To determine The nozzle diameter at the various pressures. Explanation Write the expression for the nozzle diameter. d 2 = 4 A 2 π (I) Here, the diameter of the nozzle is d 2 and the area of the nozzle at exit is A 2 . Write the expression for the area at the exit of the nozzle. A 2 = m ˙ v 2 V 2 (II) Here, the mass flow rate is m ˙ , the volume is at the exit of the nozzle V 2 and the specific volume is at exit of the nozzle v 2 . Write the expression for the volume of the nozzle at exit of the nozzle. V 2 = 2 ( h o − h 2 ) (III) Here, the specific enthalpy at stagnation pressure is h o and the specific enthalpy at exit is h 2 . Write the formula of interpolation method of two variables. y 2 = ( x 2 − x 1 ) ( y 3 − y 1 ) ( x 3 − x 1 ) + y 1 (IV) Here, the variables denote by x and y are pressure and specific enthalpy. Conclusion: Refer to table A-4 “properties of superheated water vapor” to obtain the specific enthalpy and the specific entropy at temperature T 1 = 280 ° C and pressure p 1 = 1.83 bar using interpolation as 3031.9 kJ / kg and 7.8839 kJ / kg ⋅ K respectively. Refer to table A-4 “properties of superheated water vapor” to obtain the specific enthalpy at temperature T 1 = 280 ° C and pressure p = 1.5 bar as 3032.8 kJ / kg . Refer to table A-4 “properties of superheated water vapor” to obtain the specific enthalpy at temperature T 1 = 280 ° C and pressure p = 3 bar as 3028.6 kJ / kg . Show the pressure at 3028.6 kJ / kg and 3032.8 kJ / kg as in below Table. Specific enthalpy kJ / kg Pressure bar 3028.6 kJ / kg 1.5 bar ? 1.83 bar 3032.8 kJ / kg 3 bar Substitute 1.5 bar for x 1 , 1.83 bar for x 2 , 3 bar for x 3 , 3028.6 kJ / kg for y 1 , and 3032.8 kJ / kg for y 3 in Equation (V). y 2 = { [ ( 3028.6 kJ / kg − 3032.8 kJ / kg ) ( 1.83 bar − 1.5 bar ) ] ( 3 bar − 1.5 bar ) + 3032.8 kJ / kg } = ( − 1.386 ) 1.5 kJ / kg + 3032.8 kJ / kg = − 0.924 kJ / kg + 3032.8 kJ / kg ≈ 3031.87 kJ / kg So, the specific enthalpy is 3031.87 kJ / kg . Refer to table A-4 “properties of superheated water vapor” to obtain the specific entropy at temperature T 1 = 280 ° C and pressure p = 1.5 bar as 7.9555 kJ / kg ⋅ K . Refer to table A-4 “properties of superheated water vapor” to obtain the specific entropy at temperature T 1 = 280 ° C and pressure p = 3 bar as 7.6299 kJ / kg ⋅ K . Show the pressure at 7.6299 kJ / kg ⋅ K and 7.9555 kJ / kg ⋅ K as in below Table. Specific entropy kJ / kg ⋅ K Pressure bar 7.6299 kJ / kg ⋅ K 1.5 bar ? 1.83 bar 7.9555 kJ / kg ⋅ K 3 bar Substitute 1.5 bar for x 1 , 1.83 bar for x 2 , 3 bar for x 3 , 7.6299 kJ / kg ⋅ K for y 1 , and 7.9555 kJ / kg ⋅ K for y 3 in Equation (V). y 2 = { [ ( 7.6299 kJ / kg ⋅ K − 7.9555 kJ / kg ⋅ K ) ( 1.83 bar − 1.5 bar ) ] ( 3 bar − 1.5 bar ) + 7.9555 kJ / kg ⋅ K } = ( − 1.074 ) 1.5 kJ / kg ⋅ K + 7.9555 kJ / kg ⋅ K = − 0.071632 kJ / kg ⋅ K + 7.9555 kJ / kg ⋅ K ≈ 7.8838 kJ / kg ⋅ K So, the specific entropy is 7.8838 kJ / kg ⋅ K . Since the stagnation pressure and stagnation temperature is equal to the temperature and pressure at the inlet of the nozzle, so the specific entropy and specific enthalpy are also same. Since the process is isentropic, so the entropy at state 2 is equal to the stagnation entropy. Refer to table A-4 “properties of superheated water vapor” to obtain the specific enthalpy at specific entropy s = 7.8052 kJ / kg ⋅ K and pressure p = 1.5 bar as 2952.7 kJ / kg . Refer to table A-4 “Properties of superheated water vapor” to obtain the specific enthalpy at specific entropy s = 7.9555 kJ / kg ⋅ K and pressure p = 3 bar as 3032.8 kJ / kg . Show the pressure at 7.6299 kJ / kg ⋅ K and 7.9555 kJ / kg ⋅ K as in below Table. Specific entropy kJ / kg ⋅ K Specific enthalpy kJ / kg 7.8052 kJ / kg ⋅ K 2952.7 kJ / kg 7.8838 kJ / kg ⋅ K ? 7.9555 kJ / kg ⋅ K 3032.8 kJ / kg Substitute 7.8052 kJ / kg ⋅ K for x 1 , 7.8838 kJ / kg ⋅ K for x 2 , 7.9555 kJ / kg ⋅ K for x 3 , 2952.7 kJ / kg for y 1 , and 3032.8 kJ / kg for y 3 in Equation (V). y 2 = { [ ( 7.8838 kJ / kg ⋅ K − 7.8052 kJ / kg ⋅ K ) ( 3032.8 kJ / kg − 2952.7 kJ / kg ) ] ( 7.9555 kJ / kg ⋅ K − 7

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118832318

Author: MORAN

Publisher: WILEY

See similar textbooks

Videos

Question

Chapter 9.14, Problem 130P

To determine

The nozzle diameter at the various pressures.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 9.14, Problem 129P

Chapter 9.14, Problem 131P

Students have asked these similar questions

Can you solve it analytically using laplace transforms and with Matlab code as well please. Thank You

Can you solve it analytically using laplace transforms and with Matlab code as well please. Thank You.

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.

Chapter 9 Solutions

Fundamentals of Engineering Thermodynamics

Ch. 9.14 - Prob. 1E Ch. 9.14 - Prob. 2E Ch. 9.14 - Prob. 3E Ch. 9.14 - Prob. 4E Ch. 9.14 - Prob. 5E Ch. 9.14 - 6. What is the purpose of a rear diffuser on a...Ch. 9.14 - 7. What is the meaning of the octane rating that...Ch. 9.14 - Prob. 8E Ch. 9.14 - Prob. 9E Ch. 9.14 - 10. What is the purpose of the gas turbine–powered...

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.