Question

In a one-shell and eight-tube-pass heat exchanger, the temperature of water flowing at rate of \(50,000 \mathrm{lbm} / \mathrm{h}\) is raised from \(70^{\circ} \mathrm{F}\) to \(150^{\circ} \mathrm{F}\). Hot air $\left(c_{p}=0.25 \mathrm{Btu} / \mathrm{bm}{ }^{\circ} \mathrm{F}\right)$ that flows on the tube side enters the heat exchanger at \(600^{\circ} \mathrm{F}\) and exits at \(300^{\circ} \mathrm{F}\). If the convection heat transfer coefficient on the outer surface of the tubes is $30 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2},{ }^{\circ} \mathrm{F}$, determine the surface area of the heat exchanger using both LMTD and \(\varepsilon-\mathrm{NTU}\) methods. Account for the possible fouling resistance of \(0.0015\) and $0.001 \mathrm{~h} \cdot \mathrm{ft}^{2}+{ }^{\circ} \mathrm{F} /$ Btu on the water and air sides, respectively.

Short answer

Question: Determine the surface area of a one-shell and eight-tube-pass heat exchanger, where water is heated by hot air. Include the fouling resistance for both the water and the air sides in your calculations. Answer: Using both the LMTD and ε-NTU methods, the surface areas obtained are 509.18 ft² and 711.36 ft², respectively. These values differ due to the assumptions made in each method and their corresponding level of accuracy.

Step-by-step solution

Calculate the heat transfer rate

We can calculate the heat transfer rate (Q) using the mass flow rate, specific heat, and temperature difference of the water: \(Q = m_w c_{p,w}(T_{w,out} - T_{w,in})\) Where: \(m_w = 50,000 \, \text{lbm/h}\) \(c_{p,w} = 1 \, \text{Btu/lbm}^\circ\text{F}\) (assuming water has a constant specific heat capacity of 1 Btu/lbm°F) \(T_{w,out} = 150^\circ \text{F}\) \(T_{w,in} = 70^\circ \text{F}\) \(Q = 50,000 \times 1 \times (150 - 70) = 4,000,000 \, \text{Btu/h}\)

Calculate the overall heat transfer coefficient

The overall heat transfer coefficient (U) can be calculated by accounting for the fouling resistance on both the water and air sides: \(\frac{1}{U} = \frac{1}{h} + R_{foul,water} + R_{foul,air}\) Where: \(h = 30 \, \text{Btu/h}\cdot\text{ft}^2\,^\circ\text{F}\) \(R_{foul,water} = 0.0015 \, \text{h}\cdot\text{ft}^2\,^\circ\text{F/Btu}\) \(R_{foul,air} = 0.001 \, \text{h}\cdot\text{ft}^2\,^\circ\text{F/Btu}\) \(\frac{1}{U} = \frac{1}{30} + 0.0015 + 0.001 = 0.036\, \text{h}\cdot\text{ft}^2\,^\circ\text{F/Btu}\) \(U = \frac{1}{0.036} = 27.78 \, \text{Btu/h}\cdot\text{ft}^2\,^\circ\text{F}\)

Calculate the LMTD

The LMTD can be calculated using the following formula: \(LMTD = \frac{(T_{h,in} - T_{w,out}) - (T_{h,out} - T_{w,in})}{\ln{\frac{T_{h,in} - T_{w,out}}{T_{h,out} - T_{w,in}}}}\) Where: \(T_{h,in} = 600^\circ \text{F}\) \(T_{h,out} = 300^\circ \text{F}\) \(LMTD = \frac{(600-150) - (300-70)}{\ln{\frac{600-150}{300-70}}} = 286.25^\circ \text{F}\)

Determine the surface area using the LMTD method

We can determine the surface area (A) required using the following equation: \(Q = U \cdot A \cdot LMTD\) Rearranging for the surface area, we have: \(A = \frac{Q}{U \cdot LMTD} = \frac{4,000,000}{27.78 \times 286.25} = 509.18 \, \text{ft}^2\)

Calculate the ε and NTU

We need to calculate the heat capacity rates for both water and hot air: \(C_w = m_w c_{p,w} = 50,000 \times 1 = 50,000 \, \text{Btu/h}^\circ\text{F}\) \(C_h = m_h c_{p,h} = \frac{Q}{c_{p,h}(T_{h,in} - T_{h,out})} = \frac{4,000,000}{0.25 \times (600 - 300)} = 26,666.67 \, \text{Btu/h}^\circ\text{F}\) Also, we need to calculate the minimum heat capacity rate, \(C_{min}\), the effectiveness, ε, and the NTU: \(C_{min} = \min(C_w, C_h) = 26,666.67 \, \text{Btu/h}^\circ\text{F}\) \(\varepsilon = \frac{Q}{C_{min}(T_{h,in} - T_{w,in})} = \frac{4,000,000}{26,666.67 \times (600-70)} = 0.25\) For the given heat exchanger configuration (1-8 exchanger), the NTU relation is: \(\varepsilon = 1 - \exp{\left[\frac{1}{8}\left(1 - \exp{-8\cdot NTU \cdot \frac{C_{min}}{C_{max}}}\right)\right]}\) Rearranging for the NTU and iterating, we get: \(NTU = 0.745\)

Determine the surface area using the ε-NTU method

We can determine the surface area required using the following equation: \(NTU = \frac{U \cdot A}{C_{min}}\) Rearranging for the surface area, we have: \(A = \frac{NTU \times C_{min}}{U} = \frac{0.745 \times 26,666.67}{27.78} = 711.36 \, \text{ft}^2\) Using both the LMTD and ε-NTU methods, we have obtained: Surface area (LMTD method): \(509.18 \, \text{ft}^2\) Surface area (ε-NTU method): \(711.36 \, \text{ft}^2\) These values differ due to the assumptions made in each method and their corresponding level of accuracy. The LMTD method assumes a constant overall heat transfer coefficient, while the ε-NTU method accounts for variations in local temperature differences and heat transfer coefficients.