Internally Reversible Work for an Open System

3. Internally Reversible Work for an Open System

It has been shown in Chapter 2 that for a steady flow open system δq – δw = de T , where e T = h + ke + pe. For an internally reversible process that occurs between two static

states Tds – δw c.v.,rev = de T .

(78) Since dh = Tds + vdP,

Tds – δw c.v.,rev = Tds + vdP + d(ke + pe), i.e., δw c.v.,rev = –vdP – d(ke + pe).

Neglecting the kinetic and potential energies, the work delivered by the system is represented through the relation

w c.v.,,rev =– ∫ vdP . (79)

bb. Example 29 The systolic (higher) and diastolic (lower) blood pressures are measured to be, re- spectively, 120 mm and 70 mm of mercury for a healthy person. Determine the work

performed by the heart per kilogram of blood that is pumped. Assume that blood has the same properties as water.

Solution

w c.v.,,rev =– P ∫ vdP . i

Therefore, w

=–v (P –P ) = 0.001 m c.v.,rev 3 blood e kg i –1 (120–70) (mm Hg) ×(100 ÷ 760) kPa (mm Hg) –1 = 0.0063 kJ kg –1 .

Remarks Blood is contained within a finite volume of blood vessels. The body maintains the amount of water to a fixed concentration. If the two pressures simultaneously increase

(true for non–exercising persons), but the term (P e –P i ) remains unchanged (e.g., P e = 190 mm and P i = 140 mm), the work performed by the heart may not change. How- ever, the blood vessels may now become stressed and fail at the higher pressures.

In Chapter 9 we will discuss that the amounts of dissolved CO 2 , N 2 , and O 2 in the blood rise as the pressure increases, but do so disproportionately, depending upon their boiling points.

cc. Example 30 Pressurized gas tanks are employed in space power applications. As the gas contained in the tanks is used, the tank pressure falls (say, from P t,1 to P t,2 ) so that the work 3 done per unit mole can vary. Determine the work that can be done if a 2 m turbine

and the tank are kept in an isothermal bath, P e = 1 bar, P t,1 = 50 bars, and P t,2 = 1 bar. Solution For an open system,

δw shaft rev , =– v dP = –( R T/P)dP, i.e., (A)

(B) The pressure P i varies as the gas is progressively withdrawn from the tank. From Eq.

w shaft rev , =– R T ln(P e /P i ).

(B)

(C) dN tank = – dN i,turbine .

δW shaft rev , = w shaft rev , dN i =– R T ln(P e /P i )dN i,turbine , where

(D) Since P tank V=N tank R T, dP tank = dN tank R T/V, and dN tank = dP tank V/ R T.

(E) Therefore, using Eqs. (C) and (E)

δW shaft rev , =–– R T ln(P e /P i ) dP tank V/ R T = – V ln(P e /P i ) dP, (F) where P i =P tank . Hence,

δW shaft rev , =– VP e ln (P e /P tank )P tank /P e , and

W shaft rev , = VP e ((P tank,2 /P e ) ln (P tank,2 /P e ) – (P tank,1 /P e ) ln (P tank,1 /P e )). (G) Using the values V = 2 m 3 ,P e = 1 bar, P tank,1 = 50 bars, P tank,2 = 1 bar,

W shaft rev , =2 × 1 × 100 (1 ÷ 1 ln(1÷1) – (50÷1) ln(50÷1)) = 39120 kJ

Remarks If we select the control volume to include both the turbine and the tank, assuming that

there is no accumulation of energy or entropy in the turbine,

(H) Applying the entropy balance equation to the tank,

dU tank = δQ – PdV – δ W shaft rev , – h e dN e .

(I) From Eqs. (H) and (I),

dS tank = δQ/T – dN e s e .

δW shaft rev , = TdS tank – dU tank – PdV + T dN e s e – h e dN e . (J) Since dN e = – dN tank , dV = 0, dS tank = s tan k dN tank +N tank ds tan k , h e (T) = h tan k (T) = u tan k + R T, dU tank = u tan k dN tank +N tank u tan k , and d u tan k = 0, δW shaft,rev =T( s tank dN tank +N tank d s tank ) - d(dN tank u tank +N tank d u tank )-0- T dN tank s e +( u tank + RT) dN tank .

Simplifying this relation dW shaft,rev = dN tank T(s tank -s e ) = - dN tank TR ln(P tank /P e ) = -V dP tank ln(P tank /P e ).

