fluid_mechanics
228 Chapter 5 ■ Finite Control Volume Analysis SOLUTION We use a control volume that includes the steam in the turbine from the entrance to the exit as shown in Fig. E5.21b. Applying Eq. 5.69 to the steam in this control volume we get 0 (elevation change is negligible) 0 (adiabatic flow) m # c ȟ 2 ȟ 1 V 2 2 V 2 1 2 g1z 2 z 1 2d Q # net in The work output per unit mass of steam through-flow, w shaft net in , can be obtained by dividing Eq. 1 by the mass flow rate, m # , to obtain W # shaft net in w shaft net in m # ȟ 2 ȟ 1 V 2 2 V 2 1 2 Since w shaft net out w shaft net in , we obtain W # shaft net in (1) (2) Section (1) V 1 = 30 ^ m/s h 1 = 3348 kJ/kg Thus, Control volume Steam turbine w shaft = ? F I G U R E E5.21b Section (2) V 2 = 60 ^ m/s h 2 = 2550 kJ/kg w shaft 3348 kJkg 2550 kJkg 1.35 kJkg net out 797 kJ/kg (Ans) or w shaft ȟ 1 ȟ 2 V 2 1 V 2 2 net out 2 w shaft 3348 kJkg 2550 kJkg net out 3130 m s2 2 160 ms2 2 431 J1Nm24 231 1kgm21Ns 2 2411000 JkJ2 COMMENT Note that in this particular example, the change in kinetic energy is small in comparison to the difference in enthalpy involved. This is often true in applications involving steam turbines. To determine the power output, , we must know the mass flowrate, m # W # shaft . If the flow is steady throughout, one-dimensional, and only one fluid stream is involved, then the shaft work is zero and the energy equation is V5.12 Pelton wheel turbine m # c ǔ out ǔ in a p r b a p out r b in V 2 out V 2 in 2 g1z out z in 2d Q # net in (5.70) We call Eq. 5.70 the one-dimensional, steady flow energy equation. This equation is valid for incompressible and compressible flows. For compressible flows, enthalpy is most often used in the one-dimensional, steady flow energy equation and, thus, we have m # c ȟ out ȟ in V 2 out V 2 in 2 An example of the application of Eq. 5.70 follows. g1z out z in 2d Q # net in (5.71) E XAMPLE 5.22 Energy—Temperature Change GIVEN The 420-ft waterfall shown in Fig. E5.22a involves steady flow from one large body of water to another. FIND flow. Determine the temperature change associated with this SOLUTION To solve this problem we consider a control volume consisting of a small cross-sectional streamtube from the nearly motionless surface of the upper body of water to the nearly motionless surface of the lower body of water as is sketched in Fig. E5.22b. We need to determine T 2 T 1 . This temperature change is related to the change of internal energy of the water, ǔ 2 ǔ 1 , by the relationship T 2 T 1 ǔ 2 ǔ 1 č (1)
5.3 First Law of Thermodynamics—The Energy Equation 229 Section (1) Control volume Section (2) 420 ft F I G U R E E5.22b because the flow is incompressible and atmospheric pressure prevails at sections 112 and 122. Furthermore, F I G U R E E5.22a [Photograph of Akaka Falls (Hawaii) courtesy of Scott and Margaret Jones.] where č 1 Btu1lbm # °R2 is the specific heat of water. The application of Eq. 5.70 to the contents of this control volume leads to m # c ǔ 2 ǔ 1 a p r b a p 2 r b 1 Q # net in V 2 2 V 2 1 2 g1z 2 z 1 2d We assume that the flow is adiabatic. Thus Q # net in 0. Also, (2) because the surface of each large body of water is considered motionless. Thus, Eqs. 1 through 4 combine to yield so that with V 1 V 2 0 T 2 T 1 g1z 1 z 2 2 č č 31 Btu1lbm # °R24 1778 ft # lbBtu2 3778 ft # lb1lbm # °R24 132.2 fts 2 21420 ft2 T 2 T 1 3778 ft # lb1lbm # °R24332.2 1lbm # ft21lb # s 2 24 0.540 °R (Ans) (4) a p r b a p 1 r b 2 (3) COMMENT Note that it takes a considerable change of potential energy to produce even a small increase in temperature. A form of the energy equation that is most often used to solve incompressible flow problems is developed in the next section. 