Question

Problem 7.023 SI As shown in the figure below, 1.25 kg of refrigerant R-134a is contained in a well-insulated piston-cylinder assembly fitted with an electrical resistor of negligible mass. The refrigerant is initially at 2 bar and a quality of 50%. The resistor is activated until the volume of the refrigerant doubles, while its pressure varies negligibly. Determine, in kJ (a) the amount of energy transfer by work to the piston and (b) the accompanying exergy transfer. (c) the amount of energy transfer by electrical work and (d) the accompanying exergy transfer. (e) the change in exergy of the refrigerant. (f) the amount of exergy destruction. Ignore the effects of motion and gravity and let To = 20°C, po = 1 bar. Insulation R-134a P2 = 2 bar X; = 50% V, = 2V P2P Electrical resistor

Answer 1

plz UPvote thank you

Given Dutaj P = about? V = Of + x Jeg 2, = 0.5 V, = 0.0007533 +0.5X0.099 0, 20:0503 MPIKg B U, = UG + X Ugg u = 38.28 +0.58 186.21 4 = 131.38 kJ/kg Si = Sq+X, Sfg = 0.15457 +0.5X0.78316 5,= 0.54615 kJ/kg-K P2 = abar O2 = 20, V = 2X 0:0503 U₂ = 0.1006 m 1kg > Og at P = 2 bar .: phase is superheated vapor 12= 2 bar Į 4 = 226 kJ/kg V=0.1006 ) Sq=0.943 kJ/kg-k

(9.) Amount of Energy Transfer by work ; Wpiston = medv = 1.25200(0.1006-0.0503) = 12.57kg Wpiston = 12.57 kJ (bo) Exergy Transfer :- (EW) piston = Wpiston - 1o AV = 12:57 - 100X1.95 (0.4006-0.0503) I (Ew) piston = 6.2KJ (0) Amount of Energy Transfer by Electrical work; AU = 0 (Wpiston + Were) Wele = - Wpiston - AU Welc = -12.57-1.25 (226-131.38) Were = -130.8 KJ / (d) Energy Transfers li Ew)ele = –13018 KT | (vol. change =o)

(6.) Change in exergy : - AE= m|14,-40) + f(Ve-V) -To ( 53-)] -- 131.38) +100% (0.0006-0.0503) - 293 (0.943 -0.54615 ) = -20.7 KJ IDE = - 2007 KJ (10) Amount ab Exorgy cestouchion (Ed); - | Ed = Tom (S-5,) / 0.943 - 0.54 615) Ed=293x1.25 Ed = 145.3 KJ