Explanation Write the expression for the energy balance for nozzle. 0 = Q ˙ c v − W ˙ c v + m ˙ ( h 1 − h 2 + V 1 2 − V 2 2 2 + g ( z 1 − z 2 ) ) (I) Here, the specific enthalpy at state 1 is h 1 , the specific enthalpy at state 2 is h 2 , the rate of work transfer is W ˙ c v , the mass flow rate is m ˙ ,the velocity at state 1 is V 1 , the velocity at state 2 is V 2 , the datum head at state 2 is z 2 , the datum head at state 1 is z 1 and the rate of heat transfer is Q ˙ c v . Since the device is nozzle, therefore, work and heat transfer is zero and the datum head at state 2 and the datum head at state 1 are also equal. Substitute 0 for W ˙ c v , 0 for ( z 1 − z 2 ) and 0 for Q ˙ c v in Equation (I). 0 = 0 − 0 + m ˙ ( h 1 − h 2 + V 1 2 − V 2 2 2 + g × 0 ) 0 = m ˙ ( h 1 − h 2 + V 1 2 − V 2 2 2 ) h 1 − h 2 + V 1 2 − V 2 2 2 = 0 (II) Write the expression for the change in specific enthalpy. h 1 − h 2 = h ¯ 1 − h ¯ 2 M Here, the molar specific enthalpy at state 1 is h ¯ 1 , the molar specific enthalpy at state 2 is h ¯ 2 and the molecular weight of mixture is M . Substitute h ¯ 1 − h ¯ 2 M for h 1 − h 2 in Equation (II). h ¯ 1 − h ¯ 2 M + V 1 2 − V 2 2 2 = 0 (III) Write the expression for the molecular weight of mixture. M = y H e M H e + y C O 2 M C O 2 (IV) Here, the number of mole of helium is y H e , the number of mole of carbon dioxide is y C O 2 ,the molecular weight of helium is M H e and the molecular weight of carbon dioxide is M C O 2 . Write the expression for the change in enthalpy. h ¯ 1 − h ¯ 2 = y H e ( h ¯ 1 − h ¯ 2 ) H e + y C O 2 ( h ¯ 1 − h ¯ 2 ) C O 2 (V) Here, the change in enthalpy of helium is ( h ¯ 1 − h ¯ 2 ) H e and the change in enthalpy of carbon dioxide is ( h ¯ 1 − h ¯ 2 ) C O 2 . Write the expression of molar specific heat of helium. ( C ¯ p ) H e = 5 2 R ¯ Here, the universal gas constant is R ¯ . Write the expression for the change in enthalpy of helium. ( h ¯ 1 − h ¯ 2 ) H e = ( C ¯ p ) H e ( T 2 − T 1 ) (VI) Substitute 5 2 R ¯ for ( C ¯ p ) H e in Equation (VI). ( h ¯ 1 − h ¯ 2 ) H e = 5 2 R ¯ ( T 2 − T 1 ) Substitute 5 2 R ¯ ( T 2 − T 1 ) for ( h ¯ 1 − h ¯ 2 ) H e in Equation (V). h ¯ 1 − h ¯ 2 = y H e 5 2 R ¯ ( T 2 − T 1 ) + y C O 2 ( h ¯ 1 − h ¯ 2 ) C O 2 Substitute y H e 5 2 R ¯ ( T 2 − T 1 ) + y C O 2 ( h ¯ 1 − h ¯ 2 ) C O 2 for h ¯ 1 − h ¯ 2 in Equation (III). y H e 5 2 R ¯ ( T 2 − T 1 ) + y C O 2 ( h ¯ 1 − h ¯ 2 ) C O 2 M + V 1 2 − V 2 2 2 + g ( z 1 − z 2 ) = 0 (VII) Write the expression for change in molar specific entropy. s ¯ 2 − s ¯ 1 = y H e ( s ¯ 2 − s ¯ 1 ) H e + y C O 2 ( s ¯ 2 − s ¯ 1 ) C O 2 (V) Here, the change in entropy of helium is ( s ¯ 2 − s ¯ 1 ) H e , the change in entropy of carbon dioxide is ( s ¯ 2 − s ¯ 1 ) C O 2 , the molar specific entropy at state 1 is s ¯ 1 and the molar specific entropy at state 2 is s ¯ 2 . Write the expression for change in molar specific entropy for helium. ( s ¯ 2 − s ¯ 1 ) H e = ( ( c p ) H e ln T 2 T 1 − R ¯ ln p 2 p 1 ) Here, the temperature at state 1 is T 1 , the temperature at state 2 is T 2 , the pressure at state 1 is p 1 and the pressure at state 2 is p 2 . Write the expression for change in entropy for carbon dioxide. ( s ¯ 2 − s ¯ 1 ) C O 2 = ( ( s ¯ C O 2 T 2 − s ¯ C O 2 T 1 ) − R ¯ ln p 2 p 1 ) Here, the entropy of carbon dioxide at temperature T 2 is s ¯ C O 2 T 2 and the entropy of carbon dioxide at temperature T 1 is s ¯ C O 2 T 1 . Substitute ( ( s ¯ C O 2 T 2 − s ¯ C O 2 T 1 ) − R ¯ ln p 2 p 1 ) for ( s ¯ 2 − s ¯ 1 ) C O 2 and ( ( c p ) H e ln T 2 T 1 − R ¯ ln p 2 p 1 ) for ( s ¯ 2 − s ¯ 1 ) H e in Equation (V). s ¯ 2 − s ¯ 1 = y H e ( ( c p ) H e ln T 2 T 1 − R ¯ ln p 2 p 1 ) + y C O 2 ( ( s ¯ C O 2 T 2 − s ¯ C O 2 T 1 ) − R ¯ ln p 2 p 1 ) (VI) Since the process is isentropic, therefore entropy change will be zero

