Question

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 four 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 the 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 tube's 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 \(\mathrm{NTU}\) method, determine \((a)\) effectiveness of the heat exchanger, (b) length of the tube, and (c) rate of steam condensation.

Short answer

Answer: The effectiveness of the heat exchanger is 1, the length of the tubes is 287.01 m, and the rate of steam condensation is 0.0754 kg/s.

Step-by-step solution

Find Water Temperature Range

We are given the water inlet and outlet temperature, $$T_{wi}=20^{\circ} \mathrm{C}$$ and $$T_{wo}=60^{\circ} \mathrm{C}$$. The temperature range of water is then: $$\Delta T_\mathrm{w}=T_{wo}-T_{wi}=60-20=40^{\circ} \mathrm{C}$$

Calculate Heat Exchanged

We can find the heat exchanged between the steam and water side using the mass flow rate and specific heat of water: $$Q=m_\mathrm{w} \times c_\mathrm{p, w} \times \Delta T_\mathrm{w}$$ Since there are four tubes with a mass flow rate of \(0.25 \mathrm{~kg} / \mathrm{s}\) each, the total mass flow rate is: $$m_\mathrm{total}=0.25 \times 4 = 1 \mathrm{~kg} / \mathrm{s}$$ Assuming constant specific heat for water at \(c_\mathrm{p, w} =4180 \mathrm{~J}/\mathrm{kg}\cdot\mathrm{K}\), the heat exchanged is: $$Q=1\times4180\times40 = 167,200 \mathrm{~W}$$

Calculate Overall Heat Transfer Coefficient

Using the given values, the overall heat transfer coefficient U can be found as follows: $$\frac{1}{U}=\frac{1}{h_\mathrm{s}}+R_\mathrm{fs}+R_\mathrm{ft}$$ Where $$h_\mathrm{s}=1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$$ is the steam side heat transfer coefficient, $$R_\mathrm{fs}=0.00015 \mathrm{~m}^{2}\cdot \mathrm{K} / \mathrm{W}$$ is the steam side fouling resistance, and $$R_\mathrm{ft}=0.0001 \mathrm{~m}^{2}\cdot \mathrm{K} / \mathrm{W}$$ is the tube side fouling resistance. The overall heat transfer coefficient, U, is: $$U=\frac{1}{\frac{1}{1500}+0.00015+0.0001}=1186.39 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$$

Calculate NTU and Effectiveness

Using NTU method: $$NTU=\frac{U \times A}{m_\mathrm{total} \times c_\mathrm{p,w}}$$ And for a shell and tube heat exchanger with one shell pass and an even number of tube passes, the effectiveness, \(\epsilon\), can be calculated as: $$\epsilon= \frac{1-\exp(-2NTU(\frac{m_\mathrm{min}}{m_\mathrm{max}}))}{2-(\frac{m_\mathrm{min}}{m_\mathrm{max}})\times\exp(-2NTU(\frac{m_\mathrm{min}}{m_\mathrm{max}}))}$$ Since there is no steam-side mass flow rate, we can simplify the effectiveness equation as: $$\epsilon=\frac{1-\exp(-2NTU)}{2+\exp(-2NTU)}$$ Now, also that: $$Q_\mathrm{max}=m_\mathrm{total} \times c_\mathrm{p,w} \times(\Delta T_\mathrm{w})$$ Then, the effectiveness is: $$\epsilon=\frac{Q}{Q_\mathrm{max}}=\frac{167,200}{167,200}=1$$

Calculate Length of Tubes

Now we know the effectiveness, we can find the heat exchanger area, A: $$A=\frac{m_\mathrm{total} \times c_\mathrm{p,w} \times \Delta T_\mathrm{w}}{U \times\epsilon}$$ $$A=\frac{1\times4180\times40}{1186.39\times1}=140.56 \mathrm{~m}^2$$ The tube's perimeter, P, and cross-sectional area of the flow: $$P=\pi D_\mathrm{t}=\pi\times0.0125=0.03927 \mathrm{~m}$$ $$A_\mathrm{tube}= \frac{\pi D_\mathrm{tube}^2}{4} = \frac{\pi (1.25 \times 10^{-2})^2}{4} = 1.227 \times 10^{-4} \mathrm{~m}^2$$ The total heat transfer area considering all four tubes: $$A_\mathrm{total}=4\times A_\mathrm{tube} \times L$$ Where L is the length of tubes. We can find L as follows: $$L=\frac{A_\mathrm{total}}{4 \times A_\mathrm{tube}}$$ $$L=\frac{140.56}{4\times1.227\times10^{-4}}=287.01 \mathrm{~m}$$

Calculate Rate of Steam Condensation

The rate of steam condensation, \(m_\mathrm{s}\) is given by the following equation: $$Q=m_\mathrm{s}\times h_\mathrm{fg}$$ \(h_\mathrm{fg}\) is the latent heat of condensation at the given pressure \(270.1\,\mathrm{kPa}\). By checking the steam table for the corresponding value at this pressure, we find \(h_\mathrm{fg} = 2217 \,\mathrm{kJ} / \mathrm{kg}\). Now, solving for \(m_\mathrm{s}\): $$m_\mathrm{s} =\frac{Q}{h_\mathrm{fg}}=\frac{167,200}{2217\times10^3}=0.0754\, \mathrm{kg/s}$$ The answers are: (a) Effectiveness of the heat exchanger is \(1\), (b) Length of the tubes is \(287.01\,\mathrm{m}\), and (c) Rate of steam condensation is \(0.0754\, \mathrm{kg/s}\).