SINGLE–COMPONENT INCOMPRESSIBLE FLUIDS

G. SINGLE–COMPONENT INCOMPRESSIBLE FLUIDS

For incompressible fluids the internal energy and entropy may be written in the forms u = c (T–T ref ), and s = c ln(T/T ref ). Manipulating the two relations

s = c ln ((u/c + T ref )/T ref ), i.e., (60a) s = s(u). This relation is called the entropy fundamental equation for an incompressible sin-

gle–component fluid. Likewise, expressing the internal energy as a function of the entropy, u=cT ref (exp (s/c) – 1), i.e.,

(60b) it follows that u = u(s). This relation is called the energy fundamental equation for an incom-

pressible single–component fluid. Similarly, the enthalpy fundamental equation may be ex- pressed in the form

h = u + Pv ref =cT ref (exp(s/c) –1) + Pv ref = h (s,P). This is discussed further in Chapter 7.

q. Example 17 Consider a vapor–liquid mixture in a closed system that is adiabatically and qua-

sistatically compressed. Is the process isentropic at low pressures? Assume that the mixture quality does not change significantly (cf. Figure 30 ).

(b) (a)

Figure 30: Illustration of irreversibility during compression of two phase mixture. Solution

During the process the vapor is more readily compressed, which, in turn, compresses the liquid droplets. If the process is to be isentropic, there should be no temperature difference between the vapor and liquid drops. However isentropic compression of incompressible drops cannot create a temperature rise, while it can do so for vapor. Thus the vapor must heat the drops. Therefore, even though the process is quasistatic, it is not a quasiequilibrium process, since internal temperature gradients exist during compression, which cause irreversible heat transfer between the vapor and liquid drops. Applying the First law,

–P (dV v + dV l ) = dU v +dU l , (A ) where the subscripts v and l, respectively, denote vapor and liquid. Upon compres-

sion, the increased vapor temperature causes the liquid drops to heat up, and it is usual that the liquid temperature lags behind the vapor temperature. Finally, the sys-

tem equilibrates so that T v,2 =T 2 =T l,2 . In order to simplify the problem, we assume that there is no vaporization during compression, i.e., first the vapor is compressed as though the drops are insulated from it. Since the liquid has a specific volume of 0.001

m 3 kg –1 while the vapor specific volume is of the order of 1 m 3 kg –1 we can neglect the small change in drop volume. Assuming ideal gas behavior for the vapor, we can show using Eq. (A) (or assuming an isentropic process for the vapor) that

(B) Following compression the liquid and vapor reach the equilibrium temperature T 2

T v,2 /T v,1 = (V 1 /V 2 ) (k–1) .

without any change in their respective volumes. Applying the First law,

(C) Solving for T 2 /T 1 T 2 /T 1 = (x c v0 (T v,2 /T v,1 ) + (1–x) c l )/(x c v0 + (1–x) c l )

m v c v0 (T v2 –T 2 )=m l c l (T 2 –T 1 ).

(D) where x = m v / (m l + m v ) denotes the mixture quality. Applying the entropy balance

equation S 2 –S 1 – ∫δQ/T b = σ,

s 2 –s 1 = σ/m = x (c v0 ln (T 2 /T 1 + R ln (V 2 /V 1 )) + (1–x) c l ln (T 2 /T 1 ) (E)

Using values for the compression ratio V 1 /V 2 = 2, c v0 = 1.5, c l = 4.184, and R = 0.46,

a plot of σ with respect to x with r v as a parameter can be generated ( Figure 31 ). When x = 1, the mixture is entirely vapor, and the process is reversible. When x = 0,

the mixture only contains liquid, and the process is again reversible. The entropy gen- eration term σ reaches a maxima (which can be found by differentiating Eq. (E) with

respect to x, and, subsequently, setting d σ/dx = 0) in the vicinity of x = 0.6. Remarks

We have ignored the influence of phase equilibrium and vaporization in the above analysis. As the vapor is compressed, its temperature and pressure increase according (k–1)/k

. Phase equilibrium effects induce the liquid tem- perature T (P) to increase with the pressure according to a different relationship compared to variation of vapor temperature during compression. If local phase equi- librium near the drop surface is assumed during the compression process, invariably a temperature difference exists causing irreversibility.

to the relation T sat v,2 /T 1 = (P 2 /P 1 ).

r. Example 18 Air (for which k = 1.4) is contained in an insulated piston–cylinder assembly under the conditions P = 100 kPa, V = 0.1 m 3 2 and T = 300 K. The piston is locked with a

pin and its area is 0.010 m . A weight of 2 KN is rolled onto the piston top and the pin released. Is the process reversible or irreversible? Does the relation Pv k = constant, which is valid for an isentropic process, describe the

process? Determine the final state.

Evaluate the entropy change s 2 –s 1 .

Solution When the pin is released, the molecules adjacent to the piston are immediately com-

pressed making the local gas hotter while those farther away are not. Therefore, the pressure near the piston top is higher than the cylinder bottom. This effect continues as the piston moves inward, and the system is not at a uniform state. Thus a pressure and temperature gradients are established. The process is irreversible.

0.14 0.12 Compression Ratio =

Ent.Gen., kJ/kg K 0.04

Quality, x

Figure 31: Entropy generated with respect to quality.

The relation Pv k = constant does not apply, since the process is not isen-

tropic. Using the method of Example 9 of

Chapter 2, P 2 = 2/0.010 = 200 kPa,

T 2 = 386 K, and V 2 = 0.065 m 3 .

