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Chapter 7: Problem 11

A system consists of \(2 \mathrm{~kg}\) of water at \(100^{\circ} \mathrm{C}\) and 1 bar. Determine the exergy, in kJ, if the system is at rest and zero elevation relative to an exergy reference environment for which \(T_{0}=20^{\circ} \mathrm{C}, p_{0}=1 \mathrm{bar}\).

Short Answer

Expert verified
1040.52 kJ

Step by step solution

01

Understanding Exergy

Exergy represents the maximum useful work obtainable from a system as it comes to equilibrium with its reference environment. The exergy for a system with zero kinetic and potential energy changes is given by the formula: \[ \text{Exergy} = (h - h_0) - T_0(s - s_0) \] where - \(h\) is the specific enthalpy, - \(h_0\) is the specific enthalpy at the reference state,- \(T_0\) is the reference temperature (in absolute scale),- \(s\) is the specific entropy, - \(s_0\) is the specific entropy at the reference state.
02

Convert Given Temperatures to Kelvin

Given temperatures are in Celsius. Convert them to Kelvin: \[ T = 100^{\circ}C + 273.15 = 373.15 K \] \[ T_0 = 20^{\circ}C + 273.15 = 293.15 K \]
03

Determine Properties at Initial State and Reference State

Using steam tables or other thermodynamic property tools, find the specific enthalpy and entropy at the given state (100掳C, 1 bar) and at the reference state (20掳C, 1 bar): At 100掳C, 1 bar: \[ h = 2676.1 \text{ kJ/kg} \] \[ s = 7.358 \text{ kJ/kg路K} \] At 20掳C, 1 bar: \[ h_0 = 83.91 \text{ kJ/kg} \] \[ s_0 = 0.296 \text{ kJ/kg路K} \]
04

Plug Values into Exergy Equation

Using the values obtained, the exergy can be calculated as: \[ \text{Exergy} = (h - h_0) - T_0(s - s_0) \] Plugging in the values: \[ \text{Exergy} = (2676.1 - 83.91) - 293.15(7.358 - 0.296) \]
05

Calculate

Perform the arithmetic calculations: \[ \text{Exergy} = 2592.19 - 293.15(7.062) \] \[ \text{Exergy} = 2592.19 - 2071.93 \] \[ \text{Exergy} = 520.26 \text{ kJ/kg} \] Since we have 2 kg of water: \[ \text{Total Exergy} = 520.26 \times 2 = 1040.52 \text{ kJ} \]

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Key Concepts

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

specific enthalpy
Specific enthalpy is a measure of the total energy contained in a unit mass of a substance. Think of it like the energy budget of a system. It's useful because it combines both internal energy (from the system's molecules) and the energy required to make space for it (its pressure-volume work). This makes it easier to understand the energy changes during processes like heating, cooling, or expansion. In this exercise, we used specific enthalpy values from steam tables. Steam tables are essential tools for finding properties of water and steam under various temperatures and pressures. Specific enthalpy is measured in kilojoules per kilogram (kJ/kg).
Enthalpy is denoted by the symbol 'h', and we needed to find the values for both the initial thermodynamic state of the system and the reference state. For example:
  • At 100掳C and 1 bar, the specific enthalpy of water (h) is 2676.1 kJ/kg.
  • At 20掳C and 1 bar (the reference state), the specific enthalpy (h0) is 83.91 kJ/kg.
This difference in enthalpy values is a key part of the exergy calculation.
specific entropy
Specific entropy measures the disorder or randomness in a system. Entropy is a concept from the second law of thermodynamics, illustrating how energy disperses. Higher entropy means more disorder and less available energy for doing work. Specific entropy is measured in kilojoules per kilogram per Kelvin (kJ/kg路K).
In the exercise, we used specific entropy values to help determine the exergy of our system. The specific entropy values at the given state (100掳C, 1 bar) and the reference state (20掳C, 1 bar) were critical inputs. They are denoted by 's' and 's0' respectively:
  • At 100掳C and 1 bar, the specific entropy (s) is 7.358 kJ/kg路K.
  • At 20掳C and 1 bar, the specific entropy (s0) is 0.296 kJ/kg路K.
By incorporating these entropy values into the exergy formula, we could measure the system's capacity to do useful work.
thermodynamic state
A thermodynamic state describes the condition of a system at any point, defined by properties like temperature, pressure, volume, specific enthalpy, and specific entropy. These properties provide a snapshot of the system's energy and material balance.
To fully describe the system's thermodynamic state in this exercise, we identified:
  • The temperature (100掳C) and pressure (1 bar) of water.
  • The reference environment conditions (20掳C, 1 bar).
Using these properties, we could locate the specific enthalpy and entropy of the water at both the initial and reference states from steam tables. State properties are essential for calculations in thermodynamics, as they relate and provide the necessary input values for equations like the exergy formula.
Understanding the thermodynamic state allows us to predict how the system will behave when subjected to changes like heating, cooling, compression, or expansion.

