(a) Explanation The figure below shows the T-s diagram. Figure-(1) Write the expression for interpolating temperature at point 2s. h 2 s = h 300 + h 350 − h 300 s 350 − s 300 ( s 1 − s 300 ) (I) Here, enthalpy at point 2s is, entropy at point 1 is s 1 , enthalpy at 350 °F is h 350 and the enthalpy at 300 °F is h 300 , entropy at 300 °F is s 350 and entropy at 300 °F is s 300 Write the expression for velocity at point 2s. v 2 s = v 1 2 + 2 ( h 1 − h 2 s ) (II) Here, velocity at point 2s is v 2 s , enthalpy at state 1 is h 1 and enthalpy at state 2 is h 2 . Write the expression for exit velocity. v 2 = v 2 s ( η n o z z l e ) (III) Here, exit velocity is v 2 and the nozzle efficiency is η n o z z l e . Conclusion: Refer to Table A4E, “Properties of superheated water vapor” at 100 lbf / in 2 to obtain the value of h 1 as 1279.1 Btu / lb , and s 1 as 1.7085 Btu / lb ⋅ °R . Substitute 1186.8 Btu / lb for h 300 , 1211.8 Btu / lb for h 350 , 1.7085 Btu / lb ⋅ °R for s 1 , 1.7312 Btu / lb ⋅ °R for s 350 and 1.6993 Btu / lb ⋅ °R for s 300 in Equation (I). h 2 s = 1186.8 Btu / lb + 1211.8 Btu / lb − 1186 (b) To determine The amount of entropy produced.

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118412930

Author: Michael J. Moran, Howard N. Shapiro, Daisie D. Boettner, Margaret B. Bailey

Publisher: WILEY

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Chapter 6.13, Problem 155P

(a)

To determine

The velocity of steam at the exit.

(b)

To determine

The amount of entropy produced.

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Chapter 6.13, Problem 154P

Chapter 6.13, Problem 156P

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Chapter 6 Solutions

Fundamentals of Engineering Thermodynamics

Ch. 6.13 - Prob. 1E Ch. 6.13 - Prob. 2E Ch. 6.13 - Prob. 3E Ch. 6.13 - Prob. 4E Ch. 6.13 - Prob. 5E Ch. 6.13 - 6. Is entropy produced within a system undergoing...Ch. 6.13 - 7. When a mixture of olive oil and vinegar...Ch. 6.13 - Prob. 8E Ch. 6.13 - Prob. 9E Ch. 6.13 - 10. Is Eq. 6.51a restricted to adiabatic processes...

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