HW-6 (5 points) Given: Carbon dioxide goes through an adiabatic process in a piston-cylinder assembly. provided. Find: Calculate the entropy change for each case: State data is a) Constant specific heats b) Variable specific heats c) Conclude on the accuracy of part a) and part b) EFD: Assumptions: Air can be modeled as an ideal gas. 1 Basic Equations: s = c p ln T R ln p T 1 p 1 s = s o s o 1 R ln p p 1 : 1
a) For an ideal gas with constant c p : s = c p ln T R ln p T 1 p 1 kj ( 8.314 = 0.939 kg K ln50 80 kj ) 44.01 kg K = 0.1469 kj/(kg K) ln 0 S = m s = 5 0.1469 = 0.73145 kj/k 5 b) Using the CO ideal gas tables: s o = 36.6 kj/(kmol K) s o 1 = 11.5 kj/(kmol K) s = s o s o 1 R ln p p 1 kj (36.6 11.5) = kmol K kg 44.01 kmol = 0.13534 kj/(kg K) ( 8.314 kj ) 44.01 kg K ln 0 S = m s = 5 0.13534 =.67670 kj/k 5 c) Comparing part a) and part b) 0.73145 0.67670 Relative error % = 0.67670 = 0.081 = 8.1% We observe that using constant specific heats instead of using the tables would create an error of 8.1%. Note that the tables account for the change in specific heat with temperature which is not captured by the assumption of constant specific heats. 4, may show or not a calculation
HW-7 (5 points) Given: Air and carbon monoxide are contained to opposite sides of a rigid, insulated container by a partition free to move. Find: Determine the entropy generation of the process. EFD: 1, the boundary should include both gases Assumptions: Neglect KE and P E. Both gases can be modeled as ideal gases. 1 Insulated Q = 0 1 Rigid Tank W b = 0 Basic Equations: For a closed system: S = Q T + σ s = c p ln T R ln p 1 T 1 p 1 : a) Applying the second law to the control system: 0 Q S = T + σ 1 σ = S CO + S air = (m s) CO + (m s) air 4 3
From HW 1 we know: c v,co = R co k co 1 = 0.189 1.4 1 = 0.7875 kj/kg K c v,air = R air k air 1 = 0.87 1.4 1 = 0.7175 kj/kg K Calculating the specific heats at constant pressure: c p,co = c v,co + R co = 0.7875 + 0.189 = 0.9765 kj/kg K 3 c p,air = c v,air + R air = 0.7175 + 0.87 = 1.005 kj/kg K 3 Recalling from HW 1, p final =.65 bar and T final = 419 K. We calculate the entropy change for air and CO : s CO = c pco ln T T 1 R co ln p p 1 kj = 0.9765 kg K ln419 450 0.189 kj kg K ln.65 = 0.189 kj/(kg K) 3 s air = c pair ln T T 1 R air ln p p 1 kj = 1.005 kg K ln419 350 0.87 kj kg K ln.65 5 = 0.36305 kj/(kg K) 3 Finally, the entropy generation is calculated: σ = (m s) CO + (m s) air = 4 0.189 + 0.36305 = 0.3454 kj/k 4
HW-8 (5 points) Given: Data are provided for the steady state operation in series of a boiler, turbine and condenser. Expansion valves are used between the boiler and the turbine and between the turbine and the condenser. Find: a) Locate states 1-6 on a T-s diagram. b) Calculate the power of the turbine, in kj per kg of steam flowing. c) For the valves and the turbine, evaluate the rate of entropy production, in kj/k per kg of steam flowing. d) Rank the components from the most inefficient component to the most efficient component. e) If you were to increase the efficiency of the process which component would you change. EFD: 3 Assumptions: Neglect KE and P E. 1 No heat loss from the turbine or the expansion valves. 1 Steady-State Steady Flow 1 1 DUF 5
Basic Equations: ds dt = Q T + i ṁ : i (s) i o ṁo(s) o + σ 1 a) 3 b) From the Superheated Vapor tables we obtain: h = 3517.7 kj/kg s = 7.00 kj/(kg K) h 4 = 940. kj/kg s 4 = 7.3 kj/(kg K) Applying the first law to both throttling valves we obtain: h = h 3 h 4 = h 5 6
Applying the first law to the turbine: Solving for ẇ turb : 0 0 de dt = Q Ẇ + ṁ air [ ] (h 1 h ) + KE 0 + P 0 E Ẇ turb ṁ = (h 3 h 4 ) 1 w turb = (h h 4 ) = (3517.7 940.)kJ/kg = 577.5 kj/kg c) Interpolation for s 3 : p = 40 bar p = 40 bar h 3446.0 3517.7 3537.5 s 7.09 s 3 7.08 s 3 7.09 7.08 7.09 = 3517.7 3446.0 3537.5 3446.0 s 3 = 7.09 + (7.08 7.09) s 3 = 7.189 kj/(kg K) 1 ( ) 3517.7 3446.0 3537.5 3446.0 Interpolation for s 5 : p = 1 bar p = 1 bar h 875.5 940. 954.6 s 7.836 s 5 7.996 s 5 7.836 7.996 7.836 = 940. 875.5 954.6 875.5 s 5 = 7.836 + (7.996 7.836) s 5 = 7.9669 kj/(kg K) 1 ( ) 940. 875.5 954.6 875.5 7
Applying the second law to the first throttling valve: Q S = 0 T + σ σ m = s = s 3 s = (7.189 7.00) kj/(kg K) = 0.1809 kj/(kg K) Applying the second law to the turbine: 0 Q S = T + σ σ m = s = s 4 s 3 = (7.3 7.189) kj/(kg K) = 0.0491 kj/(kg K) Applying the second law to the second throttling valve: 0 Q S = T + σ σ m = s = s 5 s 4 = (7.9669 7.3) kj/(kg K) = 0.73490 kj/(kg K) d) Ranking them from most inefficient to most efficient component: 1) Throttling valve ) Throttling valve 1 3) Turbine e) The throttling valve 1 is required for flow control and, hence, should be kept. Removing the throttling valve would allow for further expansion of the steam which would result in greater work production and less entropy generation, therefore, it is recommended to remove the throttling valve. 8