(a) Explanation The below figure shows the T-s diagram. Figure-(1) Write the expression for gas constant. R = R ¯ M (I) Here, universal gas constant is R ¯ , gas constant is R and the molecular mass is M . Write the expression for entropy production. σ ˙ c v m ˙ = s ° ( T 2 ) − s ° ( T 2 ) − R ln ( p 2 p 1 ) (II) Here, entropy production rate is σ ˙ c v m ˙ , entropy at state 2 is s ° ( T 2 ) , entropy at state 1 is s ° ( T 1 ) , pressure at state 2 is p 2 and the pressure at state 1 is p 1 . Write the expression for enthalpy at state 2. h 2 = h 1 − η t ( h 1 − h 2 s ) . (III) Here, enthalpy at state 2 is h 2 , enthalpy at point 1 is h 1 , efficiency of the turbine is η t and the enthalpy at point 2s is h 2 s Write the expression for relative pressure at point 2s. p r 2 s = p r 1 ( p 2 p 1 ) (IV) Here, relative pressure at point 2s is p r 2 s and the relative pressure at point 1 is p r 1 . Conclusion: Substitute 1.986 Btu / kmol ⋅ °R for R ¯ and 28.97 kmol / lb for M in Equation (I). R = 1.986 Btu / kmol ⋅ °R 28.97 kmol / lb = 0.06855367 Btu / lb ⋅ °R Refer to Table A-22E, “Ideal gas properties of air” at 900 °R to obtain the value of h 1 as 216.26 Btu / lb , s ° ( T 1 ) as 0.72438 Btu / lb ⋅ °R , and p r 1 as 8.411 . Substitute 8.411 for p r 1 , 25 lbf / in 2 for p 2 and 100 lbf / in 2 for p 1 in Equation (IV). p r 2 s = 8.411 ( 25 lbf / in 2 100 lbf / in 2 ) = 8.411 ( 0.25 ) = 2.10275 Calculate enthalpy at point 2s using interpolation formula. h 2 s = h L + ( h H − h L p r H − p r L ) ( p r 2 − p L ) (V) Here, higher limit of enthalpy range is h H , lower limit of enthalpy range is h L , higher limit of relative pressure is P r H , lower limit of relative pressure range is p r L and relative pressure at point 2s is p r 2 . Refer to Table A-22E, “Ideal gas properties of air” at 2.103 relative pressure to obtain the value of P r H as 2.249 p r L as 2.005 , h H as 148.28 and h L as 143.47 . Show the values of enthalpy and relative pressure as in the below Table. Relative pressure Enthalpy ( Btu / lb ) 2.005 143.47 2.103 ? 2.249 148.28 Substitute 216.26 Btu / lb for h 1 , 145.401 Btu / lb for h 2 s and 0.891 for η t in equation (III). h 2 = 216.26 Btu / lb − ( 0.891 ) ( 216.26 Btu / lb − 145.401 Btu / lb ) = 216.26 Btu / lb − 63.135 Btu / lb = 153.124 Btu / lb Calculate the entropy at state 2 by interpolation formula (b) To determine The rate of entropy production.

8th Edition

ISBN: 2818440116926

Author: MORAN

Publisher: WILEY CONS

See similar textbooks

Concept explainers

Work and Heat Transfer

It is generally related to thermal engineering which tells about the use, creation, conversion and transfer of heat energy between the systems. It is also considered that to transfer heat from one source to another, transfer of masses is also required. T…

Videos

Question

Chapter 6.13, Problem 157P

(a)

To determine

The rate of entropy production.

(b)

To determine

The rate of entropy production.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 6.13, Problem 156P

Chapter 6.13, Problem 158P

Students have asked these similar questions

In MATLAB write out a program to integrate the equations of motion of a rigid body. The inertia matrix is given by I = [125 0 0; 0 100 0; 0 0 75] which is a diagonal, where diag operator provides a matrix with given elements placed on its diagonal. Consider three cases where the body rotates 1 rad/sec about each principal axis. Integrate the resulting motion and study the angular rates and the resulting attitude (use any attitude coordinates). For each principal axis case, assume first that a pure spin about the principal axis is performed, and then repeat the simulation where a small 0.1 rad/sec motion is present about another principal axis. Discuss the stability of each motion. The code should produce a total of 6 simulations results when it is ran.

Q. A strain gauge rosette that is attached to the surface of a stressed component C). If the strain gauge rosette is of the D° gives 3 readings (a = A, b = B, &c = type (indicating the angle between each of the gauges), construct a Mohr's Strain Circle overleaf. You should assume that gauge A is aligned along the x-axis. Using the Mohr's Strain Circle calculate the: [10 marks] 100 918 ucy evods gringiz ya mwo quoy al etsede 39 926919 (i) principal strains (1, 2)? (au) oniona [5 marks] (ii) principal angles (1, 2)? You should measure these anticlockwise from the y-axis. 20 [5 marks] (iii) maximum shear strain in the plane (ymax)? Ex = Ea Ey = εc [5 marks] (epol) (apob) é Ea = A = -210 2 B=E₁ = -50 E₁ = C = 340 D = 45° bril elled ✓A bedivordan nemigas olloho shot on no eonsoup Imeneo alubom shine sail-no viss ieqse sidetiva bnat sabied 2

1) Solve and show which is converage or diyverage a = 2+(0.1)" 3 16) a = n 1-2n 2) a = In n 1+2n 17) a = n 1-5n4 3) an = n* +8n³ 18) a =√4"n n² -2n+1 n! 20) a = 4) a₁ = 10 n-1 (Ina) 5) a=1+(-1)" 21) a= 6) a 7) an = * = (12+) (1-1) 2n (-1)+1 2n-1 3n+1 22) a= 3n-1 x" 23) a= .x>0 2n+1 2n 3"x6" 8) a = 24) a = n+1 π 9) a = sin 2 sin n 10) an = n + 2 x n! 25) a = tanh(n) n² 1 26) a = -sin- 2n-1 27) a = tan(n) n n 11) a = 2" 12) a = n 13) a = 8/ +=(1+2)" 14) a = 15) a = √10n In(n+1) 29) a = n 30) an-√n²-1 1 28) a = + √2" (In n)200 n 31) a=- = 1 dx nix

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.