ME 200 L17: Conservation Laws Prime Movers
HW 6 Due today in class or before 430 PM
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HW 7 Posted Due in One Week:
This week, do some HomeWork 7 everyday!
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Nozzles
Common Steady-flow Energy Devices
Diffusers Water, Steam, Gas Turbines
Heat Exchangers and Mixers Compressors
Pump
Throttles
A turbine is a steady-flow device used to produce mechanical work (W) by reducing the internal &/or kinetic &/or potential energy of the working fluid.
• For gas turbines, the fluid drives rotating blades while the υ increases from inlet to exit as the working fluid expands (or the p drops).
Fluid expanding as it passes the blades imparts a force on the blades, which in turn produces a torque on the shaft causing rotation.
A compressor is a steady-flow device that consumes shaft work from rotating blades and in turn compresses the fluid.
• υ decreases from inlet to exit as the working fluid is compressed and the pressure rises .
A pump is a steady-flow device that consumes shaft work from rotating blades that compress the fluid.
• Compressors are used for gas systems, pumps for liquids so operating assumptions are similar.
Liquid specific volume does not change significantly with pressure
GE LM5000 gas turbine: 6.2 m long, 12.5 tons, 3600 rpm, produces 55.2 MW
•
D ~ 8 ft
•
L ~ 12 ft
•
62,000 lbf
– Boeing 747-400
–
Airbus A300-610
•
122,000 lbf
– Boeing 777, etc.
cv
cv
e
e
h e
V e
2
2
gz e
i
i
h i
V i
2
2
gz i
•
Given in problem statement.
•
Any rotating shafts, electricity, or moving boundaries? Certainly! W ≠ 0
• The Δke is often small compared to the Δh, so Δke
≈ 0.
•
No appreciable elevation change, so Δpe ≈ 0.
•
For only one inlet and one outlet.
cv
cv
cv
e h e
m
h
1
h
2
i h i
• Analysis works for aircraft gas turbines, stationary utility power gas turbines, steam turbines etc.
i
m
1
dm cv dt
cv
i
i
e
e
cv
e
V
i
V
1
Air
T
1
=410 K
P
1
=2.3 bars
A
1
=200 cm 2
cv
e
h e e
m
2
V e
2
2
2
1
W
cv
T
2
=290 K
P
2
=1.0 bars
A
2
=280 cm 2
V
2 v
2
A
2
V
2 v
2
A
1
V
1 v
1
gz e
i
• Relatively low pressure ratio of 2.8 and relatively low temperature T
1 may just indicate that this is the last stage of expansion.
i
h i
V i
2
2
gz i
i
m
1
V
i
V
1
T
1
=410 K
P
1
A
1
=2.3 bars
=200 cm 2
Air
cv
28.33
kW
cv
e
T
2
=290 K
P
2
=1.0 bars
A
2
=280 cm 2
2
1
2
Table A-22
State
V
2 v
2
T, K
290
410
A
2
V
2 v
2
A
1
V
1 v
1 u, kJ/kg h, kJ/kg
206.91
290.16
293.43
411.12
V
2 v
2
V
2
T
2
P
2
R
300
60
100
( 290 )(.
287 ) m
3
( min
/ min
) sec
KPa
( K )( KJ / kg
K )
6 .
007 kg / s
Q
Turbine
W
Turbine
1
V
1
2
2
h
2
V
2
2
2
1700
W
Turbine
60
W
Turbine
153.48
2
2000
673.23
kW
Q
Turbine
( in
h exit
)
m (
V
2 in
2000
)
m (
V
2 out
2000
)
290.16
178.6
2
2000
6 Kg/s of steam at an inlet velocity of 75 m/s enter a turbine stage at 3 MPa and
400 o C and exit at a velocity of 125 m/s at a pressure of 2 Mpa at 360 o C. Find: (a) the power developed by this steam turbine stage, (b) the percentage change in the steam density across the turbine and (c) the percentage change in the flow area. The heat loss through the casing is 33 kW.
1
1
6 kg
/
cv
33 s kW
6 kg / s
cv
Q
Turbine
W
Turbine
1
V
2
1
2 h
2
V
2
2
2
m e
m
2
m
V
2
A V
2 2
AV
1 1 v
2
v
2 v
1
33
W
Turbine
6 3230.9
2
2000
3159.3
2000
2
366.6
kW
Turbine
( )
2 2
/ v
2
AV / v
1 1 1
125 A
2
/ 0.141
75 A
1
/ 0.094
A
2
/
1
) /
( v
1
v
2
) / v
2
A
1
(0.094 0.141) / 0
(75 / 125) * (0.141 / 0.094)
0.9
A pump is a steady-flow device that consumes shaft work from rotating blades that compress the fluid.
• Compressors are used for gas systems, pumps for liquids so operating assumptions are similar.
September 17th, 2010 ME 200 17
At steady state, a well-insulated compressor takes in air at 60 ºF, 14.2 psi, with a volumetric flow rate of
1200 ft 3 /min, and compresses it to 500 ºF, 120 psi.
Kinetic and potential energy changes from inlet to exit can be neglected. Determine the compressor power, in hp, and the volumetric flow rate at the exit, in ft 3 /min.
Find
– W cv
– A
2
V
2
= ? in hp
= ? in ft 3 /min
Assumptions
• The control volume is at steady state; the flow is steady
• Q, Δke, and Δpe are negligible.
• T he air is an ideal gas.
System ( air flowing through compressor )
1 compressor
2
Basic Equations dm cv dt
i
i
e
e P
RT air
P
1
T
1
= 14.2 psi
= 60 ºF
A
1
V
1
= 1200 ft 3 /min
P
2
T
2
= 120 psi
= 500 ºF dE cv dt
cv
cv
i
i
h i
V i
2
2
gz i
e
e
h e
V e
2
2
gz e
Solution dm cv dt
i
i
e
e 1
m
2
dE cv dt
cv
cv
cv
i
i
h i
V i
2
2
m
h
1
h
2
gz i
e
e
h e
V e
2
2
gz e
m
AV
RT
P m
A
1
V
1
P
1
RT
1
m
14.2
psi
1200 ft
3
in
1540
f
520
R
1 h 1 ft
2
28.9
lb m
R
2
cv
m
h
1
h
2
From Table A-22E h
1
124 Btu lb m h
2
231 Btu lb m
W cv
lb
5310 124 231 h m
Btu lb m
1 hp
2540 Btu h
5310 lb m h
W cv
223 hp
A
2
V
2
2
A
2
V
2
m RT
2
P
2
A V
2
5310 lb m h
1540
28.9
lbf
ft lb m
R
120 psi
960
R
1 ft
2
1 h in
2
144 60 min
A
2
V
2
262 ft
3 min
In ME 300 and ME 525 we learn about the actual combustion process including pollutant reduction, efficiency enhancement, fuels, alternate
Fuels etc. For now we will assume that fuel is burned and the energy released in the combustion process is available to us as heat to calculate how much the compressed air can be heated or how much water can be boiled etc.
cv
cv
e
e
h e
V e
2
2
gz e
i
i
h i
V i
2
2
gz i
fuel
air
( h e
h i
)
Fuel
( H .
V .)
air
( h e
h i
)
H. V. means heating value of the fuel
Summary
•
Control volume energy and mass conservation equations that we learned are applicable to many practical energy devices and equipment.
•
Important learning comes from application of appropriate assumptions, considering the appropriate working substances and their properties in the proper range of operation to estimate different energy quantities.