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Tutorial for thermodynamics, Study notes of Thermodynamics

University of Southeastern Philippines (USEP)Thermodynamics

How to solve some thermodynamics problems

Typology: Study notes

2023/2024

Uploaded on 05/20/2024

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meldred-alvarado 🇵🇭

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College of Engineering
Spring Session- 2016 THERMODYNAMICS II - ME 272
Dr. Saeed J. Almalowi, smalowi@taibahu.edu.sa
College of Engineering-Taibah University
Problem 10.44 Statement
A steam power plant operates on an ideal regenerative Rankine cycle. Steam enters the turbine at
6MPa and 450C and is condensed in the condenser at 20 kPa. Steam is extracted from the turbine
at 0.4 MPa to heat the feed water in an open feed water heater. Water leaves the feed water
heater as a saturated liquid. Show the cycle on T-s diagram, and determine
(a) The net work output per kilogram of steam flowing through the boiler.
(b) The thermal efficiency of the cycle.
[Ans. (a) 1017, (b) 37.8 percent]
SOLUTION
m6h6+ m2h2= m3h3, m5= m2+ m6, y = m6
m5
→
pf3

Partial preview of the text

Download Tutorial for thermodynamics and more Study notes Thermodynamics in PDF only on Docsity!

Spring Session- 2016 THERMODYNAMICS II - ME 272 Dr. Saeed J. Almalowi, smalowi@taibahu.edu.sa Problem 10.44 Statement

A steam power plant operates on an ideal regenerative Rankine cycle. Steam enters the turbine at 6MPa and 450C and is condensed in the condenser at 20 kPa. Steam is extracted from the turbine at 0.4 MPa to heat the feed water in an open feed water heater. Water leaves the feed water heater as a saturated liquid. Show the cycle on T-s diagram, and determine

(a) The net work output per kilogram of steam flowing through the boiler. (b) The thermal efficiency of the cycle.

[Ans. (a) 1017, (b) 37.8 percent]

SOLUTION

m 6 h 6 + m 2 h 2 = m 3 h 3 , m 5 = m 2 + m 6 , y =

m 6 m 5

→

Spring Session- 2016 THERMODYNAMICS II - ME 272 Dr. Saeed J. Almalowi, smalowi@taibahu.edu.sa

m 6 h 6 + h 2 m 5 − h 2 m 6 = m 5 h 3 →

m 6 m 5

h 6 −

m 6 m 5

h 2 = (h 3 − h 2 )

y =

m 6 m 5

h 3 − h 2 h 6 − h 2

qout = (1 − y) × (h 7 − h 1 ) = (1 − 0.1463) × (2213 − 251.3) = 1674.

kJ kg

qin = (h 5 − h 4 ) = 3302 − 610.9 = 2691.

kJ kg

wnet = qin − qout = 2691.1 − 1674.7 = 1016.

kJ kg

ηth = 1 −

qout qin

Problem 11.20 Statement

Refrigeration- 134a enters the compressor of a refrigerator at 140 kPa and -10C at a rate of 0. m3/min and leaves at 1 MPa. The isentropic efficiency of the compressor is 78 percent. The refrigerant enters the throttling valve at 0.95 MPa and 30C and leaves the evaporator as saturated vapor at -18.5C. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the power input to the compressor, (b) the rate of heat removal from the refrigerated space, and (c) the pressure drop and rate of heat gain in the line between the evaporator and the compressor.

[Ans. (a) 1.88 kW, (b) 4.99 kW, (c) 1.65 kPa,0.241 kw]

SOLUTION