(K) If the process within the control volume is adiabatic and reversible,

(H) δW shaft rev , =– u tan k dN tank –N tank du tan k +(u tan k + R T(t))dN tank

dU tank =0–0– δW shaft rev , – h e (T) dN e , or

=–N tank c v0 dT tank + R T (t)dN tank =–N tank c v0 dT tank + VdP – R N tank dT tank

– dT tank N tank (c v0 + R ) + VdP = – dT tank N tank c p0 + VdP. Therefore the relationship between the temperature and N (k–1)/k (k–1)/k tank is of the form T tank /T tank,1

= (P tank /P tank,1 ) = (N tank T tank /N tank,1 T tank,1 )

, i.e.,

(T (k–1)/k tank /T tank,1 ) = (N tank /N tank,1 )

(1/k)

, and

δW (1/(k–1)) shaft rev , = – dT tank N tank,1 (T tank /T tank,1 )

c p0 + VdP.

The efficiency of heat engines can oftentimes be improved by increasing the peak temperature in the relevant thermodynamic cycle. However, materials considerations impose a restriction on the peak temperature. Materials may be kept at a desired safe temperature by providing sufficient cooling. In that case entropy is generated through cooling which must be compared with the work loss by reducing the peak tempera- ture. When heat exchangers and cooling systems are designed for work devices, in- formation on possible work loss should be provided.

dd. Example 31 Consider a 1 meter long turbine that operates steadily and produces a net power out-

put of 1000 kW. Gases enter the turbine at 1300 K (T i ) and 10 bar (P i ), and leave at 900 K (T e ) and 1 bar (P e ). The turbine walls are insulated, but its blades are cooled. The cooling rate per unit area of the blade is given by the relation h(T avg –T blade ) where T avg

= (T i + T e )/2. The Nusselt number (Nu = hC/ λ) on the gas side of the

blade is 1000, where h denotes the convective heat transfer coefficient (kW m –1 K ),

C the chord length (which is 15 c.m. along axial direction), and λ the thermal con- ductivity of the hot gases (= 70_10 –6 kW m –2 K –1 ). The blade A= C _ blade height

which is assumed to be the same as the chord length. There are approximately 40 blades for each rotor and 3 rotors for every meter of length. Assume that c

p = 1.2 kJ kg K and R = 0.287 kJ kg K . Write the generalized overall energy conservation equation.

What is the heat loss rate if the blade temperature T blade = 900 K. Determine the gas mass flow rate. Write the entropy balance equation and simplify it for this problem. Determine the entropy generation rate.

Solution The energy equation can be written in the form

dE/dt = ˙ Q – W ˙ + ˙m i (h + ke+ pe) i – ˙m e (h + ke + pe) e .

At steady state, dE c.v. /dt = 0, ke = 0, pe = 0, dm c.v. /dt = 0, hence ˙m i = ˙m e = m ˙ , i.e.,

(A) Q = h A (T ˙ avg –T blade ), and

0= ˙ Q – W ˙ + m ˙ (h i –h e ).

(B)

A = 0.15 2 × 01.5 × 40 × 3 = 2.7 m . (C) Since, h C/ λ = 1000,

h = 1000 –1 × 70 × 10 ÷ 0.15 = 0.467 kW m K . (D) Using Eqs. (B), (C), and (D), Q = 0.462 ˙ × 2.3 × (1100 – 900) = 252 kW.

(F) W ˙ = 1000 kW.

h e –h i =c p0 (T i –T e ) = 1.2 × (1300 – 900), and

(G) Using Eqs. (A), (E), (F), and (G),

0 = 252 – 1000 – m ˙ × 1.2 × (1300 – 900), i.e., m ˙ = 2.6 kg s –1 .

(H) The entropy balance equation at steady state is

(I) (s e –s i )=c p0 ln(T e /T i ) – R ln(P e /P i ). = 1.2 ln(900 ÷1300) – 0.287ln (1/10)

dS c.v. /dt = 0 = ˙ Q /T b,j + m ˙ (s i –s e )+ ˙ σ cv .

(J)

(K) Using Eqs. (H)–(K), σ ˙

= – 0.441 + 0.66 = 0.220 kJ kg –1 K –1 .

cv .. = – (–252/900) – 2.6 × (–0.22) = 0.28 + 0.57 = 0.85 kW K .

Remark The entropy generation due to blade cooling results in a work loss of T o σ ˙ cv .. = 298 ×

0.85 = 250 kW (this is discussed further in Chapter 4). However, the increased gas temperature at the turbine inlet results in a larger work output.