5.3.3 Comparison of the Energy Equation with the Bernoulli Equation When the one-dimensional energy equation for steady-in-the-mean flow, Eq. 5.67, is applied to a flow that is steady, Eq. 5.67 becomes the one-dimensional, steady-flow energy equation, Eq. 5.70. The only difference between Eq. 5.67 and Eq. 5.70 is that shaft power, W # shaft net in, is zero if the flow is steady throughout the control volume 1fluid machines involve locally unsteady flow2. If in addition to being steady, the flow is incompressible, we get from Eq. 5.70 m # c ǔ out ǔ in p out r p in r V 2 out V 2 in g1z 2 out z in 2d Q # net in Dividing Eq. 5.72 by the mass flowrate, m # , and rearranging terms we obtain p out r V 2 out 2 gz out p in r V 2 in 2 gz in 1ǔ out ǔ in q net 2 in (5.72) (5.73)
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228 Chapter 5 ■ Finite Control Volume Analysis<br />
SOLUTION<br />
We use a control volume that includes the steam in the turbine<br />
from the entrance to the exit as shown in Fig. E5.21b. Applying<br />
Eq. 5.69 to the steam in this control volume we get<br />
0 (elevation change is negligible)<br />
0 (adiabatic flow)<br />
m # c ȟ 2 ȟ 1 V 2 2 V 2 1<br />
2<br />
g1z 2 z 1 2d Q # net<br />
in<br />
The work output per unit mass of steam through-flow, w shaft net in , can<br />
be obtained by dividing Eq. 1 by the mass flow rate, m # , to obtain<br />
W # shaft<br />
net in<br />
w shaft <br />
net in m # ȟ 2 ȟ 1 V 2 2 V 2 1<br />
2<br />
Since w shaft net out w shaft net in , we obtain<br />
W # shaft<br />
net in<br />
(1)<br />
(2)<br />
Section (1)<br />
V 1 = 30<br />
^<br />
m/s<br />
h 1 = 3348 kJ/kg<br />
Thus,<br />
Control volume<br />
Steam turbine<br />
w shaft = ?<br />
F I G U R E E5.21b<br />
Section (2)<br />
V 2 = 60<br />
^<br />
m/s<br />
h 2 = 2550 kJ/kg<br />
w shaft 3348 kJkg 2550 kJkg 1.35 kJkg<br />
net out<br />
797 kJ/kg<br />
(Ans)<br />
or<br />
w shaft ȟ 1 ȟ 2 V 2 1 V 2 2<br />
net out<br />
2<br />
w shaft<br />
3348 kJkg 2550 kJkg<br />
net out<br />
3130 m s2 2 160 ms2 2 431 J1Nm24<br />
231 1kgm21Ns 2 2411000 JkJ2<br />
COMMENT Note that in this particular example, the change<br />
in kinetic energy is small in comparison to the difference in enthalpy<br />
involved. This is often true in applications involving steam<br />
turbines. To determine the power output, , we must know<br />
the mass flowrate, m #<br />
W # shaft<br />
.<br />
If the flow is steady throughout, one-dimensional, and only one <strong>fluid</strong> stream is involved, then the<br />
shaft work is zero and the energy equation is<br />
V5.12 Pelton wheel<br />
turbine<br />
m # c ǔ out ǔ in a p r b a p<br />
out<br />
r b in<br />
V 2 out V 2 in<br />
2<br />
g1z out z in 2d Q # net<br />
in<br />
(5.70)<br />
We call Eq. 5.70 the one-dimensional, steady flow energy equation. This equation is valid for incompressible<br />
and compressible flows. For compressible flows, enthalpy is most often used in the<br />
one-dimensional, steady flow energy equation and, thus, we have<br />
m # c ȟ out ȟ in V 2 out V 2 in<br />
2<br />
An example of the application of Eq. 5.70 follows.<br />
g1z out z in 2d Q # net<br />
in<br />
(5.71)<br />
E XAMPLE 5.22<br />
Energy—Temperature Change<br />
GIVEN The 420-ft waterfall shown in Fig. E5.22a involves<br />
steady flow from one large body of water to another.<br />
FIND<br />
flow.<br />
Determine the temperature change associated with this<br />
SOLUTION<br />
To solve this problem we consider a control volume consisting of<br />
a small cross-sectional streamtube from the nearly motionless<br />
surface of the upper body of water to the nearly motionless surface<br />
of the lower body of water as is sketched in Fig. E5.22b. We<br />
need to determine T 2 T 1 . This temperature change is related to<br />
the change of internal energy of the water, ǔ 2 ǔ 1 , by the relationship<br />
T 2 T 1 ǔ 2 ǔ 1<br />
č<br />
(1)