Evaluation of Entropy for a Control Volume

2. Evaluation of Entropy for a Control Volume

Recall that for closed systems we evaluate entropy by connecting a reversible path between two given states and then use dS = dU/T + PdV/T along the reversible path in order to determine the entropy change. We need to obtain a corresponding relation for a control vol- ume. For instance, let us say that we wish to find the entropy change of the air in a tire when

air is pumped into it. Assume that the initial (T 1 ,P 1 ,V 1 ,N 1 ) and final states (T 2 ,P 2 ,V 2 ,N 2 ) are known. The difference N 2 –N 1 represents the number of moles that are pumped into the tire. The process may or may not be reversible. We will now show that for a single component un- dergoing a reversible process in an open system

dS = dU/T – PdV/T + (µ/T)dN. (71) where the chemical potential µ = g = h – Ts. z. Example 26

Consider a balloon that is charged with gaseous nitrogen. The balloon is kept in a room whose pressure can be closely matched to that of the balloon. The moles within the balloon and the volume increase. Assume the process to be reversible. Show that the entropy change when dN moles of a pure component are pumped into it is repre- sented by Eq.(71).

Solution Boundary work is performed as the balloon expands. Heat transfer may also occur across the boundary. Since matter enters the balloon but does not leave it during the charging process, over a small time period

(A) The form of energy conservation has been previously discussed in Chapter 2. For en-

dE = dU = dN i h i + δQ – δW.

tropy balance, we can apply Eq. (65) with a uniform boundary temperature, i.e.,

(B) The mass or mole conservation relation is

dS = dN i s i + δQ/T b + δσ c.v. .

dN = dN i . (C) Since the processes in the balloon are internally reversible, the pressure of the gas in

the balloon is almost equal to the outside pressure so that the deformation work W d = PdV is reversible and δσ c.v. = 0, implying that T b = T. Dropping the subscript i in Eqs.

(A) and (B)

(D) dS = dN s + δQ/T.

dU = dN h – PdV + δQ, and

(E)

Equation (E) reveals that even if an open system is adiabatic and reversible dS can be nonzero since moles enter the system. Eliminating δQ from Eqs. (D) and (E), the

resulting expression is (in mass form) dU = dNh – PdV + TdS – TsdN, where µ=g= (h– Ts) is the chemical potential or Gibbs function of a species entering the system.

Therefore, U = U(S,V,N). Since δW = PdV, the third term on the right hand side of the resultant equation, i.e., µdN, can be viewed as the reversible chemical work, δW chem,rev = –µdN. The negative sign occurs, since the chemical work input when

moles are added to a system is negative according to sign convention for work. Therefore, the change of internal energy of the matter in the balloon equals the energy transfer into the control volume due to reversible heat addition less the energy out- flow through the deformation work, but in addition to the energy transfer due to the chemical work. If a gas is pumped into a rigid volume (e.g., rigid tank) then P dV = 0, and if the en- tropy is kept fixed (e.g., if the volume is cooled while matter is introduced into it so that s final N final = s initial N initial ), the energy change dU = – δW chem,rev = µdN is the chemi-

cal work performed to adiabatically pump the incremental number of moles dN. Pressure potential causes the volume to change and perform deformation (boundary) work, temperature potential causes the entropy change through the heat transfer proc- ess, and the chemical potential causes the matter to move into or out of the control volume. Rewriting Eq. (F) in the form

dS = dU/T + (P/T)dV + (µ/T)dN. (H) which represents the change of entropy along a reversible path. As before, S =

S(U,V,N). Note that for a closed system containing inert matter, N is fixed and hence S = S(U,V).

Remarks In Example (26), we determine dS for a control volume. With this expression we can integrate Eq. (G) between two states to obtain the resultant entropy change. If a multi–component gas mixture (e.g., air) is reversibly pumped into a uniform tem- perature volume, Eqs. (A), (B), and (D) are modified as (with subscripts k denoting species)

(A’) dS = ΣdN k s k + δQ/T, and

dU = ΣdN k h k + δQ – P dV,

(B’) dU = ΣdN k (h k –Ts k ) – PdV + TdS, or

(D’) dU = TdS – PdV + Σµ k dN k .

