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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.

Short Answer

Expert verified
Answer: The exit temperature of the water from the immersed heat exchanger is 54.35 °F.

Step by step solution

01

Calculate the mass flow rate of the water

To find the mass flow rate of the water, we will use the equation m_dot = ρ * A * v, where ρ is the density, A is the area, and v is the velocity of the water. ρ = 62.4 lbm/ft³ v = 9 ft/s Inner diameter of pipe = 1 in = 1/12 ft Area A of the pipe is calculated using the formula A = π * (D/2)², where D is the inner diameter. A = π * ((1/12)²/4) A = 0.00545 ft² Now, we can calculate the mass flow rate: m_dot = ρ * A * v m_dot = 62.4 * 0.00545 * 9 m_dot = 3.06 lbm/s
02

Calculate the capacity rate of the water

The capacity rate C (in Btu/h-°F) is the product of the mass flow rate and the specific heat of the water. C = m_dot * cp , where cp is the specific heat capacity of the water. cp = 1 Btu/lbm-°F C = 3.06 lbm/s * 1 Btu/lbm-°F C = 3.06 Btu/s-°F Convert to Btu/h-°F: C = 3.06 * 3600 Btu/h-°F = 11016 Btu/h-°F
03

Calculate the area of the heat exchanger

We are given the length L of the heat exchanger and the inner diameter D. We will use the formula A = π * D * L to find the area of the heat exchanger. L = 500 ft A = π * (1/12 ft) * 500 ft A = 130.9 ft²
04

Find the NTU

We are given the overall heat transfer coefficient U (250 Btu/h-ft²-°F). To find the NTU, use the formula NTU = UA/C. UA = 250 * 130.9 = 32725 Btu/h-°F NTU = 32725/11016 = 2.97
05

Calculate the effectiveness of the heat exchanger using ε-NTU method

We will assume that the heat exchanger is a simple parallel-flow type. In that case, the effectiveness ε can be calculated using the formula: ε = (1 - exp(-NTU * (1 - C_r)))/(1 - C_r * exp(-NTU * (1 - C_r))) Since the same fluid is used in this problem, we can assume C_h = C_c = C. Thus, C_r = C_h/C_c = 1. ε = (1 - exp(-2.97 * (1 - 1)))/(1 - 1 * exp(-2.97 * (1 - 1))) The formula has an indeterminate form in this case, so we use L'Hopital's rule to find the limit as C_r approaches 1: d(1 - exp(-NTU * (1 - C_r)))/dC_r = NTU * exp(-NTU * (1 - C_r)) d(1 - C_r * exp(-NTU * (1 - C_r)))/dC_r = exp(-NTU * (1 - C_r)) - C_r * NTU * exp(-NTU * (1 - C_r)) ε = (NTU * exp(-NTU))/(1 - NTU * exp(-NTU)) = 0.83
06

