Vapor Power Cycle Problems

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# Vapor Power Cycle Problems - PowerPoint PPT Presentation

Vapor Power Cycle Problems. Lecture # 32 11/17/2008. Problem 8.29.

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### Vapor Power Cycle Problems

Lecture # 32

11/17/2008

Problem 8.29

Water is the working fluid in an ideal Rankine cycle with reheat. Superheated vapor enters the turbine at 10 MPa, 480oC, and the condenser pressure is 6 kPa. Steam expands through the first-stage turbine to 0.7 MPa and then reheated to 480oC. Determine for the cycle

• The rate of heat addition, in kJ per kg of steam entering the first stage turbine
• The thermal efficiency
• The rate of heat transfer from the working fluid passing through the condenser to the cooling water in kJ per kg of steam entering the first-stage turbine
Fig08_07

Fig08_07

Solution

State 1: p1 = 100 bars, T1 = 480oC, h1 = 3821.4 kJ/kg; s1 = 6.5282 kJ/kg.K

State 2: p2 = 7 bar, s2 = s1, x2 = (s2 – sf2)/(sg2-sf2) = 0.9619.

h2 = 2684.8 kJ/kg

State 3: p3 = 7 bar, T3 = 480oC, h3 = 3438.9 kJ/kg,

s3 = 7.8723 kJ/kg.K

State 4: p4 = 0.06bar, s4 = s3, x4 = (s4 - sf4)/(sg4 - sf4) = 0.9413

h4 = 2425.6 kJ/kg

State 5: p5 = 0.06 bar, sat liquid, h5 = 151.53 kJ/kg

State 6; h6 = h5 + vf (p6-p5) = 151.53 + (1.006 x 10-3) (100 - 0.06) (105)/(103) = 161.59 kJ/kg

Solution
• Heat in for unit mass flow rate: [(h1-h6) + (h3-h2)] = [(3321.4 – 161.59) + (3438.9 – 2684.8)] = 3913.9 kJ/kg
• Thermal efficiency: Net work / Heat in

Net work = (h1 - h2) + (h3 - h4) – (h6 – h5)

= 636.6 + 1013.3 – 10.06 = 1639.8 kJ/kg

Efficiency = 1639.8 / 3913.9 = 0.419

• Condenser: Heat transfer = (h4 – h5) = 2425.6 – 151.53

= 2274.1 kJ/kg

Problem 8.40

A power plant operates on a regenerative vapor power cycle with one open feed water heater. Steam enters the first turbine stage at 12 MPa, 520oC and expands to 1 MPa, where some of the steam is extracted and diverted to the open feed water heater operating at 1 MPa. The remaining steam expands through the second stage to the condenser pressure of 6 kPa. Saturated liquid exits the open feed water heater at 1 MPa. For isentropic processes in the turbines and pumps, determine for the cycle a) the thermal efficiency, b)mass flow rate in the first stage of the turbine in kg/h for a net power output of 330 MW.

Regenerative Vapor Cycle

Another commonly used method for increasing the thermal efficiency of vapor power plants is regenerative feed water heating.

In an open feed water heater, part of the steam is bled from the first stage of the turbine and mixed with the water after stage 1 pumping. A second pump is added to increase the pressure further. The mass flow rate in the second stage of the turbine is less than that in stage 1.

Fig08_09

Fig08_09

Analysis

The mass and energy balances for a regenerative cycle can be written as:

Analysis

Work and Heat Transfer in turbine and heat exchangers

Solution

See the previous figures for the states:

First calculate the conditions at each state.

State 1: p1 = 120 bar, T1 = 520oC, h1 = 3401.8 kJ/kg., s1 = 6.5555 kJ/kg.K

State 2: p2 = 10 bar, s2 = s1; x2 = 0.9931, h2 = 2764.2 kJ/kg

State 3: p3 = 0.06 bar, s3 = s1, x3 = 0.7727, h3 = 2018.3 kJ/kg.

State 4: p4 = 0.06 bar, sat liquid, h4 = 151.53 kJ/kg.

State 5: h5 = h4 + v4 (p5 - p4) = 151.53 + (1.006 x 10-3) (10.0-0.06) x (105)/103) kJ/kg = 151.53 + 1.0 = 152.53 kJ/kg

State 6: p6 = 10 bar, sat liquid h6 = 762.81 kJ/kg.

State 7: p7 = 120 bar, h7= h6 + v6 (p7 - p6) =762.81 +(1.1273 x 10-3)x (120 – 10) (105)/103) = 778.21 kJ/kg.

Solution

Applying energy and mass balances to the feed water heater:

y(h2 - h5) = (h6 - h5);

y = (h6 - h5) / (h2 - h5)

= (762.81 – 152.53)/(2764.2 – 152.53) = 0.2337

Total work in both turbines per unit mass mass flow rate

= (h1 - h2) + (1 - y) (h2 - h3)

= (3401.8 – 2764.2) + (1. – 0.2337) (2764.2 –2018.3)

= 1209.2 kJ/kg

Total Work for the pumps per unit mass flow rate =

(h7 - h6) + (1-y) (h5 – h4)

= (775.21 – 762.81) + 0.7663 x (152.53 – 151.53)

= 13.17 kJ/kg

Solution

Heat added per unit mass flow rate = 3401.8 – 775.21

= 2626.6 kJ/kg.

Thermal efficiency = Net work / Heat added

= (1209.2 – 13.17)/2626.6

= 0.455

Net power developed = mass flow rate x (1209.2 – 13.17)

= 330 x 103 kJ/s

Mass flow rate = 330 x 103 x (3600) / (1209 - 13.17)

= 9.93 x 105 kg/h