(72a) Here, µ k = g k = (h k – Ts k ) (a more detailed discussion is contained in Chapter 8).

The first term on the RHS of Eq. (72a) represents the reversible heat transfer and the last term is the accumulation due to the exchange of matter. In a closed system con- taining inert components there is no change in the number of moles, and Eq. (72a) can

be written in the form dU = TdS – PdV. Adding d(PV) to both sides of Eq. (72a) we have

dH = dU + d(PV) = TdS + VdP + Σµ k dN k . (72b) Subtracting d(TS) from both sides of Eq. (72a) dA = dU – d(TS) = – SdT – PdV + Σµ k dN k

(72c) (72c)

dG = dH – d(TS) = – SdT + VdP + Σµ k dN k , (72d) where G = H–TS is the Gibbs function.

It is clear from the energy fundamental equation U = U(S,V,N 1 ,N 2 , ... ,N k ),

(73) that k + 2 properties are required to determine the extensive state of an open system.

For the same of illustration, assume a 9 m 3 room containing 0.09 kmole of O

2 (com- ponent 1) and 0.36 kmole of N 2 (component 2). Fresh warm air is now pumped into the room (N 1,i = 0.1 kmole and N 2,i = 0.376 kmole) and some air leaves the room (N 1,e = 0.05 and N 2,e = 0.188). Hence, N 1 = 0.09 + 0.1 – 0.05 = 0.14 kmole, N 2 = 0.36 + 0.376 – 0.188 = 0.548 kmole. The entropy increases due to the temperature rise as well as the larger number of moles of gas in the room, while the gas volume remains the same. If the room is divided into three equal parts A, B, and C, then VA = VB = VC = V/3 =

9/3 m 3 , and likewise S A =S B =S C = S/3, N OA 2 , = ... = N O 2 /3, and U A =U B =U C = U/3. Therefore, U(S A ,V A , ...) = U(S/3,V/3, N O 2 /3, …) = 1/3 U(S,V, N O 2 , …). In gen- eral, if the room is divided into λ´ parts,

U(S/ λ´, V/λ´, … = 1//λ´ U(S,V, ...), or if 1/ λ´ = λ then U(λS, λV, ..) = λU(S,V, ...) which is a homogeneous function of

degree 1. Note that intensive properties are independent of the extent of the system and are homogeneous functions of degree 0. Since T = ∂U/∂S, and the property in

any section is 1/ λ´ that of the original extensive property, i.e., T= ∂U A / ∂S A = ∂(U/λ´)/∂(S/λ´) = ∂U/∂S. Since the internal energy is an extensive property (or a homogeneous function of de-

gree 1), it must satisfy the Euler equation (cf. Chapter 1), namely,

∂U/∂S + V ∂U/∂V + N 1 ∂U/∂N 1 + ... = 1

(74a) However, ∂U/∂S = T, ∂U/∂V = – P, and ∂U/∂N 1 = µ 1 so that Eq. (74a) assumes the

form

(74b) which, upon simplification, may be written as U+ PV – TS = H – TS = G = Σµ k N o .

U = ST – PV + µ 1 N 1 +µ 2 N 2 + ...,

(74c) Likewise, applying the Euler equation to evaluate the property H(S,P,N 1 ,..) (cf. Eq.

(72b)),

H (S,P,N 1 ..) = TS + Σµ k N o , (74d) Similarly, the Helmholtz and Gibbs functions can be written in the forms

A (T,V,N 1 , …) = –PV + Σµ k N o , and (74e)

G (T,P,N 1 , ...) = Σµ k N o . (74f) Equation (73) implies that

(75a) Subtracting Eq. (72a) from Eq. (75a) we obtain

Equation (75b), which is also known as the Gibbs–Duhem equation, implies that

(76) The temperature, which is an intensive property, is a function k+1 intensive variables.