Therefore,

s 2 –s 1 = c p0 ln (T 2 /T 1 )–R

ln(P 2 /P 1 ) = ln (386÷300)–0.286 ln(200÷100)

= 0.0538 kJ kg –1 K . s. Example 19

Consider an idealized air condition- Figure 32: T-s diagram illustrating water in a ing cycle used for storage tank ap- storage tank. plications. The objective is to cool

the water stored in a tank from 25ºC (T t,1 ) to make ice at 0ºC (T t,2 ) by circulating cold Freon inside the tank. The ambient

temperature is 25ºC (T 0 ). Determine the minimum work required for every kg of wa-

ter contained in the storage tank. The heat of fusion for water ( h sf ≈ u sf ) is 335 kJ

kg –1 (cf. Figure 32 ). Solution

This example is similar to that contained in Example 7. We assume a Carnot refrig- eration cycle discarding heat to a variable low temperature reservoir. The cycle oper-

ates at a fixed higher temperature T 0 . First, the water is cooled from the initial state (state 1) to the melting point (MP) of ice, and then frozen at that temperature (state 2). As shown in Example 7,

W min = – (U t,2 –U t,1 )+T 0 (S t,2 –S t,1 ),

W min = – (mc(T 2 –T 1 )–mu sf )+T 0 (mc ln (T 2 /T 1 )–mu sf /T freeze ), or w min =W min /m = – (4.184 ×(0–25)–335)+298×(4.184×ln(273÷298)–335÷273) = 439.6 – 298 –1 × 1.594 = –35.41 kJ kg .

Remarks Figure 32 contains a representation of the process on a T–s diagram. The area under the path in the figure represents the reversible heat absorbed from the tank. We have assumed that the low temperature (of Freon) during the refrigeration cycle is exactly equal to the storage tank temperature. However, in practice it is not possible to trans- fer heat in the absence of a temperature gradient without inducing some irreversibility between the Freon and water contained in the tank.

t. Example 20

0.5 kg of coffee is contained in a cup at a temperature of 370 K. The cup is kept in an insulated room containing air at a temperature of 300 K so that, after some time, it cools to 360 K. The air mass is 100 kg. Assume the properties of coffee to be the same as those of water, and determine the following: The change in internal energy dU (= dU coffee + dU air ). The initial entropy of the coffee and air if it is assumed that both subsystems exist at an equilibrium state. The heat transfer across the cup boundary δQ.

The temperature change of the air. The entropy change of coffee dS coffee . The entropy change of air dS air . The entropy generated.

Solution Consider an isolated composite system consisting of two subsystems, i.e., coffee and air. In the absence of external interactions, isolated systems attain a stable equilibrium state. The internal energy change dU = 0 by applying the First law for a combined system, i.e.,

dU coffee = – dU air . (A ) S –1

coffee = m s = 0.5 ×s f,370 K = 0.5 × 1.25 = 0.625 kJ K . (B) S –1

air = 100 ×s 300 K = 100 × 1.7 = 170 kJ K . (C) Therefore,

S = 170.625 kJ K –1 . (D) For the coffee δQ – δW = dU coffee = m c dT coffee = 0.5 × 4.184 × (–10) = – 20.92 kJ.

Since there is no volume change, δW = 0, and δQ = – 20.92 kJ.

(E) Applying the First law dU coffee = – dU air (see Eq. (A)), since dU coffee = – 20.92 = –

dU air = – 100 × 1.0 × dT air , dT air = 0.21 K.

(F) The air temperature does not rise significantly.

Assuming the coffee temperature to be uniform within the cup, we will select the system so as to exclude boundaries where temperature gradients exist. Using the en- tropy balance equation for closed systems, and for internally reversible processes,

(G) Employing Eq. (B),

dS

coffee = δQ/T = – 20.92÷((360 + 370) × 0.5) = – 0.0573 kJ K .

S coffee = 0.625 – 0.0573 = 0.5677 kJ K –1 . dS air = 20.92

÷((300 + 300.21) × 0.5) = + 0.0697 kJ K –1 , and S = 170.0697 kJ K –1 air .

Forming a combined system that includes both coffee and air, δQ = 0. Applying the entropy balance equation dS – 0 = δσ,

dS = dS coffee + dS air = 0.0573 + 0.0697 = 0.0124 kJ K –1 so that δσ = 0.0124 kJ K –1 .

You will also find that δσ → 0 when the coffee temperature almost equals that of the air.

Remarks In this case, δσ > 0 during the irreversible process in the isolated system.

δσ → 0 when the coffee temperature almost equals the air temperature (or as reversi- bility is approached).

Vaporization of water into the air has been neglected. u. Example 21

A gas undergoes an expansion in a constant diameter horizontal adiabatic duct. As the pressure decreases, the temperature can change and the velocity increases, since the

gas density decreases. What is the maximum possible velocity?

Solution From mass conservation

d(V/v) = 0, (A) where V denotes velocity. Therefore, dV/v + V d(1/v) = 0. Applying energy conser-

vation d(h + V 2 /2) = 0, and

(B) utilizing the entropy equation

dh = T ds + v dP, or ds = dh/T – v dP/T. (C) Using Eqs. (A) and (B),

dh = – V dV = – V 2 dv/v. (D) From Eqs. (D) and (C), ds = – V 2 dv/(T v) – v dP/T.

(E) For an adiabatic duct, ds = δσ. Since δσ ≥0 for an irreversible process,

ds ≥0. (F)

2 2 Using Eqs. (F) and (E), V 2 ≤–v ( ∂P/∂v) s . Typically ( ∂P/∂v) s < 0. Hence V > 0. For

a reversible (i.e., isentropic) process,

(G) which is the velocity of sound in the gas.

2 V 2 =–v ( ∂P/∂v)

Remarks In Chapter 7, we will discuss use of the enclosed software to determine the sound speed in pure fluids.