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1 Lecture Thermodynamics 4 Enthalpy Consider a quasistatic hydrostatic constant pressure (isobaric) process weights piston, p gas Q control surface fi, p gas U -U 1 = Q +W = Q - Ú pdv = Q - p + p fi (U + p ) - (U 1 + p ) = Q Or, since p 1 =p, we can write (U + p ) - (U 1 + p 1 ) = Q Let us call H=U+pV the enthalpy (for a hydrostatic system) So H -H 1 =DH=Q for a constant pressure process. Just like U, H is a function of state: H=H(p, V) or H=H(p, T) or H=H(T, V). More properties of enthalpy: dh = du + pdv +Vdp = d Q - pdv + pdv + Vdp 1
2 dh = d Q +Vdp Ê fi Á H ˆ Ê = Á d Q ˆ = C Ë T p Ë dt p p Ê (similar to the equation fi Á U ˆ = C Ë T V ) V For a constant pressure change: dh = d Q fi DH = H - H 1 = Q = C p dt Compare what happens for a constant volume change: dh = d Q fi DU =U -U 1 = Q = C V dt For an adiabatic change: dh = Vdp fi DH = H - H 1 = ÚV dp Sometimes it is convenient to use specific enthalpy, that is, enthalpy per unit mass h=h/m, where m is the mass. p p 1 T Ú T 1 T Ú T 1 h=u+pv, where u=u/m and v=v/m=1/r (r the density). v is called the specific volume. Or we can use the enthalpy per mole. Again we can use the same symbol h=h/n, where n is the number of moles. Applications of enthalpy to flow processes Let us consider steady flow. This is a nonequilibrium situation but in a steady state. Steady flow means that the velocity of the fluid at all points is constant in time. Examples: 1) Adiabatic flow through a throttle Method (i) Consider an adiabatic process where gas in a horizontal pipe passes through a small hole. This is a non-equilibrium process because the gas accelerates through the small hole. It is not a quasistatic (reversible process). It is an irreversible process. Assume that the pressure is a constant p 1 before the throttle and a constant p after the throttle, and that these pressures are known. thermally insulating (adiabatic) pipe walls flow of gas p 1 p pipe small hole (throttle) pipe
3 The pressure on each side will be approximately constant if the pipe is not too long. To understand this process we can replace it by the idealized system below. We introduce imaginary pistons that move with the gas. They do not disturb the gas. We calculate the changes for the system enclosed by the control surface shown by the dotted lines (as usual). initial state A p 1 A p 1, p A fi p 1 A p 1 p p A fi final state B p 1 A p, p A The process is a non-equilibrium process but the gas inside the control surface is assumed to be in equilibrium for the initial and final states (apart from the small volume inside the hole which we neglect). Assume that volumes and are known (that depend on the density of the gas before and after the throttle). Let s assume that the kinetic energy change of the gas when it goes through the nozzle is negligible. (Later we will do a better calculation.) Using the 1 st law of thermodynamics: 3
4 U -U 1 = W + Q = -Ú p 1 dv - Ú p dv 0 = -p 1 V 0 [ ] V1 = p 1 - p 0 [ ] 0 - p V Therefore U 1 + p 1 = U + p Or H 1 =H So enthalpy is conserved in this process. Since we consider a constant mass for this process, we can also write h 1 =h. Another name for this sort of expansion through a throttle is a Joule-Kelvin expansion. Energy is lost in this process because work is done against forces due to dissipation (viscosity of the gas) inside the small hole. In general > for such an expansion. For an ideal gas one can show that T 1 =T (we will show this later in the course). In that case U 1 =U because U=f(T) only for an ideal gas Therefore, for an ideal gas p 1 >p (because nrt 1 / >nrt / ). In this problem we have assumed that p 1 and p are known. In practice we have to use fluid dynamics if we want to calculate these pressures. We won t do any fluid dynamics in theses lectures though. Note that we have drawn the same diameter of pipe on the left and the right, but this was not needed in the proof. The result holds for a change in diameter of pipe too. This will be discussed below. Method (ii) Let s consider the same problem but using a control surface that is fixed during the process: initial state A p 1, v 1, T 1 h 1 p, v, T h final state B p 1, v 1, T 1 h 1 p, v, T h gas mass m enters control volume 4 gas mass m leaves control volume