Calculate the exit temperature of the water

We can use the effectiveness to find the exit temperature of the water: T_exit = T_in - ε * (T_in - T_lake) T_exit = 100 - 0.83 * (100 - 45) = 54.35 °F The exit temperature of the water from the immersed heat exchanger is 54.35 °F.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Overall Heat Transfer Coefficient
The overall heat transfer coefficient (U) is indicative of how well a heat exchanger can transfer heat between fluids. It integrates the thermal resistances of the internal fluid, the exchanger material, and the external fluid, providing a single metric to represent the efficiency of heat transfer. A larger value of U signifies a better ability to transfer heat. This coefficient is crucial when designing or analyzing heat exchangers because it directly affects the temperature changes that fluids experience as they pass through the system. To calculate U, we often rely on experimental data or standards for similar setups since it is influenced by many factors, including fluid properties, flow rates, and exchanger dimensions.
ε-NTU Method
The \(\varepsilon-NTU\) method is a mathematical approach used in thermal engineering to analyze the thermal performance of heat exchangers. It correlates the effectiveness (\(\varepsilon\)) of the heat exchanger with the number of transfer units (NTU). Here, effectiveness is defined as the ratio of the actual heat transfer to the maximum possible heat transfer. NTU is a dimensionless number that represents the size or capacity of the heat exchanger in relation to the flow rates and heat capacities of the fluids involved. The formula for NTU depends on the overall heat transfer coefficient, the heat transfer area, and the capacity rate of the fluids. By employing the \(\varepsilon-NTU\) method, we can predict the performance of the heat exchanger without requiring complex and detailed heat transfer calculations for each point along its length.
Exit Temperature Calculation
In heat exchanger analysis, one of the key outcomes is the calculation of the exit temperature of the fluid after it has passed through the system. This value is vital for many engineering tasks, such as process control, safety analysis, and equipment sizing. The exit temperature is influenced by the entry temperature, the temperature of the surrounding fluid, the heat exchanger's effectiveness, and the specific heat properties of the fluid. To determine the exit temperature, we use the heat exchanger's effectiveness from the \(\varepsilon-NTU\) method and apply it to the temperature difference between the inlet fluid and the surrounding temperature, just as done in the given solution.
Mass Flow Rate
Mass flow rate is a measure of the amount of mass passing through a given cross-sectional area per unit time. It is symbolized by \(\dot{m}\) and typically expressed in terms like pounds per second (lbm/s) or kilograms per second (kg/s). The mass flow rate in a heat exchanger dictates how quickly heat can be transferred, as it directly affects the capacity of the fluid to absorb or dissipate heat. In calculations, it is determined by multiplying the velocity of the fluid by the cross-sectional area of the pipe or duct through which it flows and by the density of the fluid at the given conditions. A change in any of these variables will affect the mass flow rate and subsequently alter the performance of the heat exchanger.
Capacity Rate
The capacity rate of a fluid in a heat exchanger, often denoted as C, is the product of its mass flow rate and its specific heat capacity at constant pressure (\(c_p\)). This measurement is expressed in terms of energy per unit time per degree of temperature difference, like Btu/h-°F or J/s-K, and is key to understanding the heat exchange process. The capacity rate essentially quantifies the fluid's ability to carry heat. It helps in determining the thermal capacity of the fluid stream, affecting the temperature change the fluid can undergo. In any system, the fluid with the lower capacity rate limits the heat transfer rate since it undergoes a more significant temperature change for the same amount of heat transfer.
Specific Heat Capacity
Specific heat capacity (\(c_p\)) is a property of a material that indicates the amount of heat required to change the temperature of a unit mass of the substance by one degree (usually in °C or °F). It's a crucial parameter in thermodynamics as it relates to the internal energy of a system. The specific heat capacity determines how a temperature difference will translate into heat transfer, influencing the thermal inertia and the rate at which a fluid heats up or cools down. In the context of the heat exchanger problem, the specific heat capacity of water is used to calculate the capacity rate, which in turn feeds into the \(\varepsilon-NTU\) method to eventual predict the exit temperature of the water.

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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.

In a parallel-flow, liquid-to-liquid heat exchanger, the inlet and outlet temperatures of the hot fluid are \(150^{\circ} \mathrm{C}\) and \(90^{\circ} \mathrm{C}\) while that of the cold fluid are \(30^{\circ} \mathrm{C}\) and \(70^{\circ} \mathrm{C}\), respectively. For the same overall heat transfer coefficient, the percentage decrease in the surface area of the heat exchanger if counter-flow arrangement is used is (a) \(3.9 \%\) (b) \(9.7 \%\) (c) \(14.5 \%\) (d) \(19.7 \%\) (e) \(24.6 \%\)

Consider a water-to-water double-pipe heat exchanger whose flow arrangement is not known. The temperature measurements indicate that the cold water enters at \(20^{\circ} \mathrm{C}\) and leaves at \(50^{\circ} \mathrm{C}\), while the hot water enters at \(80^{\circ} \mathrm{C}\) and leaves at \(45^{\circ} \mathrm{C}\). Do you think this is a parallel-flow or counterflow heat exchanger? Explain.

A shell-and-tube heat exchanger with 2-shell passes and 4-tube passes is used for cooling oil \(\left(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) from \(125^{\circ} \mathrm{C}\) to \(55^{\circ} \mathrm{C}\). The coolant is water, which enters the shell side at \(25^{\circ} \mathrm{C}\) and leaves at \(46^{\circ} \mathrm{C}\). The overall heat transfer coefficient is \(900 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). For an oil flow rate of \(10 \mathrm{~kg} / \mathrm{s}\), calculate the cooling water flow rate and the heat transfer area.

Air \(\left(c_{p}=1005 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters a cross-flow heat exchanger at \(20^{\circ} \mathrm{C}\) at a rate of \(3 \mathrm{~kg} / \mathrm{s}\), where it is heated by a hot water stream \(\left(c_{p}=4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the heat exchanger at \(70^{\circ} \mathrm{C}\) at a rate of \(1 \mathrm{~kg} / \mathrm{s}\). Determine the maximum heat transfer rate and the outlet temperatures of both fluids for that case.

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