subsequent steps.

We are now going to determine the overall heat transfer coefficient for the heat exchanger (\(U_{o}\)). We are given the fouling resistance (\(R_{f, w}\) and \(R_{f, s}\)) and the convection heat transfer coefficient (\(h_{s}\)): $$\frac{1}{U_{o}}=\frac{1}{h_{s}}+R_{f, s}+R_{f, w}$$ Substituting the given values, we get: $$\frac{1}{U_{o}}=\frac{1}{1500} + 0.00015 + 0.0001$$ Now, we can solve for the overall heat transfer coefficient: $$U_{o}=\frac{1}{\frac{1}{1500}+0.00015+0.0001} \approx 1273.24 W/m^2 \cdot K$$

In order to determine the effectiveness of the heat exchanger using the NTU method, we first need to calculate the NTU (Number of Transfer Units). The expression for NTU is: $$\text{NTU}=\frac{U_{o}A}{\dot{C}_{min}}$$ In this case, the minimum heat capacity rate is the heat capacity rate of the water: $$\dot{C}_{min}=\dot{C}_{w}=4180 W/K$$ Since we don't have the value for the area (\(A\)) at this point, the NTU is unknown. We can, however, calculate the effectiveness-NTU. For a shell-and-tube heat exchanger with one shell pass and an arbitrary number of tube passes, the effectiveness-NTU relationship is given by: \(\epsilon = \frac{1 - e^{-\text{NTU}(1-\text{CR})}}{1-\text{CR}e^{-\text{NTU}(1-\text{CR})}}\) where \(\epsilon\) is the effectiveness and CR is the capacity rate ratio: $$\text{CR}=\frac{\dot{C}_{min}}{\dot{C}_{max}}$$ Since the heat capacity rate of the steam is yet to be determined, we can't calculate the effectiveness directly. We, however, have the inlet and outlet temperatures of water, and we can use the definition of effectiveness to find the relationship between \(\dot{m}_{s}\) and \(\epsilon\). So, $$\epsilon = \frac{\dot{Q}_{act}}{\dot{Q}_{max}}$$ Where \(\dot{Q}_{act}\) is the actual heat transfer rate and \(\dot{Q}_{max}\) is the maximum heat transfer rate. Rewriting the equation in terms of steam and water flow rates, $${\epsilon} = \frac{\dot{m}_{s}(h_v-h_f)}{\dot{m}_{w} C_{p,w}(T_{2w} - T_{1w})}$$ Where \(T_{1w}\) and \(T_{2w}\) are the inlet and outlet temperatures of the water, respectively. Substituting the known values, $$\epsilon=\frac{\dot{m}_s(2706.7-698.1)}{1\times 4180\times (60-20)}$$

Now, we can calculate the actual rate of heat transfer (\(\dot{Q}_{act}\)) using the outlet and inlet temperature of the water and the heat capacity rate of the water: $$\dot{Q}_{act}=\dot{C}_{w}(T_{2w} - T_{1w})$$ Substituting the known values, we get: $$\dot{Q}_{act}=4180 \times (60-20) = 167200 W$$

Now, we can determine the length of the tube (\(L\)) by using the heat transfer rate, the overall heat transfer coefficient, the temperature difference between the steam and water, and the area of the tube (\(A\)). The expression is: $$\dot{Q}_{act}=U_{o}A\Delta{T}_{lm}$$ Where \(\Delta{T}_{lm}\) is the log mean temperature difference, calculated as: $$\Delta{T}_{lm}=\frac{(\Delta{T}_1-\Delta{T}_2)}{\ln{(\Delta{T}_1/\Delta{T}_2)}}$$ In this case, \(\Delta{T}_1=T_{s}-T_{1w}\) and \(\Delta{T}_2=T_{s}-T_{2w}\). Assume \(T_s\) is the saturation temperature of steam at the given pressure, approximately \(134.2^{\circ}C\). Then, $$\Delta{T}_{lm}=\frac{(134.2-20)-(134.2-60)}{\ln{((134.2-20)/(134.2-60))}}\approx 69.77^{\circ}C$$ Now, we can solve for the area of the tube (\(A\)): $$A=\frac{\dot{Q}_{act}}{U_{o}\Delta{T}_{lm}}=\frac{167200}{1273.24\times 69.77} \approx 1.870 m^2$$ We can calculate the area of a single tube using the diameter of the tube \(D_{t}\). $$A_t=\pi D_{t}L$$ Since there are 4 tubes in total, the total area is: $$A=4A_t$$ Now we can solve for the length of the tube: $$L=\frac{A}{4\pi D_{t}}=\frac{1.870}{4\pi \times 0.0125}\approx 9.96 m$$

Finally, we can calculate the rate of steam condensation (\(\dot{m}_s\)) using the rate of heat transfer and the enthalpy difference between the steam and water: $$\dot{m}_s=\frac{\dot{Q}_{act}}{h_v-h_f}=\frac{167200}{2706.7-698.1}\approx 0.086 kg/s$$ Now, we can substitute this value back into the expression for the effectiveness to find the actual effectiveness: $$\epsilon=\frac{\dot{m}_s(2706.7-698.1)}{1\times 4180\times (60-20)}$$ $$\epsilon=\frac{0.086\times (2706.7-698.1)}{4180\times 40}\approx 0.712$$ In summary, the results are as follows: (a) The effectiveness of the heat exchanger is approximately \(0.712\). (b) The length of the tube is approximately \(9.96 m\). (c) The rate of steam condensation is approximately \(0.086 kg/s\).