5 The diagram above shows the considered fixed control volume by the dotted lines. During the change from state A to state B a mass m of gas enters from the left hand side and the same mass leaves from the right hand side. (This must be true if the flow is steady and there are no leaks.) The process is adiabatic. The shaded region in A shows the gas in the control volume. The shaded region in B shows where this same gas is at this later time. (Some has left the control volume on the right hand side.) The control volume is fixed so the internal energy inside the control volume (dotted lines) must be the same before and after, so DU=0. The things trying to change the internal energy in the control volume are: 1) Energy in the form of mass mu 1 is transported in (u is the internal energy per unit mass) ) Energy in the form of mass mu 1 is transported out 3) Work p 1 mv 1 is done by the fluid outside the control surface on the fluid inside. * Here v is the specific volume or the volume per unit mass (=1/r, where r is the density). 4) Work p mv is done by the fluid inside the control surface on the fluid outside We need to count contributions 1) and 3) as positive and ) and 4) as negative in the 1st law: So m(u 1 -u +p 1 v 1 -p v )=0 Therefore u 1 +p 1 v 1 =u +p v or h 1 =h as before. In this second method, that used a fixed control surface, we have in fact used the 1st law of thermodynamics in a more general form: DU=W+Q+Wc Here Wc is called the chemical work. It represents the energy in the form of mass crossing the control surface into the control volume. In the above example we have DU=0 and Q=0, W=m(p 1 v 1 -p v ), and Wc=m(u 1 -u ). * It is a little difficult to understand how to calculate the work done by the moving fluid that is situated outside the control surface on the moving fluid inside the control surface when the fluid is crossing the fixed control surface. One can imagine it like this: at each instant in time consider that the fluid outside is pushing the fluid inside with force p 1 A, where A is the cross-sectional area of the pipe, and the fluid moves a tiny distance dx from outside to inside the control surface, where dxæ0. So dw=p 1 Adx=p 1 d. For each instant in time the fluid at the control surface is replaced by new fluid. Summing all the contributions, W =p 1 =p 1 mv 1. This work is done against forces caused by dissipation, due to viscosity. 5
6 Note that DU=0 for the above control surface. Here mu 1 is the internal energy of the gas on the left that flows into the control volume during the change AÆB. (Note that it is not the total internal energy of the gas in the control volume on the left.) ) Adiabatic flow in a curved pipe in a gravitational field h, p, v, u here h 1, p 1, v 1, u 1 here pipe wall gas mass dm enters control volume gas mass dm leaves control volume Consider the steady flow of gas or liquid (that is, a fluid) in a pipe with variable cross-section as shown. Steady flow without turbulence (eddies) is called laminar flow. At two points on the pipe shown we have defined the enthalpy h, pressure p, specific volume v and the internal energy per unit mass u. Using the same argument as we did for the example of a throttle [method (ii) above], h 1 =h So the specific enthalpy for the flow in a curved pipe is constant. To derive a more detailed equation consider a thin pipe which varies with height x: speed V flow line pipe wall pressure p x x=0 Define the internal energy per unit mass u(x=0)=u(0) for the fluid at rest. 6
