General Expression

1. General Expression

The entropy balance equation for a closed system containing a fixed control mass assumes the form

(28) where the subscript c.m. denotes the control mass. Work does not explicitly enter into this ex-

dS c.m. – δQ/T b = σ c.m. ,

pression. For an open system which exchanges mass, heat and work with its ambient (cf. Figure 34 ), the derivation of the corresponding balance equation is similar to that of the energy conservation equation. The entropy change within a fixed mass over an infinitesimal time δt

can be written in the form dS c.m. =S c.m.,t+dt –S c.m.,t .

(62a) The control mass at time t (illustrated within the dashed boundary in Fig. 25(a) ) includes both

the control volume mass and a small mass dm i waiting to enter the control volume. After an

T b,1 δδδδQ 1

δδδδW

δδδδW

c.m. boundary c.v. boundary

Figure 34: Schematic diagram to illustrate the entropy balance equation.

infinitesimal time δt as the mass dm i enters at the inlet, a small mass dm e leaves through the control volume exit, and for the control volume Eq. (62a) may be expressed in the form

(62b) Heat transfer can occur across the control volume boundary. In general, the boundary

dS c.m. = (S c.v.,t+dt + dm e s e ) – (S c.v.,t + dm i s i ).

temperature at its inlet (but within the dashed boundary) is different from the corresponding temperature at the exit, and the heat transfer rate δQ may vary from the inlet to exit. For the

same of analysis we divide the boundary into sections such that at any section j the boundary temperature is T b,j and the heat transfer rate across the boundary is Q ˙ j . For an infinitesimally

small time period δt, the term δQ/T is given as ( δQ/T b )= Σ j Q ˙ j δt/T b,j .

(63) Using Eqs. (28), (62), and (63), expanding S c.v.,t+δt in a Taylor series around time t, and divid-

ing the resultant expression by δt, and letting when δt → 0 so that the control mass and control

volume boundaries merge (i.e., c.m. → c.v . ) , we obtain dS c.v. /dt = ˙m i s i – ˙m e s e + Σ j Q ˙ j /T b,j + ˙ σ cv .

On a mole basis Eq. (64) may be written in the form

(65) and generalizing for multiple inlets and exits

dS c.v. /dt = Ns ˙ ii – Ns ˙ ee +Σ j Q ˙ j /T b,j + ˙ σ cv ,

(66) where Q ˙ j denotes the heat interaction of the system with its surroundings at section j with a

dS c.v. /dt = Σ ˙m i s i – Σ ˙m e s e + Σ j Q ˙ j /T b,j + ˙ σ cv .

boundary temperature T b,j . The first term on the RHS of Eq. (66) represents the input entropy through the various inlets, the second term the outlet entropy, and the third term the transit entropy due to heat transfer. Equation (66) may be interpreted as follows: The rate of entropy accumulation in the control volume = entropy inflow through advection – entropy outflow through advection + change in the transit entropy through heat transfer + entropy generated due to irreversible processes. The various terms are also illustrated in Figure 35 .

Equations (64) and (65) may be, respectively, rewritten in the form

dS c.v. = dm i s i – dm e s e + Σ j δQ j /T b,j + d σ, and

(67a)

(67b) Recall that for closed system TdS = δQ + dσ. Equation (67) is the corresponding

dS c.v. = dN i s i – dN e s e + Σ j δQ j /T b,j + d σ.

equation for the open system. In case of elemental reversible processes in the presence of uni- form control volume properties (e.g., in an isothermal swimming pool)

d σ = 0, T b,j = T so that

TdS = δQ + dm i Ts i – dm e Ts e .

For an open system with property gradients within the control volume, as in a turbine, the system may be divided into small sections to apply the relation

TdS = δQ + dm i Ts i – dm e Ts e

to each subsystem within which the temperature is virtually uniform. For a closed system, Eq. (64) reduces to Eq. (31). If the process is reversible, Eq.(66) becomes

(68) Equation (66) provides information on the rate of change of entropy in a control volume. If it

dS c.v. /dt = Σ ˙m i s i – Σ ˙m e s e + Σ j Q ˙ j /T b,j .

becomes difficult to evaluate Σ j Q ˙ j /T b,j , the system boundary may be drawn so that T b,j = T 0 , i.e., all irreversibilities are contained inside the selected control volume. For instance, if in

Figure 34 the boundary is selected just outside the control volume, then T b,j =T 0 . At steady state Eq. (66) becomes

(69) For single inlet and exit at steady state s i -s e + Σq j /T bj + σ c.v =0

Σ ˙m i s i – Σ ˙m e s e + Σ j Q ˙ j /T b,j + ˙ σ cv = 0.

Where σ m , entropy generates per unit mass. If process is reversible σ c.v = 0, T b = T and hence s i -s e + Σq j /T =0

or

ds = Σ I {δq j ) rev /T ds = Σ I {δq j ) rev /T

the relevant form of Eq. (66) is dS c.v. /dt = Σ k ms ˙ kiki , , –Σ k ms ˙ keke , , +Σ j Q ˙ j /T b,j + ˙ σ cv .

