/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 Water flows through a horizontal... [FREE SOLUTION] | 91Ó°ÊÓ

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Water flows through a horizontal coil heated from the outside by high- temperature flue gases. As it passes through the coil the water changes state from liquid at \(200 \mathrm{kPa}\) and \(353.15 \mathrm{K}\left(80^{\circ} \mathrm{C}\right)\) to vapor at \(100 \mathrm{kPa}\) and \(398.15 \mathrm{K}\left(125^{\circ} \mathrm{C}\right)\). Its entering velocity is \(3 \mathrm{m} \mathrm{s}^{-1}\) and its exit velocity is \(200 \mathrm{m} \mathrm{s}^{-1}\). Determine the heat transferred through the coil per unit mass of water. Enthalpies of the inlet and outlet streams are: Inlet: \(334.9 \mathrm{kJ}\) kg-'; Outlet: \(2726.5 \mathrm{kJ} \mathrm{kg}^{-1}\).

Short Answer

Expert verified
The heat transferred per unit mass of water is 2411.6 kJ/kg.

Step by step solution

01

Understand the System

The system involves water flowing through a coil where it is heated and changes state from liquid to vapor. We are asked to find the heat transferred per unit mass of water as it flows through the coil.
02

Identify Given Data

We know the initial and final states of the water: \(P_{in} = 200 \text{kPa}, T_{in} = 353.15 \text{K}, V_{in} = 3 \text{m/s}\); \(P_{out} = 100 \text{kPa}, T_{out} = 398.15 \text{K}, V_{out} = 200 \text{m/s}\). The enthalpies are \(h_{in} = 334.9 \text{kJ/kg}\) and \(h_{out} = 2726.5 \text{kJ/kg}\).
03

Apply Energy Balance Equation

The change in energy per unit mass can be described by the energy balance equation: \[ q = h_{out} - h_{in} + \frac{V_{out}^2}{2} - \frac{V_{in}^2}{2} \].
04

Substitute Values in Energy Equation

Substitute the given values into the energy balance equation: \[ q = 2726.5 \text{kJ/kg} - 334.9 \text{kJ/kg} + \frac{200^2 \text{m}^2/\text{s}^2}{2 \times 1000} \text{kJ/kg} - \frac{3^2 \text{m}^2/\text{s}^2}{2 \times 1000} \text{kJ/kg} \].
05

Calculate Kinetic Energy Changes

Calculate the kinetic energy terms: \[ \delta KE = \frac{200^2}{2 \times 1000} - \frac{3^2}{2 \times 1000} = \frac{40000 - 9}{2000} = 19.9955 \text{kJ/kg} \].
06

Compute Total Heat Transfer

Sum the changes in enthalpy and kinetic energy to find the heat transfer per unit mass: \[ q = 2726.5 - 334.9 + 19.9955 = 2411.5955 \text{kJ/kg} \].
07

Conclusion

The total heat transferred per unit mass of water as it flows through the coil is \(2411.6 \text{kJ/kg}\), considering the significant figures.

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Key Concepts

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

Enthalpy Change
When we talk about enthalpy change in the context of heat transfer, we are discussing the change in energy specifically associated with the process of heating or cooling substances during transformations such as phase changes. In the exercise, as water transitions from liquid to vapor via heating, enthalpy change is a key factor. Enthalpy itself is a measure of the total heat content of a system:
  • Inlet enthalpy, where water is initially a liquid, is 334.9 kJ/kg.
  • Outlet enthalpy, where water becomes vapor, increases significantly to 2,726.5 kJ/kg.

This considerable increase in enthalpy reflects the energy absorbed by the water to overcome intermolecular forces and facilitate the phase transition from liquid, under specific pressure and temperature conditions, to vapor. Understanding these changes helps us quantify energy input requirements during processes that involve heating and phase changes.
Energy Balance Equation
The energy balance equation is a fundamental principle in thermodynamics. It helps account for all energy transfers in a system, ensuring energy conservation. In this situation, we consider the energy coming in and out as water moves through the coil.The equation is expressed as: \[q = h_{out} - h_{in} + \frac{V_{out}^2}{2} - \frac{V_{in}^2}{2}\]Where:
  • \( q \) is the heat transferred per unit mass.
  • \( h_{in} \) and \( h_{out} \) are the specific enthalpies at the inlet and outlet.
  • \( V_{in} \) and \( V_{out} \) are the velocities at inlet and outlet.

