Application to Open Systems

6. Application to Open Systems

The entropy generation ˙ σ ′′′ cv can be determined as a function of spatial location within

a volume by solving Eq. (83) provided that the temperature and pressure are known. In Chap- ter 4 we will discuss that the work lost due to irreversibilities is given by the product T 0 σ ˙ ′′′ cv .

Fins are used in heat exchangers to increase the heat transfer rate. We have discussed that a hot body can be cooled either by directly transferring heat to the ambient (by generating entropy) or by using a heat engine to produce reversible work (without producing entropy). The fins are entropy generators at steady state for which Eq. (83) yields

ρ V ⋅∇ = −∇ ⋅ ∇ s ( λ T /) T + ′′′ σ ˙ cv . (84) Since there is no convection heat transfer within the solid fins,

rr

−∇ ⋅ ∇ ( λ TT /) + ′′′ = σ ˙ cv 0 .

a. Steady Flow Consider the one–dimensional steady flow of a fluid with negligible temperature and velocity gradients. For this case Eq. (83) simplifies to the form

ρv ds/dx = ˙ σ ′′′ cv . For an ideal gas ρv ds/dx = ˙ σ ′′′ cv . For an ideal gas

In this case dP/dx ≤ 0. For a fixed mass flow rate, integrating between the inlet (P = P i ) and any other section where P < P i ,

σ ˙ ′′′ cv = (R ρv/x) ln (P i /P). Friction causes pressure losses so that. ˙ σ ′′′ cv > 0.

b. Solids The energy conservation for solids can be written in the form

r r ρc∂T/∂t = – ∇ ⋅ ′′ Q

(85a) Since s = c dT/T, manipulating Eq. (83)

( ρc/T)∂T/∂t = –– ∇ ⋅ ′′ ( Q /) T + ˙ σ ′′′ cv . (85b) Dividing Eq. (85a) by the temperature and subtracting the result from Eq. (85b) we obtain

( ∇ ⋅ ′′ Q )/ T + ∇ ⋅ ′′ ( Q /) T + ′′′ = σ ˙ cv 0 , i.e.,

2 rr

σ ˙ ′′′ = cv (/ 1 TQ ) ′′∇ ⋅ T , i.e., Since ˙ σ ′′′ cv > 0,

r r Q ′′ ⋅∇ T < 0.

The important implication is that Q ·> 0 if ′′ ∇T < 0, i.e., heat can only flow in a di- rection of decreasing temperature.

ee. Example 32 Consider a 5 mm thick infinitely large and wide copper plate, one surface of which is maintained at 100ºC and the other at 30ºC. The specific heat and thermal conductivity of copper are known to be, respectively, 0.385 kJ kg –1 K –1 and 0.401 kW m –1 K –1 . Determine the entropy at 100ºC and 30ºC, and the entropy production rate.

Solution s = c ln(T/T ref ).

(A) At 100ºC, s = 0.385 ln(373

÷273) = 0.1202 kJ kg –1 K .

Likewise, at 30ºC s = 0.0401 kJ kg –1 K –1 . Using Eq. (85b), for the one-dimensional conduction problem,

σ ˙ ′′′ cv = –d/dx( λ/T dT/dx). (B) From energy conservation,

– λ dT/dx = ˙q ′′ = Constant. (C) Therefore, integrating from the boundary condition, namely, T = T 0 at x = 0, T 0 –T= ˙q ′′ x/ λ.

(D) Dividng Eq. (C) by T,replacing T with Eq.(D), and then using the result in Eq. (B)

σ ˙ ′′′ cv = d/dx( ˙q ′′ /(T 0 – ˙q ′′ x/ λ)

Multiplying by dx and integrating within the limits x=0 and x= L,

q ′′

q ˙ ′′

σ ′′ cv .. () x =

(E)

T 0 −′′ qx ˙ / λ T 0

For the given problem ˙q –2 ′′ = 0.401 × (373 – 303) ÷ 0.005 = 5614 kW m , and σ ˙

cv .. ''() L =(614×(1÷(373–5614×0.005÷0.401)–(÷(373))== 3.48 kW m K –1 . Even though entropy is produced at a rate of 3.48 kW m –2 K –1 within the plate, the entropy at the edges (across the plate thickness) will not increase, since the entropy is

a property that depends only on the local temperatures. In addition, the entropy that is produced is flushed out in the form of transit entropy at x = 0, i.e., through thermal conduction. This statement can be verified by employing the entropy balance equation for a control volume around the entire plate,

dS/dt – ∫δ ˙" Q /T b = ˙ σ ′′′ cv .

Since the temperature at any location is constant, the entropy cannot accumulate at r steady state and dS/dt = 0. Therefore, ∫d Q /T ′′ b =– ˙q ′′ (1 ÷373 – 1÷303) = ˙ σ ′′′ cv = 3.48

kW m –1 K .

A similar example involves electric resistance heating for which the coil temperatures can be maintained at steady state by transferring heat out of the coils as fast as it is produced. r

The entropy flux d Q /T ′′ b acts in the same manner in order to maintain constant entropy. Summarizing this section on the entropy balance, ∆S = σ for an isolated system. dS c.v. /dt = σ ˙ c.v. , in rate form for an isolated system.

dS c.v. /dt = σ ˙ c.v. , in rate form for an adiabtic closed system.

dS c.v. /dt = Q ˙ c.v. /T b + σ ˙ c.v. , in rate form for any system.

dS c.v. /dt = Q ˙ c.v. /T b + m ˙ i s i - m ˙ e s e + σ ˙ c.v. , in rate form for an open system.