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Ideal Regenerative Cycle - Sample Problem (2)

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SAMPLE PROBLEM:
In an Ideal regenerative cycle, 54,500 kg/hr of steam generated at 7.0 MPa, 550°C is received by a regenerative
engine. Extraction for feedwater heating occurs at 3.0 MPa and again at 2.0 MPa, with the remaining steam
expanding to 20.0 kPa.
Determine:
a. Schematic and Ts diagram
b. Enthalpies, kJ/kg
c. Hourly quantities of steam extracted, kg/hr
d. Heat added and rejected, kW
e. Turbine and Total Pump work, kW
f. Net work and Energy Chargeable, kW
g. Overall cycle thermal and engine efficiency, %
SCHEMATIC DIAGRAM:
TS DIAGRAM:
At Point 1: 𝑃 = 7.0 π‘€π‘ƒπ‘Ž; 𝑇 = 550℃
Condition: Superheated Steam
π’Œπ‘±
π’Œπ’ˆ
π‘˜π½
𝑠 = 6.9486
π‘˜π‘” − 𝐾
π’‰πŸ = πŸ‘πŸ“πŸ‘πŸŽ. πŸ—
At Point 2: 𝑃 = 3.0 π‘€π‘ƒπ‘Ž; 𝑠 = 𝑠 = 6.9486
Condition: Superheated Steam
By Interpolation:
s,
T, °C
h,
6.9211
6.9486
7.0833
3230.8
β„Ž
3344.0
π’‰πŸ = πŸ‘πŸπŸ’πŸ—. πŸ—πŸ—πŸπŸ’
400
𝑇
450
π’Œπ‘±
π’Œπ’ˆ
𝑇 = 408.48℃
At Point 3: 𝑃 = 2.0 π‘€π‘ƒπ‘Ž; 𝑠 = 𝑠 = 6.9486
Condition: Superheated Steam
By Interpolation:
s,
T, °C
h,
6.7663
6.9486
6.9562
3023.5
β„Ž
3137.0
π’‰πŸ‘ = πŸ‘πŸπŸ‘πŸ. πŸ’πŸ“πŸ•πŸ”
300
𝑇
350
π’Œπ‘±
π’Œπ’ˆ
𝑇 = 348.00℃
At Point 4: 𝑃 = 0.020 π‘€π‘ƒπ‘Ž; 𝑠 = 𝑠 = 6.9486
Condition: Wet-mixture
Solving for quality using the given entropy, x:
π‘₯=
𝑠 −𝑠
𝑠
=
6.9486 − 0.8319
7.0766
π‘₯ = 0.8644
Solving for β„Ž ;
β„Ž =β„Ž
+ π‘₯β„Ž
β„Ž = [251.38 + (0.8644)(2358.3)]
π’‰πŸ’ = πŸπŸπŸ–πŸ—. πŸ–πŸ—πŸ’πŸ“
π‘˜π½
π‘˜π‘”
π’Œπ‘±
π’Œπ’ˆ
At Point 5: 𝑃 = 0.02 π‘€π‘ƒπ‘Ž
Condition: Saturated Liquid
π’Œπ‘±
π’Œπ’ˆ
π‘š
𝜐 = 0.001017
π‘˜π‘”
π’‰πŸ“ = πŸπŸ“πŸ. πŸ‘πŸ–
At Point 6: 𝑃 = 2.0 π‘€π‘ƒπ‘Ž
Condition:Compressed Liquid
Using Approximate pump work;
= 𝜐 (𝑃
−𝑃 )
π‘š
π‘˜π‘
= 0.001017
(2000 − 20)
π‘˜π‘”
π‘š
π‘˜π½
π‘Š = 2.0137
π‘˜π‘”
π‘Š
π‘Š
Solving for β„Ž ;
β„Ž =π‘Š
+β„Ž
β„Ž = (2.0137 + 251.38)
π’‰πŸ” = πŸπŸ“πŸ‘. πŸ‘πŸ—πŸ‘πŸ•
π‘˜π½
π‘˜π‘”
π’Œπ‘±
π’Œπ’ˆ
