To determine The rate of entropy production for steam generator. The rate of entropy production for turbine. The name of component that contributes more inefficient operation of the overall system. Explanation Write the expression for mass flow rate at state 1. m ˙ 1 = ( A V ) 1 ( R T 1 P 1 ) (I) Here, the mass flow rate at state 1 is m ˙ 1 , area of steam generator at state 1 is A 1 , volumetric flow at state 1 is V 1 , gas constant is R , temperature at state 1 is T 1 and pressure at state 1 is P 1 . Write the expression for rate of entropy production for steam generator. σ ˙ he = m ˙ 1 ( s 2 − s 1 ) + m ˙ 3 ( s 4 − s 3 ) (II) Here, the rate of entropy production for steam generator is σ ˙ he , entropy at state 2 is s 2 , entropy at state 1 is s 1 , entropy at state 3 is s 3 , entropy at state 4 is s 4 , mass flow rate at state 1 is m ˙ 1 and mass flow rate at state 3 is m ˙ 3 . Write the expression for rate of entropy production for turbine. σ ˙ turbine = m ˙ 3 ( s 5 − s 4 ) (III) Here, the rate of entropy production for turbine is σ ˙ turbine and entropy at state 5 is s 5 . Conclusion: The atmospheric pressure is 14.7 lbf / in 2 and gas constant for air is 53.33 ft ⋅ lbf / lb ⋅ ° R . Convert the temperature at state 1 from Fahrenheit to Rankine. T 1 = 400 ° F = ( 400 ° F + 460 ) ° R = 860 ° R Substitute ( 2 × 10 5 ft 3 / min ) for ( A V ) 1 , 53.33 ft ⋅ lbf / lb ⋅ ° R for R , 14.7 lbf / in 2 for P 1 and 860 ° R for T 1 in Equation (I). m ˙ 1 = ( 2 × 10 5 ft 3 / min ) ( 53.33 ft ⋅ lbf / lb ⋅ ° R ) ( 860 ° R ) 14.7 lbf / in 2 = ( 2 × 10 5 ft 2 / min ) ( 144 in 2 1 ft 2 ) ( 14.7 lbf / in 2 ) ( 45863.8 lbf / lb ) = 9230.809 lb / min Refer to Table A-22E “Ideal gas properties of air” to obtain enthalpy ( h 1 ) at state 1 as 206.46 Btu / lb corresponding to temperature 860 ° R and enthalpy ( h 2 ) at state 2 as 172.39 Btu / lb . Refer to Table A-2E “Properties of saturated water” to obtain enthalpy ( h 3 ) as 70 Btu / lb corresponding to state 3. Write the expression for enthalpy at state 4. h 4 = h 3 + ( m ˙ 1 m ˙ 3 ) ( h 1 − h 2 ) (IV) Here, enthalpy at state 4 is h 4 . Substitute 70 Btu / lb for h 3 , 9230.809 lb / min for m ˙ 1 , 206.46 Btu / lb for h 1 , 172.39 Btu / lb for h 2 and 275 lb / min for m ˙ 3 in Equation (IV). h 4 = ( 70 Btu / lb ) + ( 9230.809 lb / min 275 lb / min ) ( 206.46 Btu / lb − 172.39 Btu / lb ) = ( 70 Btu / lb ) + ( 1143.6133 Btu / lb ) = 1213.613 Btu / lb Refer to Table A-4E “Properties of superheated water vapor” to obtain pressure P 4 as 40 lbf / in 2 corresponding to enthalpy 1213.613 Btu / lb

8th Edition

ISBN: 2818440116926

Author: MORAN

Publisher: WILEY CONS

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 6.13, Problem 109P

To determine

The rate of entropy production for steam generator.

The rate of entropy production for turbine.

The name of component that contributes more inefficient operation of the overall system.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 6.13, Problem 108P

Chapter 6.13, Problem 110P

Students have asked these similar questions

CORRECT AND DETAILED SOLUTION WITH FBD ONLY. I WILL UPVOTE THANK YOU. CORRECT ANSWER IS ALREADY PROVIDED. I REALLY NEED FBD. The cantilevered spandrel beam shown whose depth tapers from d1 to d2, has a constant width of 120mm. It carries a triangularly distributed end reaction.Given: d1 = 600 mm, d2 = 120 mm, L = 1 m, w = 100 kN/m1. Calculate the maximum flexural stress at the support, in kN-m.2. Determine the distance (m), from the free end, of the section with maximum flexural stress.3. Determine the maximum flexural stress in the beam, in MPa.ANSWERS: (1) 4.630 MPa; (2) 905.8688 m; (3) 4.65 MPa

CORRECT AND DETAILED SOLUTION WITH FBD ONLY. I WILL UPVOTE THANK YOU. CORRECT ANSWER IS ALREADY PROVIDED. I REALLY NEED FBD A concrete wall retains water as shown. Assume that the wall is fixed at the base. Given: H = 3 m, t = 0.5m, Concrete unit weight = 23 kN/m3Unit weight of water = 9.81 kN/m3(Hint: The pressure of water is linearly increasing from the surface to the bottom with intensity 9.81d.)1. Find the maximum compressive stress (MPa) at the base of the wall if the water reaches the top.2. If the maximum compressive stress at the base of the wall is not to exceed 0.40 MPa, what is the maximum allowable depth(m) of the water?3. If the tensile stress at the base is zero, what is the maximum allowable depth (m) of the water?ANSWERS: (1) 1.13 MPa, (2) 2.0 m, (3) 1.20 m

CORRECT AND DETAILED SOLUTION WITH FBD ONLY. I WILL UPVOTE THANK YOU. CORRECT ANSWER IS ALREADY PROVIDED. I NEED FBD A short plate is attached to the center of the shaft as shown. The bottom of the shaft is fixed to the ground.Given: a = 75 mm, h = 125 mm, D = 38 mmP1 = 24 kN, P2 = 28 kN1. Calculate the maximum torsional stress in the shaft, in MPa.2. Calculate the maximum flexural stress in the shaft, in MPa.3. Calculate the maximum horizontal shear stress in the shaft, in MPa.ANSWERS: (1) 167.07 MPa; (2) 679.77 MPa; (3) 28.22 MPa

Chapter 6 Solutions

FUNDAMENTALS OF ENGINEERING THERMODYNAM

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...

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.