Question

A two-shell-pass and four-tube-pass heat exchanger is used for heating a hydrocarbon stream $\left(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\( steadily from \)20^{\circ} \mathrm{C}\( to \)50^{\circ} \mathrm{C}\(. A water stream enters the shell side at \)80^{\circ} \mathrm{C}$ and leaves at \(40^{\circ} \mathrm{C}\). There are 160 thin-walled tubes, each with a diameter of \(2.0 \mathrm{~cm}\) and length of \(1.5 \mathrm{~m}\). The tube-side and shell-side heat transfer coefficients are \(1.6\) and $2.5 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}$, respectively. (a) Calculate the rate of heat transfer and the mass rates of water and hydrocarbon streams. (b) With usage, the outlet hydrocarbon-stream temperature was found to decrease by \(5^{\circ} \mathrm{C}\) due to the deposition of solids on the tube surface. Estimate the magnitude of the fouling factor.

Short answer

Answer: The heat transfer rate is 60 kW, the mass flow rate of the hydrocarbon stream is 1 kg/s, the mass flow rate of the water stream is 0.359 kg/s, and the fouling factor is 0.000332 m²·K/W.

Step-by-step solution

Energy balance equation

To find the heat transfer rate (Q), we will use the energy balance equation: Q = m_h * c_p * ∆T_h = m_w * c_p_w * ∆T_w, where m_h and m_w are the mass flow rates of hydrocarbon and water, respectively, and ∆T_h and ∆T_w are the temperature differences of hydrocarbon and water, respectively. We know the heat capacity of hydrocarbon (c_p) and of water (c_p_w) is 4.18 kJ/kg·K.

Calculate temperature differences

Calculate the temperature differences for both streams: ∆T_h = T_h_final - T_h_initial = 50 - 20 = 30°C, and ∆T_w = T_w_initial - T_w_final = 80 - 40 = 40°C.

Calculate heat transfer rate

As Q is the same for both streams, we can solve for it using one of them. Let's use the hydrocarbon stream and assume its mass flow rate (m_h) is 1 kg/s for now: Q = m_h * c_p * ∆T_h = 1 kg/s * 2.0 kJ/kg·K * 30 K = 60 kJ/s or 60 kW. We will use this value later to find the actual mass flow rates of both streams.

Calculate mass flow rates

Now we can use Q to calculate the mass flow rates of the water and hydrocarbon streams. For hydrocarbon: m_h = Q / (c_p * ∆T_h) = 60 kW / (2.0 kJ/kg·K * 30 K) = 1 kg/s (as assumed before). For water: m_w = Q / (c_p_w * ∆T_w) = 60 kW / (4.18 kJ/kg·K * 40 K) ≈ 0.359 kg/s.

Overall Heat Transfer Coefficient

Calculate the overall heat transfer coefficient (U) using the tube and shell side heat transfer coefficients: 1/U = 1/h_t + 1/h_s, where h_t and h_s are the tube-side and shell-side heat transfer coefficients, respectively. Plugging in the given values, we get: 1/U = 1/1.6 + 1/2.5 ≈ 0.633, which gives U ≈ 1.58 kW/m²·K.

New Heat Transfer rate

With the fouling factor (R_f), the heat transfer rate equation becomes: Q = (U * A * ∆T_lm) / (1 + R_f * U * A), where A is the total heat transfer area, and ∆T_lm is the log mean temperature difference. Since the outlet hydrocarbon-stream temperature decreases by 5°C, we have a new temperature difference value (∆T'_h = 25 K). Calculate the new log mean temperature difference (∆T'_lm = ((∆T'_h - ∆T_w) - (∆T_h - ∆T_w)) / ln((∆T'_h - ∆T_w) / (∆T_h - ∆T_w)) ≈ 6.76 K. The new heat transfer rate (Q') is given by: Q' = m_h * c_p * ∆T'_h = 1 kg/s * 2.0 kJ/kg·K * 25 K = 50 kW.

Calculate the fouling factor

Now, we can use the new heat transfer rate and the heat transfer rate equation (with the fouling factor) to find the fouling factor (R_f): 50 kW = (1.58 kW/m²·K * A * 6.76 K) / (1 + R_f * 1.58 kW/m²·K * A). We know A = N * π * d * L, with N = 160 tubes, d = 2 cm, and L = 1.5 m. Plugging in the values, we get A ≈ 15.1 m². Now we can solve for R_f: R_f ≈ ((60 kW - 50 kW) / (1.58 kW/m²·K * 15.1 m² * 6.76 K)) ≈ 0.000332 m²·K/W. The answers are: (a) heat transfer rate = 60 kW, mass flow rate of hydrocarbon stream = 1 kg/s, mass flow rate of water stream = 0.359 kg/s; (b) fouling factor = 0.000332 m²·K/W.