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Chapter 7: Problem 8

When matter flows across the boundary of a control volume, an energy transfer by work, called flow work, occurs. The rate is \(\dot{m}(p v)\) where \(\dot{m}, p\), and \(v\) denote the mass flow rate, pressure, and specific volume, respectively, of the matter crossing the boundary (see Sec. 4.4.2). Show that the exergy transfer accompanying flow work is given by \(\dot{m}\left(p v-p_{0} v\right)\), where \(p_{0}\) is the pressure at the dead state.

Short Answer

Expert verified
\[\dot{m}(pv - p_{0}v)\].

Step by step solution

01

Understand the given formula

The given problem states that the work done per unit mass when matter flows across the boundary of a control volume is given by the rate \(\dot{m}(p v)\) where \(\dot{m}\) is the mass flow rate, \(p\) is the pressure, and \(v\) is the specific volume.
02

Define exergy transfer

Exergy transfer refers to the maximum useful work possible during a process that brings the system into equilibrium with a heat reservoir. In this problem, it deals with the flow work part.
03

Identify pressure at dead state

The dead state refers to a state where the system is in equilibrium with its surroundings. Here, the pressure at the dead state is denoted as \(p_{0}\).
04

Derive exergy transfer formula

The exergy transfer associated with flow work is found by considering the difference between the actual pressure \(p\) and the pressure at the dead state \(p_{0}\). Therefore, the exergy transfer rate is given by \[\dot{m}(pv - p_{0}v)\].
05

Simplify the exergy transfer formula

Rewriting the above, we get the exergy transfer as \[ \dot{m}(p v - p_{0} v) = \dot{m} v(p - p_{0}). \]

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Key Concepts

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

Control Volume
In thermodynamics, a control volume refers to a specific region or space within which we analyze the mass and energy flows. Imagine it as an invisible boundary that surrounds a system or part of a system. Matter and energy can cross this boundary. The control volume helps in analyzing processes where fluid enters and leaves the system. It is extremely useful in practical applications, such as engineering, where we deal with engines, turbines, pumps, and other systems where mass flow occurs. By focusing on the control volume, we can easily track the exchange of energy and matter, leading to better understanding and efficient design of thermal systems.
Mass Flow Rate
Mass flow rate, denoted as \(\backslash \backslashdot{m}\), is a measure of the amount of mass passing through a boundary or cross-section per unit of time. In our context, it refers to the flow of matter across the boundary of a control volume. The formula for mass flow rate is:

\[ \text{Mass Flow Rate} = \dot{m} = \frac{dm}{dt} \]

Where \(dm\) is the mass passing through and \(dt\) is the time interval. Understanding mass flow rate is crucial because it allows us to calculate the energy transfer, like flow work, and exergy transfer in systems where fluid is involved. It's a key factor in various thermodynamic calculations as it connects the mechanical properties of the flow to the energy interactions.
Dead State Pressure
The dead state refers to a condition where a system is in thermodynamic equilibrium with its surroundings. It means that the system has reached a state where no further work can be extracted from it. Dead state pressure, denoted as \(\backslash p_0\), is the pressure at this equilibrium state. This concept is significant because it defines a reference point for evaluating the maximum useful work (exergy) that can be obtained from a system. In our problem, we look at exergy transfer relative to the dead state pressure to determine how much useful energy the system can deliver during the flow work process.
Specific Volume
Specific volume, denoted as \(\backslash v\), is the volume occupied by a unit mass of a substance. It is the reciprocal of density and is an essential property in thermodynamics. The formula is as follows:

\[ v = \frac{V}{m} \]

Where \(V\) is the volume, and \(m\) is the mass. In the context of our exercise, understanding specific volume is vital because it directly affects the calculation of flow work and exergy transfer. It forms part of the formula \(\backslashdot{m}(pv)\) and helps give us the comprehensive picture of energy transfer in processes involving fluids. Specific volume changes with the state of the substance, which means it will be different at different pressures and temperatures.

