We next consider a special case—fl ow processes that are at steady-state and revers- ible, with one stream in and one stream out. We wish to come up with an expression to evaluate the work in such a process. The fi rst-law balance, in differential form, is given by:
05 2n#cdah1MWVS2
2 1MWgzb d 1 dQ#
sys1 dW#
s (3.26)
where MW is the molecular weight. The second law is:
n#ds1 dQ#
surr
T 50
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3.8 The Mechanical Energy Balance and the Bernoulli Equation ◄ 161 However, since:
dQ#
surr5 2dQ#
sys
Equation (3.27) can be substituted into (3.26) to give:
05 2n#cdah1MWVS2
2 1MWgzb d 1n#Tds1 dW#
s
Rearranging to solve for work:
dW#
s
n# 5 cdh2Tds1MWdaVS2
2b 1MWgdz§ (3.28) We can simplify Equation (3.28) even further. For a reversible process, we can write:
du5 dqrev1 dwrev5Tds2Pdv If we add d(Pv) to both sides, we get:
dh5d1u1Pv2 5Tds1vdP Solving for 1dh2Tds2 and plugging into Equation (3.28) gives:
dW#
s
n# 5 cvdp1MWdaVS2
2b 1MWgdz§ (3.29)
Equation (3.29) is termed the differential mechanical energy balance. It is a useful form, since the work is written in terms of the measured properties P and v as well as bulk potential and kinetic energy. It is applicable to reversible, steady-state processes with one stream in and one stream out.
If we integrate Equation (3.29), we get:
W#
s/n# 5 3
2
1
vdP1MWc 1VS222VS122/2d 1MWg1z22z12 (3.30)
There are two frequent cases where Equation (3.30)is applied:
Case 1: No Work (Nozzles, Diffusers)
05 3
2
1
vdP1MW ¢VS222VS21
2 ≤ 1MWg1z22z12 (3.31)
Equation (3.31) is the celebrated Bernoulli equation.
Case 2: No eK, eP (Turbines, Pumps) W#
s
n# 5 3
2
1
vdP (3.32)
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162 ► Chapter 3. Entropy and the Second Law of Thermodynamics
An ideal gas enters a turbine with a fl ow rate of 250 mol/s at a pressure of 125 bar and a specifi c volume of 500 cm3/mol. The gas exits at 8 bar. The process operates at steady-state. Assume the process is reversible and polytropic with:
Pv1.55const Find the power generated by the turbine.
SOLUTION Since this process is at steady-state and is reversible, we can use Equation (3.32):
W#
s
n# 5 3
P2
P1
vdP (E3.12A)
Since both v and P vary during the process, we must write v in terms of P to perform the integral in Equation (E3.12A). This polytropic process is described by:
Pv1.55const5P1v11.5
where we have written the constant in terms of state 1, since we know both v and P. Solving for v gives:
v5¢P1v11.5
P ≤
2/3
(E3.12B) Substituting Equation (E3.12B) into (E3.12A) and integrating leaves:
W#
s
n# 51P120.6667v13
P2
P1
P212/32dP531P120.6667v13P0.33334P1
P2
Plugging in numerical values, we get:
W# s
n# 5311.2531073Pa4 20.6667a5310243m3/mol4b a"83 31053Pa42 "1.253 31073Pa4b 5 211,250 3J/mol4
Finally, solving for power gives:
W#
s5 1250 3mol/s4 2 1211,250 3J/mol4 25 22.8 3MW4
The sign is negative, since we get useful work out of the turbine. Note, for comparison, that a coal-fi red power plant generates on the order of 1000 MW.
EXAMPLE 3.12 Power Generated by a Turbine
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3.8 The Mechanical Energy Balance and the Bernoulli Equation ◄ 163
In an actual expansion through the turbine of Example 3.12, 22.1 3MW4 of power is obtained.
What is the isentropic effi ciency, hturbine, for the process? The isentropic effi ciency is given by:
hturbine5 1W#
s2real
1W#
s2rev
where 1W#
s2rev represents the power obtained in the reversible process from the same inlet state and outlet pressure and 1W#
s2real is the power of the actual process.
SOLUTION We run into several different defi nitions of effi ciency, depending on the context.
Isentropic efficiencies compare the actual performance of a process operation with the per- formance it would obtain if it operated reversibly; that is, we compare the given process to the best it could do. For a turbine, the same inlet state and same exit pressure are used in the calculation. Figure E3.13 shows both the actual process (solid line) and the ideal, reversible process (dashed line) on a Ts diagram. Both processes start at the same state and end at the same pressure. The fi nal temperature of the actual process is higher than the reversible pro- cess, since less energy is removed via work.
The isentropic effi ciency is calculated to be:
hturbine522.13MW4 22.83MW450.75 We obtain an isentropic effi ciency of 75%.
We can use a similar approach to determine isentropic effi ciencies of other unit processes, such as pumps or nozzles. The isentropic effi ciency of a pump compares the minimum work needed from the same inlet state and an outlet at the same pressure to the actual work:
hpump5 1W#
s2rev
1W#
s2real
Can you draw a fi gure analogous to Figure E3.13 for a pump? The isentropic effi ciency of a nozzle compares the actual exit kinetic energy to that the fl uid would obtain in a reversible process. Values between 70% and 90% are typical for turbines and pumps, while nozzles typically obtain 95% or better.
EXAMPLE 3.13 Correction for Effi ciency in a Real Turbine
P = 125 bar
P = 8 bar Inaccessible States
Actual process
2real
s2,real
T2,real T2,rev
s1 = s2,rev
2rev Isobars
T1
T
Δs = 0
1
s
Figure E3.13 A Ts diagram llustrating the states between which the isentropic efficiency is calculated. The actual process is shown by the solid line, between states 1 and 2real, while the ideal, reversible process is shown by the dashed line between states 1 and 2rev.
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164 ► Chapter 3. Entropy and the Second Law of Thermodynamics