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Chapter 11: Problem 149

Hot water at \(60^{\circ} \mathrm{C}\) is cooled to \(36^{\circ} \mathrm{C}\) through the tube side of a 1-shell pass and 2-tube passes heat exchanger. The coolant is also a water stream, for which the inlet and outlet temperatures are \(7^{\circ} \mathrm{C}\) and \(31^{\circ} \mathrm{C}\), respectively. The overall heat transfer coefficient and the heat transfer area are \(950 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(15 \mathrm{~m}^{2}\), respectively. Calculate the mass flow rates of hot and cold water streams in steady operation.

Short Answer

Expert verified
Answer: The mass flow rate of hot water (m_h) is 15.579 kg/s, and the mass flow rate of cold water (m_c) is 18.096 kg/s.

Step by step solution

01

Calculate the log mean temperature difference (ΔT_lm)

To find the log mean temperature difference, we use the given temperatures of the hot and cold water streams. We have the following temperatures: - Hot water inlet: 60°C - Hot water outlet: 36°C - Cold water inlet: 7°C - Cold water outlet: 31°C We can calculate the log mean temperature difference using the formula: ΔT_lm = (ΔT_1 - ΔT_2) / ln(ΔT_1 / ΔT_2), where ΔT_1 and ΔT_2 are the temperature differences at the two ends of the heat exchanger. Here, ΔT_1 = 60 - 7 = 53°C, and ΔT_2 = 36 - 31 = 5°C. So, we get: ΔT_lm = (53 - 5) / ln(53 / 5) ≈ 21.65°C.
02

Calculate the heat transfer rate (Q)

Now, we can find the heat transfer rate (Q) using the formula: Q = U * A * ΔT_lm Here, U = 950 W/m²·K, A = 15 m², and ΔT_lm = 21.65°C. So, we get: Q = 950 * 15 * 21.65 ≈ 308875 W = 308.875 kJ/s.
03

Calculate the mass flow rate of hot water (m_h)

To find the mass flow rate of hot water, we can use the energy balance equation for the hot water stream: Q = m_h * Cp * (T_inlet - T_outlet) Here, Cp ≈ 4.18 kJ/kg·K, T_inlet = 60°C, and T_outlet = 36°C. Rearranging the equation for m_h, we get: m_h = Q / (Cp * (T_inlet - T_outlet)) Now substituting the values, we get: m_h = 308.875 / (4.18 * (60 - 36)) ≈ 15.579 kg/s.
04

Calculate the mass flow rate of cold water (m_c)

Similarly, to find the mass flow rate of cold water, we can use the energy balance equation for the cold water stream: Q = m_c * Cp * (T_outlet - T_inlet) Here, T_outlet = 31°C and T_inlet = 7°C. Rearranging the equation for m_c, we get: m_c = Q / (Cp * (T_outlet - T_inlet)) Now substituting the values, we get: m_c = 308.875 / (4.18 * (31 - 7)) ≈ 18.096 kg/s. Finally, we have calculated the mass flow rates for both hot and cold water streams: - Mass flow rate of hot water (m_h) = 15.579 kg/s - Mass flow rate of cold water (m_c) = 18.096 kg/s

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Most popular questions from this chapter

Can the temperature of the cold fluid rise above the inlet temperature of the hot fluid at any location in a heat exchanger? Explain.

Saturated liquid benzene flowing at a rate of \(5 \mathrm{~kg} / \mathrm{s}\) is to be cooled from \(75^{\circ} \mathrm{C}\) to \(45^{\circ} \mathrm{C}\) by using a source of cold water \(\left(c_{p}=4187 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) flowing at \(3.5 \mathrm{~kg} / \mathrm{s}\) and \(15^{\circ} \mathrm{C}\) through a \(20-\mathrm{mm}-\) diameter tube of negligible wall thickness. The overall heat transfer coefficient of the heat exchanger is estimated to be \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the specific heat of the liquid benzene is \(1839 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and assuming that the capacity ratio and effectiveness remain the same, determine the heat exchanger surface area for the following four heat exchangers: \((a)\) parallel flow, \((b)\) counter flow, \((c)\) shelland-tube heat exchanger with 2 -shell passes and 40-tube passes, and \((d)\) cross-flow heat exchanger with one fluid mixed (liquid benzene) and other fluid unmixed (water).

11-100 E(S) Reconsider Prob. 11-99. Using EES (or other) software, investigate the effects of the inlet temperature of hot water and the heat transfer coefficient on the rate of heat transfer and the surface area. Let the inlet temperature vary from \(60^{\circ} \mathrm{C}\) to \(120^{\circ} \mathrm{C}\) and the overall heat transfer coefficient from \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) to \(1250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Plot the rate of heat transfer and surface area as functions of the inlet temperature and the heat transfer coefficient, and discuss the results. 11-101E A thin-walled double-pipe, counter-flow heat exchanger is to be used to cool oil \(\left(c_{p}=0.525 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) from \(300^{\circ} \mathrm{F}\) to \(105^{\circ} \mathrm{F}\) at a rate of \(5 \mathrm{lbm} / \mathrm{s}\) by water \(\left(c_{p}=\right.\) \(1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\) ) that enters at \(70^{\circ} \mathrm{F}\) at a rate of \(3 \mathrm{lbm} / \mathrm{s}\). The diameter of the tube is 5 in and its length is \(200 \mathrm{ft}\). Determine the overall heat transfer coefficient of this heat exchanger using (a) the LMTD method and \((b)\) the \(\varepsilon-\mathrm{NTU}\) method.

The cardiovascular counter-current heat exchanger mechanism is to warm venous blood from \(28^{\circ} \mathrm{C}\) to \(35^{\circ} \mathrm{C}\) at a mass flow rate of \(2 \mathrm{~g} / \mathrm{s}\). The artery inflow temperature is \(37^{\circ} \mathrm{C}\) at a mass flow rate of \(5 \mathrm{~g} / \mathrm{s}\). The average diameter of the vein is \(5 \mathrm{~cm}\) and the overall heat transfer coefficient is \(125 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the overall blood vessel length needed to warm the venous blood to \(35^{\circ} \mathrm{C}\) if the specific heat of both arterial and venous blood is constant and equal to \(3475 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

Consider a water-to-water counter-flow heat exchanger with these specifications. Hot water enters at \(95^{\circ} \mathrm{C}\) while cold water enters at \(20^{\circ} \mathrm{C}\). The exit temperature of hot water is \(15^{\circ} \mathrm{C}\) greater than that of cold water, and the mass flow rate of hot water is 50 percent greater than that of cold water. The product of heat transfer surface area and the overall heat transfer coefficient is \(1400 \mathrm{~W} / \mathrm{K}\). Taking the specific heat of both cold and hot water to be \(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), determine (a) the outlet temperature of the cold water, \((b)\) the effectiveness of the heat exchanger, \((c)\) the mass flow rate of the cold water, and \((d)\) the heat transfer rate.
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