Fundamentals of Engineering Thermodynamics, Binder Ready Version

8th Edition

ISBN: 9781118820445

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

Publisher: WILEY

See similar textbooks

Concept explainers

Entropy

In physics, entropy means the amount of disorder or the degree of randomness associated with a thermodynamic system. The system is not self-organized when the entropy of the system is high. For instance, when a system is at an elevated temperature, its m…

Videos

Question

Chapter 12.9, Problem 21P

To determine

The pressure in atm at nozzle exit.

The temperature at nozzle exit in °F.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 12.9, Problem 20P

Chapter 12.9, Problem 22P

Students have asked these similar questions

Q11. Determine the magnitude of the reaction force at C. 1.5 m a) 4 KN D b) 6.5 kN c) 8 kN d) e) 11.3 KN 20 kN -1.5 m- C 4 kN -1.5 m B Mechanical engineering, No Chatgpt.

please help with this practice problem(not a graded assignment, this is a practice exam), and please explain how to use sohcahtoa

Solve this problem and show all of the work

Chapter 12 Solutions

Fundamentals of Engineering Thermodynamics, Binder Ready Version

Ch. 12.9 - Prob. 1E Ch. 12.9 - Prob. 2E Ch. 12.9 - Prob. 3E Ch. 12.9 - Prob. 4E Ch. 12.9 - Prob. 5E Ch. 12.9 - Prob. 6E Ch. 12.9 - Prob. 7E Ch. 12.9 - Prob. 8E Ch. 12.9 - Prob. 9E Ch. 12.9 - Prob. 10E

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

The pressure in atm at nozzle exit. The temperature at nozzle exit in °F .

Concept explainers

Videos

The pressure in atm at nozzle exit. The temperature at nozzle exit in °F.

Want to see the full answer?

Chapter 12 Solutions

The pressure in atm at nozzle exit.

The temperature at nozzle exit in °F.