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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°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°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°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

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

Consider a heat exchanger that has an NTU of \(0.1\). Someone proposes to triple the size of the heat exchanger and thus triple the NTU to \(0.3\) in order to increase the effectiveness of the heat exchanger and thus save energy. Would you support this proposal?

Consider the flow of saturated steam at \(270.1 \mathrm{kPa}\) that flows through the shell side of a shell-and-tube heat exchanger while the water flows through 4 tubes of diameter \(1.25 \mathrm{~cm}\) at a rate of \(0.25 \mathrm{~kg} / \mathrm{s}\) through each tube. The water enters the tubes of heat exchanger at \(20^{\circ} \mathrm{C}\) and exits at \(60^{\circ} \mathrm{C}\). Due to the heat exchange with the cold fluid, steam is condensed on the tubes external surface. The convection heat transfer coefficient on the steam side is \(1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the fouling resistance for the steam and water may be taken as \(0.00015\) and \(0.0001 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), respectively. Using the NTU method, determine \((a)\) effectiveness of the heat exchanger, \((b)\) length of the tube, and \((c)\) rate of steam condensation.

Hot oil \(\left(c_{p}=2.1 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(110^{\circ} \mathrm{C}\) and \(8 \mathrm{~kg} / \mathrm{s}\) is to be cooled in a heat exchanger by cold water \(\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) entering at \(10^{\circ} \mathrm{C}\) and at a rate of \(2 \mathrm{~kg} / \mathrm{s}\). The lowest temperature that oil can be cooled in this heat exchanger is (a) \(10.0^{\circ} \mathrm{C}\) (b) \(33.5^{\circ} \mathrm{C}\) (c) \(46.1^{\circ} \mathrm{C}\) (d) \(60.2^{\circ} \mathrm{C}\) (e) \(71.4^{\circ} \mathrm{C}\)

Explain how the maximum possible heat transfer rate \(\dot{Q}_{\max }\) in a heat exchanger can be determined when the mass flow rates, specific heats, and the inlet temperatures of the two fluids are specified. Does the value of \(\dot{Q}_{\max }\) depend on the type of the heat exchanger?

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