At Point 7: 𝑃 = 2.0 π‘€π‘ƒπ‘Ž
Condition: Saturated Liquid
π’Œπ‘±
π’Œπ’ˆ
π‘š
𝜐 = 0.001177
π‘˜π‘”
π’‰πŸ• = πŸ—πŸŽπŸ–. πŸ•πŸ•
At Point 8: 𝑃 = 3.0 π‘€π‘ƒπ‘Ž
Condition:Compressed Liquid
Using Approximate pump work;
= 𝜐 (𝑃
−𝑃 )
π‘š
π‘˜π‘
= 0.001177
(3000 − 2000)
π‘˜π‘”
π‘š
π‘Š
π‘Š
π‘Š
= 1.177
π‘˜π½
π‘˜π‘”
Solving for β„Ž ;
β„Ž =π‘Š
+β„Ž
β„Ž = (1.177 + 908.77)
π’‰πŸ– = πŸ—πŸŽπŸ—. πŸ—πŸ’πŸ•
π‘˜π½
π‘˜π‘”
π’Œπ‘±
π’Œπ’ˆ
At Point 9: 𝑃 = 3.0 π‘€π‘ƒπ‘Ž
Condition: Saturated Liquid
π’Œπ‘±
π’Œπ’ˆ
π‘š
𝜐 = 0.001216
π‘˜π‘”
π’‰πŸ— = πŸπŸŽπŸŽπŸ–. πŸ’πŸ
At Point 10: 𝑃 = 7.0 π‘€π‘ƒπ‘Ž
Condition:Compressed Liquid
Using Approximate pump work;
= 𝜐 (𝑃 − 𝑃 )
π‘š
π‘˜π‘
= 0.001216
(7000 − 3000)
π‘˜π‘”
π‘š
π‘˜π½
π‘Š = 4.864
π‘˜π‘”
π‘Š
π‘Š
Solving for β„Ž ;
β„Ž
β„Ž
=π‘Š +β„Ž
= (4.864 + 1008.41)
π’‰πŸπŸŽ = πŸπŸŽπŸπŸ‘. πŸπŸ•πŸ’
C. MASS OF STEAM EXTRACTED, kg/hr
π’Œπ‘±
π’Œπ’ˆ
π‘˜π½
π‘˜π‘”
By Mass and Energy Balance at OH1 and OH2;
For OH1;
π‘€π‘Žπ‘ π‘  𝐼𝑛 = π‘€π‘Žπ‘ π‘  𝑂𝑒𝑑
π‘š +π‘š = π‘š
πΈπ‘›π‘’π‘Ÿπ‘”π‘¦ 𝐼𝑛 = πΈπ‘›π‘’π‘Ÿπ‘”π‘¦ 𝑂𝑒𝑑
π‘š β„Ž +π‘š β„Ž =π‘š β„Ž
For OH2;
π‘€π‘Žπ‘ π‘  𝐼𝑛 = π‘€π‘Žπ‘ π‘  𝑂𝑒𝑑
π‘š +π‘š =π‘š
πΈπ‘›π‘’π‘Ÿπ‘”π‘¦ 𝐼𝑛 = πΈπ‘›π‘’π‘Ÿπ‘”π‘¦ 𝑂𝑒𝑑
π‘š β„Ž +π‘š β„Ž =π‘š β„Ž
Solving for the mass of the steam extracted;
For OH1;
Since;
π‘š +π‘š = π‘š
Let;
π‘š =π‘š −π‘š
Substituting;
π‘š β„Ž + (π‘š − π‘š )β„Ž = π‘š β„Ž
π‘š β„Ž +π‘š β„Ž −π‘š β„Ž = π‘š β„Ž
π‘š (β„Ž − β„Ž ) = π‘š (β„Ž − β„Ž )
Then;
π‘š =π‘š
π‘š = 54,500
β„Ž −β„Ž
β„Ž −β„Ž
π‘˜π‘” 1008.41 − 909.947
β„Žπ‘Ÿ 3249.9924 − 909.947
π’ŽπŸ = πŸπŸπŸ—πŸ‘. πŸπŸπŸ•πŸ•
Also;
π’Œπ’ˆ
𝒉𝒓
π‘š =π‘š −π‘š
π‘š = (54,500 − 2293.2177)
π‘š = 52,206.7823
For OH2;
Since;
π‘š +π‘š =π‘š
Let;
π‘˜π‘”
β„Žπ‘Ÿ
π‘˜π‘”
β„Žπ‘Ÿ
π‘š =π‘š −π‘š
Substituting;
π‘š β„Ž + (π‘š − π‘š )β„Ž = π‘š β„Ž
π‘š β„Ž +π‘š β„Ž −π‘š β„Ž =π‘š β„Ž
π‘š (β„Ž − β„Ž ) = π‘š (β„Ž − β„Ž )
Then;
π‘š =π‘š
π‘š = 52,206.7823
β„Ž −β„Ž
β„Ž −β„Ž
π‘˜π‘”
908.77 − 253.3937
β„Žπ‘Ÿ 3132.4576 − 253.3937
π’ŽπŸ = 𝟏𝟏, πŸ–πŸ–πŸ’. πŸπŸŽπŸπŸ”
Also;
π’Œπ’ˆ
𝒉𝒓
π‘š =π‘š −π‘š
π‘š = (52,206.7823 − 11,884.1016)