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Most popular questions from this chapter

A pump operating at steady state takes in saturated liquid water at \(65 \mathrm{lbf} / \mathrm{in}^{2}\) at a rate of \(10 \mathrm{lb} / \mathrm{s}\) and discharges water at \(1000 \mathrm{lbf} / \mathrm{in}^{2}\). The isentropic pump efficiency is \(80.22 \%\). Heat transfer with the surroundings and the effects of motion and gravity can be neglected. If \(T_{0}=75^{\circ} \mathrm{F}\), determine for the pump (a) the exergy destruction rate, in Btu/s (b) the exergetic efficiency.

Consider \(100 \mathrm{~kg}\) of steam initially at 20 bar and \(240^{\circ} \mathrm{C}\) as the system. Determine the change in exergy, in \(\mathrm{kJ}\), for each of the following processes: (a) The system is heated at constant pressure until its volume doubles. (b) The system expands isothermally until its volume doubles. Let \(T_{0}=20^{\circ} \mathrm{C}, p_{0}=1\) bar and ignore the effects of motion and gravity.

The sun shines on a \(300-\mathrm{ft}^{2}\) south-facing wall, maintaining that surface at \(98^{\circ} \mathrm{F}\). Temperature varies linearly through the wall and is \(77^{\circ} \mathrm{F}\) at its other surface. The wall thickness is 6 inches and its thermal conductivity is \(0.04 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}{ }^{\circ} \mathrm{R}\). Assuming steady state, determine the rate of exergy destruction within the wall, in Btu/h. Let \(T_{0}=77^{\circ} \mathrm{F}\).

Air enters a turbine operating at steady state at a pressure of \(75 \mathrm{lbf} / \mathrm{in} .^{2}\), a temperature of \(800^{\circ} \mathrm{R}\), and a velocity of \(400 \mathrm{ft} / \mathrm{s}\) At the exit, the conditions are \(15 \mathrm{lbf}^{2} .^{2}, 600^{\circ} \mathrm{R}\), and \(100 \mathrm{ft} / \mathrm{s}\). There is no significant change in elevation. Heat transfer from the turbine to its surroundings at a rate of 10 Btu per lb of air flowing takes place at an average surface temperature of \(700^{\circ} \mathrm{R}\). (a) Determine, in Btu per lb of air passing through the turbine, the work developed and the exergy destruction rate. (b) Expand the boundary of the control volume to include both the turbine and a portion of its immediate surroundings so that heat transfer occurs at a temperature \(T_{0}\). Determine, in Btu per lb of air passing through the turbine, the work developed and the exergy destruction rate. (c) Explain why the exergy destruction rates in parts (a) and (b) are different. Let \(T_{0}=40^{\circ} \mathrm{F}, p_{0}=15 \mathrm{lbf} / \mathrm{in}^{2}\)

A concrete slab measuring \(0.3 \mathrm{~m} \times 4 \mathrm{~m} \times 6 \mathrm{~m}\), initially at \(298 \mathrm{~K}\), is exposed to the sun for several hours, after which its temperature is \(301 \mathrm{~K}\). The density of the concrete is 2300 \(\mathrm{kg} / \mathrm{m}^{3}\) and its specific heat is \(c=0.88 \mathrm{~kJ} / \mathrm{kg}+\mathrm{K}\). (a) Determine the increase in exergy of the slab, in kJ. (b) To what elevation, in \(\mathrm{m}\), would a \(1000-\mathrm{kg}\) mass have to be raised from zero elevation relative to the reference environment for its exergy to equal the exergy increase of the slab? Let \(T_{0}=298 \mathrm{~K}\), \(p_{0}=1 \mathrm{~atm}, g=9.81 \mathrm{~m} / \mathrm{s}^{2}\).
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