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Most popular questions from this chapter

A steam turbine operating at steady state develops \(9750 \mathrm{hp}\). The turbine receives 100,000 pounds of steam per hour at \(400 \mathrm{lbf} / \mathrm{in} .^{2}\) and \(600^{\circ} \mathrm{F}\). At a point in the turbine where the pressure is \(60 \mathrm{lbf} / \mathrm{in}^{2}\). and the temperature is \(300^{\circ} \mathrm{F}\), steam is bled off at the rate of \(25,000 \mathrm{lb} / \mathrm{h}\). The remaining steam continues to expand through the turbine, exiting at \(2 \mathrm{lbf} / \mathrm{in}^{2}\) and \(90 \%\) quality. (a) Determine the rate of heat transfer between the turbine and its surroundings, in Btu/h. (b) Devise and evaluate an exergetic efficiency for the turbine. Kinetic and potential energy effects can be ignored. Let \(T_{0}=\) \(77^{\circ} \mathrm{F}, p_{0}=1 \mathrm{~atm}\).

A system undergoes a refrigeration cycle while receiving \(Q_{\mathrm{C}}\) by heat transfer at temperature \(T_{\mathrm{C}}\) and discharging energy \(Q_{\mathrm{H}}\) by heat transfer at a higher temperature \(T_{\mathrm{H}}\) There are no other heat transfers. (a) Using energy and exergy balances, show that the net work input to the cycle cannot be zero. (b) Show that the coefficient of performance of the cycle can be expressed as $$ \beta=\left(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}-T_{\mathrm{C}}}\right)\left(1-\frac{T_{\mathrm{H}} \mathrm{E}_{\mathrm{d}}}{T_{0}\left(Q_{\mathrm{H}}-Q_{\mathrm{C}}\right)}\right) $$ where \(\mathrm{E}_{\mathrm{d}}\) is the exergy destruction and \(T_{0}\) is the temperature of the exergy reference environment. (c) Using the result of part (b), obtain an expression for the maximum theoretical value for the coefficient of performance.

Two kilograms of carbon monoxide in a piston-cylinder assembly, initially at 1 bar and \(27^{\circ} \mathrm{C}\), is heated at constant pressure with no internal irreversibilities to a final temperature of \(227^{\circ} \mathrm{C}\). Employing the ideal gas model, determine the work, the heat transfer, and the amounts of exergy transfer accompanying work and heat transfer, each in kJ. Let \(T_{0}=300 \mathrm{~K}, p_{0}=1\) bar and ignore the effects of motion and gravity.

Air enters a turbine operating at steady state with a pressure of \(75 \mathrm{lbf} / \mathrm{in}^{2}\), a temperature of \(800^{\circ} \mathrm{R}\), and a velocity of 400 ft/s. At the turbine exit, the conditions are \(15 \mathrm{lbf} / \mathrm{in}^{2}\), \(600^{\circ} \mathrm{R}\), and \(100 \mathrm{ft} / \mathrm{s}\). Heat transfer from the turbine to its surroundings takes place at an average surface temperature of \(620^{\circ} \mathrm{R}\). The rate of heat transfer is 2 Btu per lb of air passing through the turbine. For the turbine, determine the work developed and the exergy destruction, each in Btu per lb of air flowing. Let \(T_{0}=40^{\circ} \mathrm{F}, p_{0}=15 \mathrm{lbf} / \mathrm{in}^{2}{ }^{2}\)

Determine the specific flow exergy, in Btu/lbmol and Btu/ lb, at \(440^{\circ} \mathrm{F}, 73.5 \mathrm{lbf} / \mathrm{in}^{2}\). for (a) nitrogen \(\left(\mathrm{N}_{2}\right)\) and (b) carbon dioxide \(\left(\mathrm{CO}_{2}\right)\), each modeled as an ideal gas, and relative to an exergy reference environment for which \(T_{0}=77^{\circ} \mathrm{F}, p_{0}=\) \(14.7 \mathrm{lbf} / \mathrm{in}^{2}\) Ignore the effects of motion and gravity.
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