(70) where s k denotes the entropy of the k–th component in the mixture. For ideal gas mixtures

s 0 k (T,P) = s k (T) – ln(PX k ). x. Example 24

Water enters a boiler at 60 bar as a saturated liquid (state 1). The boiler is supplied with heat from a nuclear reactor maintained at 2000 K while

me se

the boiler interior walls are at 1200 K. The reactor transfers about 4526

dS/dt

kJ of heat to each kg of water. De- termine the entropy generated per

kg of water for the following cases F assuming the system to be in steady

state:

For control surface 1 shown in mi si

Q/Tb

Figure 36 with P 2 = 60 bars and T 2

= 500°C. For control surface 2 which in- cludes the nuclear reactor walls

with P 2 = 60 bars and T 2 = 500°C.

Figure 35: Illustration of the entropy band dia- For control surface 1 with P 2 = 40 gram.

bars and T2 = 500°C. Solution

At steady state m ˙ (s i –s e )+ Q ˙ j /T b,boiler + σ ˙ = 0, i.e.,

(A)

(B) Using the standard Steam tables ( –1

σ = (s 2 –s 1 )–q 12 /T b,boiler .

–1 A-4A )s 1 = 3.03 kJ kg at P 1 = 60 bar for the satu- rated liquid and s –1 2 = 6.88 kJ kg at P 2 = 60 bars for steam at 500°C. The heat transfer

q 12 = 4526 kJ kg of water and T b,boiler = 1200 K. Substituting the data in Eq.(B), σ = (6.88 – 3.03) – 4526÷1200 = 0.07833 kJ kg –1 of water.

For control surface 2, Eq.(B) may be written in the form

(C) Using the Steam tables ( A-4 ) data

σ=s 2 –s 1 –q 12 /T b,reactor .

σ = (6.88 – 3.03) – 4526 ÷ 2000 = 1.587 kJ kg –1 of water. For this case s 2 = 7.0901 at P 2 = 40 bars and T 2 = 500 C, and

σ = 7.09 – 3.03 – 4526 ÷ 1200 = 0.288 kJ kg –1 of water. Remarks

The difference in the value of s between cases (a) and (b), (i.e., 1.587 – 0.0783 = 1.509 kJ kg –1 of water) is due to the irreversible heat transfer between the two control surfaces 1 and 2. Therefore,

σ q = q 12 (1/T b,boiler – 1/T b,reactor ), σ q = q 12 (1/T b,boiler – 1/T b,reactor ),

Power plant systems are designed to minimize the generated entropy. If the heat transfer q

12 for the first case is given as 5000 kJ, σ = –0.32 kJ kg . Is this possible or is the heat

transfer value incorrect? y. Example 25

Consider a human being who weighs 70 kg. At 37ºC the typical heat loss is 100 W. The person is injected with glucose (s = 1.610 kJ kg –1 K –1 ) at the rate of 0.54 kg day –1 . Air at 27ºC is inhaled at the rate –1 of 0.519 kg hr . Assume steady state, no excretion, and for the products to possess the same proper- ties as air. The products are exhausted through the Figure 36: Illustration for Example 24. nose at 37ºC. Select the c.v. boundary so that it lies just below human skin.

a. Write the mass conservation and entropy balance equations for the system and sim- plify the equations.

b. Determine the exhaust mass flow rate (e.g., through the human nose).

c. What is the entropy generation rate per day?

d. What is the entropy generation rate per day per unit mass?

e If the entropy generation during a species’ life cannot exceed 10,000 kJ kg –1 K –1 , what is this human being’s life span?

Solution

dm cv /dt = m ˙ i – m ˙ e , and dS cv /dt = Σ ˙Q cv,j /T b,j + Σ m ˙ i s i – Σ m ˙ e s e + σ ˙ cv .

At steady state

m ˙ i – m ˙ e = 0, and Σ ˙Q cv,j /T b,j + Σ m ˙ i s i – Σ m ˙ e s e + σ ˙ cv = 0.

From mass conservation m ˙ i = m ˙ e = m ˙ = 0.519 kg hr –1 . Therefore, σ ˙ cv = –100 ×3600×24/(310×1000)+0.54×1.610+12.456×1.702–

(0.54+ 0.519 –1 ×24)×1.735 = 28.35 kJ K day .

Per unit mass σ ˙

cv = 28.35/70 =0.405 kJ kg day . The life span is 10000/(365 ×0.405) = 68 years.

Remarks In Chapter 11, the irreversibility due to metabolism will be considered. From Example 12 in Chapter 2 we see that -0.33 ˙q

G ∝m b while empirical results sug- gest that ˙q G (kW/kg) = 0.003552m -0.26 b . Part (e) of the problem can be mathemati-

cally expressed as ≈ σ ˙ /m = σ ˙ m ≈ ( ˙q G /T b ) equals the specific meatbolic rate ÷T b .

Likewise, from part (c), the life span of a species ≈ CT b / ˙q G where C ≈ 10,000 kJ kg –1 K –1 .

The metabolic rate during the lifetime of an organism varies, with the highest meta- bolic rate being for a baby and the lowest for an older person. The minimum meta- bolic rate for maintaining bodily functions is of the order of 1 W. The expression in part d is based on an average metabolic rate. The entropy change in the environment can be obtained by considering the atmos- phere as the system. There are no gradients in the environment outside of the skin. Therefore, the entropy generation is zero. The entropy growth rate in the environment is

dS –1 env /dt = 0 + ˙ Q /T ∞ + 0 = 100/300 = 0.33 W K per human being.