Question # 1Water is the working fluid in an ideal Rankine cycle with reheat. Superheated vapor enters the turbine at10 MPa, 6500C, and the condenser pressure is (60 kPa =0.6 bar). Steam expands through the first-stageturbine to 0.7 MPa and then is reheated to 650oC. Determine for the cycle650°c10 MPALO0 bar7bar2.60kPa=0.6 bara. The rate of heat addition,per kg of steam entering the first-stage turbine.b. Pump workc. Turbine work in stage 1 and stage 2The thermal efficiencyd.

Answered: 650°c 10 MPA 0gmpa 7 bar 60kPa=0.6 bar…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Step 1

A Rankine reheat cycle is given in which steam is first heated at 10 MPa and 650 ^oC and then it is expanded to 0.7 MPa. After the first expansion the steam is again heated to 650 ^oC. Then it is again expanded to 60 kPa.

Obtain the following data from steam table,

$A t P = 10 M P a a n d T = 650 ° C, h_{1} = 3747.954 k J / k g, s_{1} = 7.041 k J / k g \cdot K, A t P = 0.7 M P a, h_{f} = 697 k J / k g, h_{g} = 2762.749 k J / k g s_{f} = 1.992 k J / k g \cdot K s_{g} = 6.707 k J / k g \cdot K A t P = 0.7 M P a a n d 650 ° C, h_{3} = 3812.232 k J / k g s_{3} = 8.322 k J / k g A t P = 60 k P a, h_{f} = 359.837 k J / k g h_{g} = 2652.853 k J / k g s_{f} = 1.145 k J / k g \cdot K s_{g} = 7.531 k J / k g \cdot K v_{f} = v_{5} = 0.001 m^{3} / k g$

Step 2

From steam table obtain the value of enthalpy at state 2 and 4.

$P r o c e s s 1 - 2 i s i s e n t r o p i c e x p a n s i o n, s_{1} = s_{2} = 7.041 A t P = 0.7 M P a a n d s = 7.041 k J / k g \cdot K h_{2} = 2932.7 k J / k g \cdot K P r o c e s s 3 - 4 i s i s e n t r o p i c e x p a n s i o n, s_{3} = s_{4} = 8.322 k J / k g \cdot K A t P = 0.06 M P a a n d s = 8.322 k J / k g \cdot K h_{4} = 3000.5 k J / k g$

(a) Calculating rate of heat addition form the first stage of turbine:

For process 5-6,

$P u m p w o r k = v d P W_{p} = v_{5} (P_{1} - P_{5}) h_{6} - h_{5} = 0.001 \times (10 - 0.06) \times 10^{3} h_{6} - 359.837 = 99.400 h_{6} = 459.237 k J / k g$

The rate of heat addition is given as,

$Q_{6 - 1} = h_{1} - h_{6} Q_{6 - 1} = 3747.954 - 459.237 Q_{6 - 1} = 3288.717 k J / k g$

(b) The pump work is:

$W = v d P W = v_{5} (P_{1} - P_{5}) W = 0.001 (10 - 0.06) \times 10^{3} W = 99.4 k J / k g$

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