Work and Heat
In this chapter we will discuss the two quantities that result from energy transfer across the boundary of a system: work and heat. This will lead into the first law of thermodynamics. Work will be calculated for several common situations. Heat transfer, often simply called “heat,” however, is a quantity that requires substantial analysis for its calculation. Heat transferred by conduction, convection, or radiation to systems or control volumes will either be given or information will be provided that it can be determined in our study of thermodynamics; it will not be calculated from temperature information, as is done in a heat transfer course.
Our definition of work includes the work done by expanding exhaust gases after combustion occurs in the cylinder of an automobile engine, as shown in Fig. 3.1. The energy released during the combustion process is transferred to the crankshaft by means of the connecting rod, in the form of work. Thus, in thermodynamics, work can be thought of as energy being transferred across the boundary of a system, where, in this example, the system is the gas in the cylinder.
Figure 3.1 Work being done by expanding gases in a cylinder.
Figure 3.2 Work being done by electrical means.
Work, designated W, is often defined as the product of a force and the distance moved in the direction of the force: the mechanical definition of work. A more general definition of work is the thermodynamic definition: Work is done by a system if the sole external effect on the surroundings could be the raising of a weight. The magnitude of the work is the product of that weight and the distance it could be lifted. The schematic of Fig. 3.2a qualifies as work in the thermodynamic sense since it could lift the weight of Fig. 3.2b.
The convention chosen for positive work is that if the system performs work on the surroundings it is positive.1 A piston compressing a fluid is doing negative work on the system, whereas a fluid expanding against a piston is doing positive work. The units of work are quickly observed from the units of a force multiplied by a
distance: in the SI system, newton-me.ters (N · m) or joules (J).
The rate of doing work, designated W, is called power. In the SI system, power has
units of joules per second (J/s), or watts (W). We will use the unit of horsepower because of its widespread use in rating engines. To convert we simply use 1 hp = 746 W.
The work associated with a unit mass will be designated w:
w = W/m (3.1)
A final general comment concerning work relates to the choice of the system. Note that if the system in Fig. 3.2 included the entire battery-resistor setup in part (a), or the entire battery-motor-pulley-weight setup in part (b), no energy would cross the system boundary, with the result that no work would be done. The identification of the system is very important in determining work.
There are a number of work modes that occur in various engineering situations. These include the work needed to stretch a wire, to rotate a shaft, to move against friction, or to cause a current to flow through a resistor. We are primarily concerned with the work required to move a boundary against a pressure force.
Consider the piston-cylinder arrangement shown in Fig. 3.3. There is a seal to contain the gas in the cylinder, the pressure is uniform throughout the cylinder, and there are no gravity, magnetic, or electrical effects. This assures us of a quasiequilibrium process, one in which the gas is assumed to pass through a series of equilibrium states. Now, allow an expansion of the gas to occur by moving the piston upward a small distance dl. The total force acting on the piston is the pressure times the area of the piston. This pressure is expressed as absolute pressure since pressure is a result of molecular activity; any molecular activity will yield a pressure that will result in work being done when the boundary moves. The infinitesimal work
Figure 3.3 Work due to a moving boundary.
that the system (the gas) does on the surroundings (the piston) is then the force multiplied by the distance:
δW = PAdl (3.2)
The symbol δ W will be discussed shortly. The quantity Adl is simply dV, the differential volume, allowing Eq. (3.2) to be written in the form
δW = PdV (3.3)
As the piston moves from some position 1 to another position 2, the above expression can be integrated to give
where we assume the pressure is known for each position as the piston moves from volume V to volume V . Typical pressure-volume diagrams are shown in Fig. 3.4. The work W1-2is the area under the P-V curve.The integration process highlights two important features in Eq. (3.4). First, aswe proceed from state 1 to state 2, the area representing the work is very dependent on the path that we follow. That is, states 1 and 2 in Fig. 3.4a and b are identical, yet the areas under the P-V curves are very different; work depends on the actual path that connects the two states. Thus, work is a path function, as contrasted to a point function (that is dependent only on the end points). The differential of a path function is called an inexact differential, whereas the differential of a point function is an exact differential. An inexact differential will be denoted with the symbol δ.
Figure 3.4 Work depends on the path between two states.