π‘š = 40,322.6807
π‘˜π‘”
β„Žπ‘Ÿ
π‘˜π‘”
β„Žπ‘Ÿ
D. HEAT ADDED AND REJECTED, kW
Solving for heat added in the boiler;
𝑄
𝑄
= 54,500
= π‘š (β„Ž − β„Ž )
π‘˜π‘”
π‘˜π½ 1 β„Žπ‘Ÿ
( 3530.9 − 1013.274)
β„Žπ‘Ÿ
π‘˜π‘” 3600 𝑠
𝑸𝑨𝑩 = πŸ‘πŸ–, πŸπŸπŸ’. πŸŽπŸ”πŸŽπŸ‘
π’Œπ‘±
𝒐𝒓 π’Œπ‘Ύ
𝒔
Solving for heat rejected, 𝑄 in kW
𝑄 = π‘š (β„Ž − β„Ž )
𝑄 = 40,322.6807
π‘˜π‘”
π‘˜π½ 1 β„Žπ‘Ÿ
(2289.8945 − 251.38)
β„Žπ‘Ÿ
π‘˜π‘” 3600 𝑠
𝑸𝑹 = 𝟐𝟐 πŸ–πŸ‘πŸ. πŸ–πŸ–πŸŽπŸ’
π’Œπ‘±
𝒐𝒓 π’Œπ‘Ύ
𝒔
D. TURBINE AND PUMP WORK, kW
Solving for Turbine work in kW;
π‘Š =π‘š β„Ž −π‘š β„Ž −π‘š β„Ž −π‘š β„Ž
π‘Š = [(54500)(3530.9) − (2293.2177)(3249.9924) − (11884.1016)(3132.4576) − (40322.6807)(2289.8945)]
βˆ™
π‘˜π½
1 β„Žπ‘Ÿ
π‘₯
π‘˜π‘” 3600 𝑠
π’Œπ‘±
𝒐𝒓 π’Œπ‘Ύ
𝒔
𝑾𝑻 = πŸπŸ“, πŸ‘πŸ—πŸ’. πŸ’πŸ‘πŸ—πŸ
Solving for the total Work pump in kW;
π‘Š
π‘Š
π‘Š
π‘Š
=
=π‘Š
π‘Š
+π‘Š
+π‘Š
=π‘š π‘Š +π‘š π‘Š
+π‘š π‘Š
= [(54500)(4.864) + (52206.7823)(1.177) + (40322.6807)(2.0137)]
𝑾𝑷𝑻 = πŸπŸπŸ‘. πŸπŸ“πŸ—πŸ
π‘˜π‘” π‘˜π½
1 β„Žπ‘Ÿ
βˆ™
π‘₯
β„Žπ‘Ÿ π‘˜π‘” 3600 𝑠
π’Œπ‘±
𝒐𝒓 π’Œπ‘Ύ
𝒔
E. NET WORK AND ENERGY CHARGEABLE, kW
Solving for work net in kW;
π‘Š
π‘Š
=π‘Š −π‘Š
= (15,394.4391 − 113.2592)π‘˜π‘Š
𝑾𝑡𝑬𝑻 = πŸπŸ“, πŸπŸ–πŸ. πŸπŸ•πŸ—πŸ— π’Œπ‘Ύ
Solving for Energy Chargeable in kW;
Since;
𝐸𝑐 = πΈπ‘›π‘‘β„Žπ‘Žπ‘™π‘π‘¦ π‘œπ‘“ π‘ π‘‘π‘’π‘Žπ‘š
π‘“π‘Ÿπ‘œπ‘š π‘π‘œπ‘–π‘™π‘’π‘Ÿ π‘’π‘›π‘‘π‘’π‘Ÿπ‘–π‘›π‘” π‘‘β„Žπ‘’ π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’ − πΈπ‘›π‘‘β„Žπ‘Žπ‘™π‘π‘¦ π‘œπ‘“ π‘“π‘’π‘’π‘‘π‘€π‘Žπ‘‘π‘’π‘Ÿ π‘™π‘’π‘Žπ‘£π‘–π‘›π‘” π‘‘β„Žπ‘’ π‘™π‘Žπ‘ π‘‘ β„Žπ‘’π‘Žπ‘‘π‘’π‘Ÿ
𝐸𝑐 = π‘š (β„Ž − β„Ž )
𝐸𝑐 = 54500
π‘˜π‘”
π‘˜π½ 1 β„Žπ‘Ÿ
(3530.9 − 1008.41)
π‘₯
β„Žπ‘Ÿ
π‘˜π‘” 3600𝑠
𝑬𝒄 = πŸ‘πŸ–, πŸπŸ–πŸ•. πŸ”πŸ—πŸ“πŸ–
π’Œπ‘±
𝒐𝒓 π’Œπ‘Ύ
𝒔
π‘˜π‘”
β„Žπ‘Ÿ
F. OVERALL CYCLE THERMAL AND ENGINE EFFICIENCY, %
Solving for 𝑒 ;
𝑒 =
𝑒 =
π‘Š
𝑄
× 100
15,281.799
× 100
38,114.0603
𝒆π‘ͺ = πŸ’πŸŽ. πŸŽπŸ—πŸ‘πŸ‘%
Solving for 𝑒 ;
𝑒 =
𝑒 =
π‘Š
× 100
𝐸𝑐
15,394.4391
× 100
38,187.6958
𝒆𝑬 = πŸ’πŸŽ. πŸ‘πŸπŸπŸ”%
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