7 Then u=u(0)+ PE m + KE m, where PE=mgx=gravitational potential energy, and KE= 1 m =kinetic energy. (Here, g is the acceleration due to gravity, and V is the speed of the fluid in the pipe.) Therefore u=u(0)+gx+ 1 The law, h 1 =h, means that h=constant. Therefore, u(0)+gx+ 1 +pv=u(0)+gx+ 1 + p =constant (1) r If in addition the fluid is incompressible, then its volume will not change, and so it s temperature will not change in an adiabatic flow. In that case we can put u(0)=constant. So for incompressible fluids, gx p r =constant Bernouilli s theorem This is called Bernoulli s theorem. Since liquids are approximately incompressible, it applies to liquids. [Note gases are not incompressible so for gases we have to use Eq. (1) instead]. Strictly this theorem applies along any flow line (or streamline), such as shown by the dotted line in the above diagram. It applies to any thin tube of flow (see below), even in thick pipes. ) Applications of Bernouilli s theorem (i) Incompressible horizontal flow in a pipe pipe wall tube of flow The diagram above shows the flow in a horizontal pipe of non-uniform cross section. We concentrate on part of the flow shown by the imaginary tube of flow: 7
8 L 1 L speed speed dm area A 1 area Adm tube of flow p 1 high p low Note that no fluid crosses the walls of the tube of flow (since by definition these walls are parallel to the fluid flow). Consider what happens to a small element of fluid of mass dm as shown. If the mass is incompressible then the volume will be the same after moving along the tube of flow (r=constant), so L 1 A 1 =L A. Therefore L L 1 = A 1 A = Apply Bernoulli s theorem: 1 + p =constant (since gx is constant) r Therefore 1 + p 1 r = 1 + p r () If A 1 >A then >, and therefore from Eq. () we have p 1 >p. This Eq. () is used in a device called a Venturi tube, shown here for a liquid flow. Venturi Tube area A 1 area A flow line p 1 p liquid Hg in tube Applying Bernoulli s theorem: 8
9 1 + p 1 r = 1 + p r = 1 A 1 A + p r fi p 1 >p (as before). We can measure if we know r, p 1, p, A 1 and A : = ( p 1 - p )A È Í Î 1. r( A 1 - A ) None of these problems require that we know the viscosity of the fluid. Note that in a pipe of constant cross-section the pressure will decrease as a function of distance because of viscosity. We have ignored such effects in the above treatment. We have assumed that the pipes are not too long. (ii) Non-adiabatic flow in a device If there is a steady flow of some fluid through a device we can call it a steady state steady flow process. p 1, v 1, u 1, control surface x 1 p, v, u, x w =work done on unit mass of fluid by the surroundings q=heat into unit mass of fluid from the surroundings We are interested in the work done by such a device for every kg of fluid that flows through it. Apply the general form of the first law to this problem: DU=W+Q+W c Or, per unit mass, Du=w+q+w c Because of the fixed control surface we have DU=Du=0. We know that w=w +p 1 v 1 -p v (see p. 5) 9
10 We also know that w c =u 1 -u =u 1 (0)+gx u (0)-gx - 1 (see p. 7) (This is the chemical work per unit mass, that is, the net energy transported into the control volume in the form of mass, kinetic energy or potential energy. * ) The 1st law therefore gives: Du=w+q+w c =0=w +p 1 v 1 -p v +q+u 1 (0)+gx u (0)-gx - 1 The work out of the device per unit mass, -w, is given by -w =h 1 -h + 1 ( - )+g(x 1 -x )+q (where h=u(0)+pv) We can apply this equation to various situations. The throttle, revisited Previously we assumed that the kinetic energy was negligible. We do not have to assume this. Assume only that w =q=0 and x 1 =x : Then h 1 -h = 1 ( - ) (because> > ) We see that the specific enthalpy h drops by across the throttle. Turbine A turbine is designed to do work. Assume q=0 and x 1 =x, and ignore the fluid kinetic energy ( and relatively small): Then the work out -w =h 1 -h * The derivation here assumes that the internal energy U and the chemical work W c are each made up of three contributions: U=U(0)+U PE +U KE, and W c =W c (0)+W cpe +W cke.. Here, U(0), U PE and U KE refer to energy stored in a control volume in the form of molecular internal energy, gravitational potential energy, and kinetic energy respectively. W c (0), W cpe and W cke refer to energy crossing a control surface associated with the mass, with gravitational potential energy and with kinetic energy, respectively. In problems with no flow, we have just used U=U(0), the molecular internal energy. This is the thermal energy associated with molecular motion. 10
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