The integral of δ W is W1-2, where the subscript emphasizes that the work is associated with the path as the process passes from state 1 to state 2; the subscript may be omitted, however, and work done written simply as W. We would never write W or
W2, since work is not associated with a state but with a process. Work is not a property. The integral of an exact differential, for example dT, would be
The second observation to be made from Eq. (3.4) is that the pressure is assumed to be constant throughout the volume at each intermediate position. The system passes through each equilibrium state shown in the P-V diagrams of Fig. 3.4. An equilibrium state can usually be assumed even though the variables may appear to be changing quite rapidly. Combustion is a very rapid process that cannot be modeled as a quasiequilibrium process. The other processes in the internal combustion engine⎯expansion, exhaust, intake, and compression⎯can be assumed to be quasi- equilibrium processes; they occur at a relatively slow rate, thermodynamically.
As a final comment regarding work we may now discuss what is meant by a simple system, as defined in Chap. 1. For a system free of surface, magnetic, and electrical effects, the only work mode is that due to pressure acting on a moving boundary. For such simple systems only two independent variables are necessary to establish an equilibrium state of the system composed of a homogeneous substance. If other work modes are present, such as a work mode due to an electric field, then additional independent variables would be necessary, such as the electric field intensity.
EXAMPLE 3.1
One kilogram of steam with a quality of 20 percent is heated at a constant pressure of 200 kPa until the temperature reaches 400°C. Calculate the work done by the steam.
Solution
The work is given by
To evaluate the work we must determine v1 and v 2. Using Table C.2 we find
From the superheat table C.3 we locate state 2 at T = 400°C and P2 = 0.2 MPa:
Since the pressure has units of kPa, the result is in kJ.
EXAMPLE 3.2
A 110-mm-diameter cylinder contains 100 cm3 of water at 60°C. A 50-kg piston sits on top of the water. If heat is added until the temperature is 200°C, find the work done.
Solution
The pressure in the cylinder is due to the weight of the piston and remains constant. Assuming a frictionless seal (this is always done unless information is given to the contrary), a force balance provides
The atmospheric pressure is included so that absolute pressure results. The volume at the initial state 1 is given as
Using v1 at 60°C, the mass is calculated to be
At state 2 the temperature is 200°C and the pressure is 0.15 MPa (this pressure is within 1 percent of the pressure of 0.1516 MPa, so it is acceptable). The volume is then
W = P(V2 − V1 ) = 151.6(0.1420 − 0.0001) = 21.5 kJ
EXAMPLE 3.3
Energy is added to a piston-cylinder arrangement, and the piston is withdrawn in such a way that the temperature (i.e., the quantity PV) remains constant. The initial pressure and volume are 200 kPa and 2 m3, respectively. If the final pressure is 100 kPa, calculate the work done by the ideal gas on the piston.
Solution
The work, using Eq. (3.4), can be expressed as
where we have used PV = C. To calculate the work we must find C and V . The constant C is found from
C = P1V1 = 200 × 2 = 400 kJ
To find V2e use P2V2 = P V , which is, of course, the equation that would result from an isothermal process (constant temperature) involving an ideal gas.This can be written as
This is positive, since work is done during the expansion process by the gas.
It must be emphasized that the area on a PV diagram represents the work for a quasiequilibrium process only. For nonequilibrium processes the work cannot be calculated using the integral of PdV. Either it must be given for the particular process or it must be determined by some other means. Two examples will be given. Consider a system to be formed by the gas in Fig. 3.5. In part (a) work is obviously crossing the boundary of the system by means of the rotating shaft, yet the volume does not change. We could calculate the work input by multiplying the weight by the distance it dropped, neglecting friction in the pulley system. This would not, however, be equal to the integral of PdV, which is zero. The paddle wheel provides us with a nonequilibrium work mode.
Suppose the membrane in Fig. 3.5b ruptures, allowing the gas to expand and fill the evacuated volume. There is no resistance to the expansion of the gas at the moving boundary as the gas fills the volume; hence, there is no work done, yet there is a change in volume. The sudden expansion is a nonequilibrium process, and again we cannot use the integral of PdV to calculate the work for this nonequilibrium work mode.
EXAMPLE 3.4
A 100-kg mass drops 3 m, resulting in an increased volume in the cylinder shown of 0.002 m3. The weight and the piston maintain a constant gage pressure of 100 kPa. Determine the net work done by the gas on the surroundings. Neglect all friction.
Solution
The paddle wheel does work on the system, the gas, due to the 100-kg mass dropping 3 m. That work is negative and is
W = −F × d = −(100 × 9.81) × 3 = −2940 N ⋅ m
The work done by the system on this frictionless piston is positive since the system is doing the work. It is
W = PA × h = PV = 200 000 × 0.002 = 400 N ⋅ m
where absolute pressure has been used. The net work done is thus
Wnet = Wpaddle +wheel + Wboundary = −2940 + 400 = −2540 J
Figure 3.5 Nonequilibrium work.
Labels: Thermodynamics
