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Chapter 11: Problem 57

Geothermal water \(\left(c_{p}=1.03 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) is to be used as the heat source to supply heat to the hydronic heating system of a house at a rate of \(40 \mathrm{Btu} / \mathrm{s}\) in a double-pipe counter-flow heat exchanger. Water \(\left(c_{p}=1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) is heated from \(140^{\circ} \mathrm{F}\) to \(200^{\circ} \mathrm{F}\) in the heat exchanger as the geothermal water is cooled from \(270^{\circ} \mathrm{F}\) to \(180^{\circ} \mathrm{F}\). Determine the mass flow rate of each fluid and the total thermal resistance of this heat exchanger.

Short Answer

Expert verified
Question: Calculate the mass flow rates of the hydronic water and the geothermal water, as well as the total thermal resistance of the heat exchanger. Short Answer: The mass flow rate of the hydronic water is 0.6667 lbm/s, the mass flow rate of the geothermal water is 0.4316 lbm/s, and the total thermal resistance of the heat exchanger is 0.75 °F/Btu.

Step by step solution

01

Find the mass flow rate of hydronic water

We are given the heat transfer rate and the temperature change of the hydronic water. To find the mass flow rate of hydronic water, we use the formula: Q = m_hw * c_p_hw * (T_out_hw - T_in_hw) Where m_hw is the mass flow rate of hydronic water, c_p_hw is the specific heat capacity of hydronic water, T_out_hw is the outlet temperature of hydronic water, and T_in_hw is the inlet temperature of hydronic water. From the given information, we have: Q = 40 Btu/s c_p_hw = 1.0 Btu/(lbm * °F) T_in_hw = 140°F T_out_hw = 200°F Rearrange the formula to solve for m_hw: m_hw = Q / (c_p_hw * (T_out_hw - T_in_hw)) m_hw = 40 / (1.0 * (200 - 140)) m_hw = 40 / 60 m_hw = 0.6667 lbm/s
02

Find the mass flow rate of geothermal water

Using the energy balance principle, the heat transferred to the hydronic water must be equal to the heat extracted from the geothermal water. Therefore, we use the formula: Q = m_gw * c_p_gw * (T_in_gw - T_out_gw) Where m_gw is the mass flow rate of geothermal water, c_p_gw is the specific heat capacity of geothermal water, and T_in_gw and T_out_gw are the inlet and outlet temperatures of geothermal water. From the given information, we have: c_p_gw = 1.03 Btu/(lbm * °F) T_in_gw = 270°F T_out_gw = 180°F Rearrange the formula to solve for m_gw: m_gw = Q / (c_p_gw * (T_in_gw - T_out_gw)) m_gw = 40 / (1.03 * (270 - 180)) m_gw = 40 / 92.7 m_gw = 0.4316 lbm/s
03

Calculate the total thermal resistance

To find the total thermal resistance of the heat exchanger (R_th), we use the formula: R_th = (T_in_gw - T_out_gw) / Q - (T_out_hw - T_in_hw) / Q Where Q is the heat transfer rate, and the temperatures are defined as before. From the given information, we have: Q = 40 Btu/s T_in_gw = 270°F T_out_gw = 180°F T_in_hw = 140°F T_out_hw = 200°F Substitute the values into the formula: R_th = (270 - 180) / 40 - (200 - 140) / 40 R_th = 90 / 40 - 60 / 40 R_th = 2.25 - 1.50 R_th = 0.75 °F/Btu The mass flow rate of the hydronic water is 0.6667 lbm/s, the mass flow rate of the geothermal water is 0.4316 lbm/s, and the total thermal resistance of the heat exchanger is 0.75 °F/Btu.

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Most popular questions from this chapter

A shell-and-tube heat exchanger with 2-shell passes and 8 -tube passes is used to heat ethyl alcohol \(\left(c_{p}=2670 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) in the tubes from \(25^{\circ} \mathrm{C}\) to \(70^{\circ} \mathrm{C}\) at a rate of \(2.1 \mathrm{~kg} / \mathrm{s}\). The heating is to be done by water \(\left(c_{p}=4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the shell side at \(95^{\circ} \mathrm{C}\) and leaves at \(45^{\circ} \mathrm{C}\). If the overall heat transfer coefficient is \(950 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the heat transfer surface area of the heat exchanger.

Cold water \(\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters a cross-flow heat exchanger at \(14^{\circ} \mathrm{C}\) at a rate of \(0.35 \mathrm{~kg} / \mathrm{s}\) where it is heated by hot air \(\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the heat exchanger at \(65^{\circ} \mathrm{C}\) at a rate of \(0.8 \mathrm{~kg} / \mathrm{s}\) and leaves at \(25^{\circ} \mathrm{C}\). Determine the maximum outlet temperature of the cold water and the effectiveness of this heat exchanger.

Consider a closed loop heat exchanger that carries exit water \(\left(c_{p}=1 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right.\) and \(\left.\rho=62.4 \mathrm{lbm} / \mathrm{ft}^{3}\right)\) of a condenser side initially at \(100^{\circ} \mathrm{F}\). The water flows through a \(500 \mathrm{ft}\) long stainless steel pipe of 1 in inner diameter immersed in a large lake. The temperature of lake water surrounding the heat exchanger is \(45^{\circ} \mathrm{F}\). The overall heat transfer coefficient of the heat exchanger is estimated to be \(250 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\). What is the exit temperature of the water from the immersed heat exchanger if it flows through the pipe at an average velocity of \(9 \mathrm{ft} / \mathrm{s}\) ? Use \(\varepsilon-\mathrm{NTU}\) method for analysis.

A shell-and-tube heat exchanger is to be designed to cool down the petroleum- based organic vapor available at a flow rate of \(5 \mathrm{~kg} / \mathrm{s}\) and at a saturation temperature of \(75^{\circ} \mathrm{C}\). The cold water \(\left(c_{p}=4187 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) used for its condensation is supplied at a rate of \(25 \mathrm{~kg} / \mathrm{s}\) and a temperature of \(15^{\circ} \mathrm{C}\). The cold water flows through copper tubes with an outside diameter of \(20 \mathrm{~mm}\), a thickness of \(2 \mathrm{~mm}\), and a length of \(5 \mathrm{~m}\). The overall heat transfer coefficient is assumed to be \(550 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the latent heat of vaporization of the organic vapor may be taken to be \(580 \mathrm{~kJ} / \mathrm{kg}\). Assuming negligible thermal resistance due to pipe wall thickness, determine the number of tubes required.

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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