This formula accounts for changes in internal energy (enthalpy) and kinetic energy. Once the specific enthalpy and velocity changes are known, we can easily calculate the heat transferred. This equation highlights the direct relationship between energy changes in both heat content and motion, as seen with the speeding up of water vapor in the coil.
Thermodynamics Analysis
Thermodynamics analysis involves examining processes, like heating water in coils, from a scientific standpoint that blends both theoretical and practical concerns. In this context:
  • We begin by understanding the state changes occurring in the coil—liquid water converting to vapor.
  • The principle of energy conservation guides our analysis, particularly through the use of the energy balance equation.
This type of analysis ensures accuracy by taking into account both macroscopic observations (pressure, temperature changes) and microscopic kinetic and potential energy variations that can be summed up using equations such as the energy balance equation.
By analyzing these aspects, students can gain insights into how energy is transferred and transformed, preparing them better for practical applications in engineering and scientific fields.

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Most popular questions from this chapter

One kilogram of air is heated reversibly at constant pressure from an initial state of \(300 \mathrm{K}\) and 1 bar until its volume triples. Calculate \(\mathrm{W}, \mathrm{Q}, A U,\) and \(A H\) for the process. Assume for air that \(\mathrm{P} V / \mathrm{T}=83.14 \mathrm{bar} \mathrm{cm}^{3} \mathrm{mol}^{-1} \mathrm{K}^{-1}\) and \(C_{P}=29 \mathrm{Jmol}^{-1} \mathrm{K}^{-1}\).

The conditions of a gas change in a steady-flow process from \(293.15 \mathrm{K}\left(20^{\circ} \mathrm{C}\right)\) and \(1000 \mathrm{kPa}\) to \(333.15 \mathrm{K}\left(60^{\circ} \mathrm{C}\right)\) and \(100 \mathrm{kPa}\). Devise a reversible nonflow process (any number of steps) for accomplishing this change of state, and calculate \(A U\) and \(A H\) for the process on the basis of 1 mol of gas. Assume for the gas that \(P V / T\) is constant, \(C_{V}=(5 / 2) R,\) and \(C_{P}=(7 / 2) R\).

Comment on the feasibility of cooling your kitchen in the summer by opening the door to the electrically powered refrigerator.

A steel casting weighing \(2 \mathrm{kg}\) has an initial temperature of \(773.15 \mathrm{K}\left(500^{\circ} \mathrm{C}\right) ; 40 \mathrm{kg}\) of water initially at \(298.15 \mathrm{K}\left(25^{\circ} \mathrm{C}\right)\) is contained in a perfectly insulated steel tank weighing 5 kg. The casting is immersedin the water and the system is allowed to come to equilibrium. What is its final temperature?Ignore any effect of expansionor contraction, and assume constant specific heats of \(4.18 \mathrm{kJ} \mathrm{kg}^{-1} \mathrm{K}^{-1}\) for water and \(0.50 \mathrm{kJ} \mathrm{kg}^{-1} \mathrm{K}^{-1}\) for steel.

A nonconducting container filled with \(25 \mathrm{kg}\) of water at \(293.15 \mathrm{K}\left(20^{\circ} \mathrm{C}\right)\) is fitted with a stirrer, which is made to turn by gravity acting on a weight of mass 35 kg. The weight falls slowly through a distance of \(5 \mathrm{m}\) in driving the stirrer. Assuming that all work done on the weight is transferred to the water and that the local acceleration of gravity is \(9.8 \mathrm{m} \mathrm{s}^{-2},\) determine: (a) The amount of work done on the water. (b) The internal-energy change of the water. (c) The final temperature of the water, for which \(C_{P}=4.18 \mathrm{kJ} \mathrm{kg}^{-1}^{\circ} \mathrm{C}^{-1}\) (d) The amount of heat that must be removed from the water to return it to its initial temperature (e) The total energy change of the universe because of (1) the process of lowering the weight, (2) the process of cooling the water back to its initial temperature, and (3) both processes together.

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