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Steamulator Solved Examples
Advanced Problems |
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Example 1
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Steam is
expanded from 800゜F and 250 psia to 5 psia through an 85% efficient
turbine. What is the final enthalpy of the steam? |
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Step 1 |
Since T
= 800゜F > Tcritical, the initial steam is superheated.
Enthalpy and entropy at this stage is obtained by:
Sub Region: Superheated Steam
Units: US
Given: T = 800゜F & P = 250 psia
Find: H
Answer: H = 1423.4082 BTU/lbm
Find: S
Answer: S = 1.7405 BTU/lbm/゜R |
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Step 2 |
If the
expansion is isentropic (100% efficient), then the enthalpy at the
end of the expansion can be found by:
Sub Region: Water-Steam Mixture
Units: US
Given: P = 5 psia & S = 1.7405 BTU/lbm/゜R
Find: H
Answer: H = 1066.5264 BTU/lbm |
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Step 3 |
The
energy extracted by the ideal turbine is:
H1-H2 = 1423.4082-1066.5264 = 356.8818
The energy extracted by the real turbine is:
85%(H1 - H2) = 0.85*356.8818 = 303.3495
The final enthalpy is:
H2= H1-303.3495 = 1120.0587 BTU/lbm |
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Example
2
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A boiler
feed pump increases the pressure of 15 psia 100゜F water to 180 psia.
What is the final water temperature if the pump has an isentropic
efficiency of 80%? |
Step 1
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Sub
Region: Compressed Water
Units: US
Given: T = 100゜F & P = 15 psia
Find: H
Answer: H = 68.0360 BTU/lbm
Find: S
Answer: S = 0.1295 BTU/lbm/゜R
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Step 2 |
If the
compression is isentropic (100% efficient), then the enthalpy at the
end of the compression can be found by:
Sub Region: Compressed Water
Units: US
Given: P = 180 psia & S = 0.1295 BTU/lbm/゜R
Find: H
Answer: H = 68.5101 BTU/lbm
The energy added to the water by an ideal pump is:
H2 -H1 = 68.5101 - 68.0360 = 0.4741 BTU/lbm |
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Step 3 |
Since
the pump is not 100% efficient, not all of the 0.4741 BTU/lbm
enthalpy increase goes into raising the pressure. Therefore, more
energy must be added to the water:
(H2 -H1)/80%=0.4741/0.80=0.5926 BTU/lbm
The final enthalpy is:
H2 = H1 + 0.5926 = 68.6286 BTU/lbm
The final temperature can be found by:
Sub Region: Compressed Water
Units: US
Given: P = 180 psia & H = 68.6286 BTU/lbm
Find: T
Answer: T = 100.1576゜F |
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Example
3
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A 800kW
steam turbine has a water rate of 25 lbm/kW-hr when expanding 75゜F
superheat steam from 150 psig to 25" Hg absolute. Find the quantity
of 70゜F cooling water needed for a mixing condenser if the turbine
operates straight condensing. |
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Step 1 |
Find S
of the superheated steam.
P=150 psig = 150 + 14.7 = 164.7 psia
Sub Region: Saturated Steam
Units: US
Given: P = 164.7 psia
Find: T
Answer: T = 365.8694゜F
Temperature of 75゜F superheat steam
= 365.8694 + 75 = 440.8694゜F
The entropy of the superheated steam can be found by:
Sub Region: Superheated Steam
Units: US
Given: T = 440.8694゜F & P = 164.7 psia
Find: H
Answer: H = 1240.2779 BTU/lbm
Find: S
Answer: S = 1.6137 BTU/lbm/゜R |
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Step 2 |
Find H
and T at the end of the condensing.
Water Rate = 3413/(H1 - H2)
H2 = H1 - 3413/WR = 1240.2779 - 3413/25
= 1103.7579 BTU/lbm
Sub Region: Water-steam Mixture
Units: P: in Hg; T: F; H: BTU/lbm; S: BTU/lbm/゜R
Given: P = 25 in Hg & H=1103.7579 BTU/lbm
Find: T
Answer: T = 203.0774゜F
Find: X
Answer: X = 0.9556 |
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Step 3 |
Find the
required cooling water quantity.
Assume after cooling, the water becomes saturated:
Sub Region: Saturated Water
Units: P: in Hg; T: F; H: BTU/lbm; S: BTU/lbm/゜R
Given: P = 25 in Hg
Find: T
Answer: T = 203.0774゜F
Find: Hf
Answer: Hf = 171.1819 BTU/lbm
The enthalpy change during the cooling is:
H2-Hf = 1103.7579 - 171.1819 = 932.576 BTU/lbm
The mass flow rate is:
WR*kW = 25 lbm/kW-hr * 800kW = 20,000 lbm/hr
Total enthalpy change = 20,000 * 932.576
= 1.865152 x 107 BTU/hr
From heat balance:
Cooling water rate = 1.865152 x 107 /(203.0774-70)
=1.402 x 105 lbm/hr |
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