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Chapter 11: Problem 166

Saturated water vapor at \(40^{\circ} \mathrm{C}\) is to be condensed as it flows through the tubes of an air-cooled condenser at a rate of \(0.2 \mathrm{~kg} / \mathrm{s}\). The condensate leaves the tubes as a saturated liquid at \(40^{\circ} \mathrm{C}\). The rate of heat transfer to air is (a) \(34 \mathrm{~kJ} / \mathrm{s}\) (b) \(268 \mathrm{~kJ} / \mathrm{s}\) (c) \(453 \mathrm{~kJ} / \mathrm{s}\) (d) \(481 \mathrm{~kJ} / \mathrm{s}\) (e) \(515 \mathrm{~kJ} / \mathrm{s}\)

Short Answer

Expert verified
a) 402 kJ/s b) 450 kJ/s c) 470 kJ/s d) 481 kJ/s Answer: d) 481 kJ/s

Step by step solution

01

Determine the enthalpies

First, we need to find the enthalpies of the saturated water vapor and the saturated liquid at the given temperature. From the steam tables, we can look up these enthalpy values at 40掳C: Enthalpy of saturated vapor, \(h_\text{v}\): \(2574.2\,\mathrm{kJ/kg}\) Enthalpy of saturated liquid, \(h_\text{l}\): \(167.57\,\mathrm{kJ/kg}\)
02

Calculate the enthalpy change

We will now calculate the change in enthalpy during the condensation process using the found enthalpies for the saturated vapor and the saturated liquid: Enthalpy change, \(\Delta h = h_\text{v} - h_\text{l}\) \(\Delta h = 2574.2\,\mathrm{kJ/kg} - 167.57\,\mathrm{kJ/kg} = 2406.63\,\mathrm{kJ/kg}\)
03

Calculate the rate of heat transfer

The rate of heat transfer is the product of the mass flow rate and the change in enthalpy: Rate of heat transfer, \(Q = \text{mass flow rate} \times \Delta h\) We are given the mass flow rate as 0.2 kg/s: \(Q = (0.2\,\mathrm{kg/s}) \times 2406.63\,\mathrm{kJ/kg} = 481.33\,\mathrm{kJ/s}\)
04

Find the answer in the given options

We need to compare our result to the given options to find which one is correct. Our calculated rate of heat transfer is: \(Q = 481.33\,\mathrm{kJ/s}\) Comparing this to the given options, we see that option (d) is the closest: (d) \(481\,\mathrm{kJ/s}\) Therefore, the correct answer is (d) \(481\,\mathrm{kJ/s}\).

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Key Concepts

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

Saturated Vapor
Saturated vapor is a term used in thermodynamics to describe a state where a vapor is completely saturated with its corresponding liquid, meaning they are in equilibrium. In this state, if you attempt to cool the vapor, even just a little, condensation occurs, converting some of the vapor back to liquid form. This point is crucial because it marks the transition from vapor to liquid without further cooling the actual temperature. In the exercise, the saturated vapor of water is at a specific temperature of 40掳C. At this temperature, the steam tables tell us the enthalpy of the saturated vapor, which quantifies the energy contained within the vapor. Understanding these properties from steam tables is fundamental to solving problems that involve phase changes like condensation.
Enthalpy Change
Enthalpy change is a critical concept in understanding energy transfer during physical processes, such as phase changes occurring in heating or cooling systems. When substances undergo phase changes, like from vapor to liquid or vice versa, the total enthalpy of the system changes. This is because energy is either absorbed or released during the process.In the exercise, the specific task is to calculate the enthalpy change when a saturated vapor condenses into a saturated liquid at 40掳C. The enthalpy change, denoted as \( \Delta h \), is calculated using the formula: \\[\Delta h = h_\text{v} - h_\text{l}\]where \( h_\text{v} \) is the enthalpy of the saturated vapor and \( h_\text{l} \) is the enthalpy of the saturated liquid. This difference in enthalpy represents the energy per kilogram transferred out of the vapor during condensation.
Condensation Process
The condensation process is a phase change where vapor turns into liquid. This is an exothermic reaction, meaning it releases heat. When saturated vapor condenses, energy is transferred from the water vapor to the surrounding environment, typically as heat, causing the vapor to become a liquid.In such a process, the vapor needs to lose a significant amount of energy, known as the heat of condensation. This is captured in the problem as the rate of heat transfer (measured in kJ/s), which is calculated by multiplying the mass flow rate by the enthalpy change, \( \Delta h \). The formula used is:\[Q = \text{mass flow rate} \times \Delta h\]This calculation gives the rate of heat released to the environment during the condensation process. Understanding these calculations is essential in designing efficient cooling systems, where proper heat removal is necessary to maintain functional temperatures.

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Most popular questions from this chapter

Consider a double-pipe heat exchanger with a tube diameter of \(10 \mathrm{~cm}\) and negligible tube thickness. The total thermal resistance of the heat exchanger was calculated to be \(0.025 \mathrm{~K} / \mathrm{W}\) when it was first constructed. After some prolonged use, fouling occurs at both the inner and outer surfaces with the fouling factors \(0.00045 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) and \(0.00015 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), respectively. The percentage decrease in the rate of heat transfer in this heat exchanger due to fouling is (a) \(2.3 \%\) (b) \(6.8 \%\) (c) \(7.1 \%\) (d) \(7.6 \%\) (e) \(8.5 \%\)

A shell-and-tube heat exchanger with 2-shell passes and 12 -tube passes is used to heat water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) with ethylene glycol \(\left(c_{p}=2680 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\). Water enters the tubes at \(22^{\circ} \mathrm{C}\) at a rate of \(0.8 \mathrm{~kg} / \mathrm{s}\) and leaves at \(70^{\circ} \mathrm{C}\). Ethylene \(\mathrm{glycol}\) enters the shell at \(110^{\circ} \mathrm{C}\) and leaves at \(60^{\circ} \mathrm{C}\). If the overall heat transfer coefficient based on the tube side is \(280 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer and the heat transfer surface area on the tube side.

There are two heat exchangers that can meet the heat transfer requirements of a facility. One is smaller and cheaper but requires a larger pump, while the other is larger and more expensive but has a smaller pressure drop and thus requires a smaller pump. Both heat exchangers have the same life expectancy and meet all other requirements. Explain which heat exchanger you would choose and under what conditions.

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 \(\left(c_{p}=0.25 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) 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 \(0.001 \mathrm{~h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F} /\) Btu on water and air side, respectively.

Cold water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) leading to a shower enters a thin-walled double-pipe counter-flow heat exchanger at \(15^{\circ} \mathrm{C}\) at a rate of \(0.25 \mathrm{~kg} / \mathrm{s}\) and is heated to \(45^{\circ} \mathrm{C}\) by hot water \(\left(c_{p}=4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters at \(100^{\circ} \mathrm{C}\) at a rate of \(3 \mathrm{~kg} / \mathrm{s}\). If the overall heat transfer coefficient is \(950 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer and the heat transfer surface area of the heat exchanger using the \(\varepsilon-\mathrm{NTU}\) method.
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