T = T(P,,µ 1 ,µ 2 , ..., µ k ).

Equation (76) is also known as the intensive equation of state. Applying Eq. (72a) to examine the fluid discharge from a rigid control volume we obtain the relation

dU = T dS – Σµ k dN k , (77a) which describes the internal energy change due to the change in the number of moles

(i.e., dN k ). Even if s k is constant, the change in entropy can be nonzero, since in this case dS = d(ΣN k s k )=Σs k dN k so that

(77b) Therefore, the internal energy change when matter is isentropically discharged equals

dU = TΣ s k dN k – Σµ k dN k = TΣ s k dN k – Σ( h k –T s k ) dN k = –Σ h k dN k .

the enthalpy of the matter leaving the system (Example 16, Chapter 2, gas discharge from tank).

We may rewrite Eq. (72a) in the form of the entropy fundamental equation,

dS = dU/T + (P/T)dV - Σ {µ k /T) dN k .

i.e., S = S(U,V,N 1 ,...).

aa. Example 27 Nitrogen is pumped into a 0.1 m 3 rigid tank. The initial state of the gas is at 300 K

and 1 atm. Determine the chemical work done to isentropically pump 0.016 kmole of the gas into the tank to a 10 bar pressure.

Solution Applying the ideal gas law, N 1 =1 × 0.1 ÷ (0.08314 × 300) = 0.004 kmole.

Therefore, the entropy change dS = d( s N) = 0, i.e., d s / s = dN/N, or

s 2 /s 1 =N 1 /N 2 = 0.004 ÷ 0.020 = 0.2.

3 Now, v –1

2 = 0.1 ÷ 0.02 = 5 m kmole , and v 1 = 0.1 ÷ 0.004 = 25 m kmole so that s 2 /s 1 =(c v0 ln(T 2 /T ref )+ R ln( v 2 /v ref ))/( c v0 ln(T 1 /T ref )+ R ln( v 1 /v ref ))

Using the values T –1

ref = 273 K, v ref =1m kmole , c v0 = 20 kJ kmole K , and R =

8.314 kJ kmole –1 K , T 2 = 186 K.

The chemical work, W chem,rev =– ∫µdN = –(U 2 –U 1 ). Now,

U 2 –U 1 = 0.02 × 20 × (186 – 273) – 0.004 × 20 × (300 – 273) = –36.96 kJ, i.e., W chem,rev = +36.96 kJ.

a. Example 28 Determine the chemical potential of pure O 2 at T = 2000 K and P = 6 bar, and O 2 pre- sent in a gaseous mixture at T = 2000 K and P = 6 bar, and X O2 = 0.3, assuming the mixture to behave as an ideal gas.

Solution µ= g =( h –T s ), where for ideal gases s = s 0 – R ln(P/P ref ). Using values from ta- bles ( Table A-19 ),

× (268.655 kJ kmole K – 8.314 kJ kmole –1 K –1 × ln(6 bar ÷1 bar)) = – 439,636 kJ kmole –1 .

µ = 67881 kJ kmole –1 2 – 2000 K

For an ideal gas mixture according to the Gibbs–Dalton law,

s –1

k (T,p k )= s 0 k (T) – ln(p k /P ref ) = (268.655 kJ kmole K – 8.314 kJ kmole K ×

×ln ((6 bar × 0.3) ÷1 bar) = 272.747 kJ kmole –1 K . Since µ –1

k =(h k –Ts k ), µ O 2 = – 477,613 kJ (kmole O 2 in the mixture) . Remarks

The chemical potential of O 2 decreases as its concentration is reduced from 100% (pure gas) to 30%. You will later see that the chemical potential plays a major role in determining the di- rection of chemical reactions (in Chapter 10) and of mass transfer (in this chapter), just as temperature determines the direction of heat transfer.