University:
North Idaho College
Course:
ENGR-123 | Introduction to Engineering
Academic year:

2023
Views:

200

Pages:

8
Author:

Hani A.

Example 8 Five kilograms of R-134a is Compressed Adiabatically from Saturated vapor at -10°C to a final pressure of 1.2 MPa and Temperature of 80°C. a) Draw a schematic of the system. b) Determine the actual work c) Energy Destruction Solution: Introduction: material and Energy Balance Equation: Qin + Win + Hin + Epin + Ekin = = Qout + Wout + Hout + Epout + Ekout [((Qout - Qin) + (Wout - Win) + (Hout - Hin) + (Epout - Epin) + (Ekout - Ekin)] = 0 ΔQ + ΔW = m[(ΔH + ΔEK + ΔEP)] ΔH = ΔU + Δ(PV) ΔQ + ΔW = m[(ΔU + Δ(PV)) + ΔEK + ΔEP] ΔQ = Qout - Qin Qout: Heat transfer from the system Qin: Heat transfer to the system. ΔW = Wout - Win Wout : work Done From the system Win : work Done on the system ΔH = ΔU + Δ(PV) m: Mass Flow rate of Refrigerant R-134a ΔEk: change in the Kinetic Energy ΔEp: change in the Potential Energy Adiabatic process → Constant Heat Capacity ΔQ = zero → No Heat transfer dT/T = -R.dv/Cv.V T2/T1 = (V1/V2)^(R/Cv) P2/P1 = (V1/V2)^γ T.V^(γ-1) = Constant T.P^(-1) = Constant P.V^γ = Constant Y → activity coefficient Y = Cp/Cv Cw = R/(γ-1) W = Cw.ΔT = R.ΔT/(γ-1) R.T₁ = P₁.V₁ R.T₂ = P₂.V₂ R.T₂ - R.T₁ = P₂.V₂ - P₁.V₁ W = R.ΔT/(γ-1) = P₁.V₁[(P₂/P₁)^(γ-1) - 1] = R. Ans: for second question. 5 5 kg of R-134a is Compressed Adiabatically from Saturated Vapor at -10C to Final Pressure of 1.2 MPa and Temperature of 80C Draw a schematic of the System. Answer: R-134a m=5kg T₁=-10°C= 263 K" Psat? P2=1.2 M Compressor Satulated DQ=0 T2=353 Vapor No Heat loss at T₁=-10°C = 263K Psat = 0.32 Κρα Sanninted Vapo し Determine the actual work. Answer: 80 actual work W = - P.V. ((P2/P1)^γ - 1) = - R.T. ((P2/P1)^γ - 1) P = 0.32 KPa T = 263 K° ΔQ = 0 m = 5 Kg we found P. From Appendit Table For Saturated Vapor at T=-10c R-134a Tetra Flouro Ethand poly atomic gas and For polyatomic Gas γ=1.3 W = R.T. [(P2/P)^(γ-1)/γ - 1] = [(8.314 J.K^-1) (263 K )] x { (1200/0.32)^(1.3-1)/1.3 - 1 } 1200 0.32 1.3 15 W = (8.314 J.mol^-1.K^-1) (263K)^((1.3-1)/(1.3-1)) (1200/0.32)^((1.3-1)/1.3-1) (7288.606) (6.676-1) 41,370.12 J MOL 7 = For Refrigerant R-134a -CH2 FC F3 1,1,1,2 Tetrafluoro ethane Molecular weight ≈ 101.969/mol if we have 5 kg of R-134a So 49.04 mol R-134a So actual work W W = (41,370.12 J/mol) x (49.04 mol) = 2,028,790.68 J = 2.028 MJ Energy Destruction Answer u From Material and Energy Balance ΔQ + ΔW = m(ΔH + ΔEK + ΔEP) ΔEK = 0 → No Change in The Kinetic ΔEP = o → No Change in The Potential ΔQ = 0 oT ΔW = ΔH + o + o oT ΔW = ΔH → We - Wi = Hz - Hi = ΔH ΔH = -Wi = -2.0.28 MJ ΔH = -2.0.28MJ