Question

We = in (haahi) Qinani Chahu) ME 3210 Thermodynamics I Quiz 8 Summer 2020 Name: 1. R134a in an ideal vapor compression heat pump generates 35 kW to heat a room. A mixture of 95% quality enters the compressor at 0.6 bar and saturated liquid exits the condenser at 14 bar. Determine: a) Mass flow rate (kg/hr) th Compressor power (kJ/min) c) Heat transferred from outside (kW) d) Coefficient of performance

Answer 1

Given Dasta (R-134a) works on ideal vapour compression Q Heat pump = 35kW quality (2) (dryness fraction = 0.95 P compress 0.6bar "Compress = 60 kPa e condensor lebar 1400 kPa Solution - Draw vapour compression cycle for above data T Q Hearpump 2 (1-2) Is entropic Compression Pcord Yo (2-3) + (Constant pressure Heat Rejection (3-4) Isenthal fue expertson) 14-2) + (constant pressure Head addition 4 Comp. 15,=521 on (3-4) dettled line due to irreversible prouss Some of the values must be taken from R-134 (A) Pressure Table Ast (60 kPa) hf = 3.9 kJ/kg hig 223.9 kJ/ kg, 227-8 kJ 1 kg hg

g2 o.glob kJ / kg k If = 0.0164 kJ / kg k Stg 0.9481 kJ / kg k Sg = 0.9 645 kJ /ky k At (14ook Pa) ~ hf = 124. 3 Kolkg hfg = 148.9 kolky hg = 276.2 kJ / kg Sf = 0.4533 kJ Stg = 0.4523 kollege kg k mass flow rate 1kg the ) teat pump milha - Ring) [ hoge = hf at P = 1400kPa) I hg = 194.3 kolky But we do not know the value of the ), to find he (S, = 52) - de - due to isentropel process, لما so first we need for this we know that

3 S, = sf + a (stg) yat (pz Gokpal S, = 0.0164 +(-95) (0.9481) $9/kg) SA o.glfogs S2 Now from Super heated stable R-134 (9) we matches vale (si = 0.910095 ) at (P = 1.4 MPa we get value between between los 0.9003 0.92.93) d (itsat = 50°c) (its at aboch Now we interpolation method (for finding the valush h 223.40 kg / kg t = 50°C h+260°c = 28310 kolky First apply interpolation between pritroby , we get 0.9297 - 0.9003 0.9297 -0.91 togs 60 50 Go - (t) t = 55.71°C sale

We need find ha hH = 55.710) Again using interpolation between enthalpies 283.10 - 273.90 28 3.10 h 60 -50 160-55.al = 278.93 kJ / kg 4-55-22 as 255.7°C 19/g ( NOU this h = 55-010 | is actually our hy valus So, Now wsing a = in 1 h ₂ ha) mi 248.93 278 , 93 - 127.5) 35 =m in = 0.23 08 (kg / sec ) m = 0.2308 X 60X Go (kg/ Rr m 830.922 yeter Ang

5 (b) Compressor work (Win) Win in Chahil In this first we need to find the values h = hf + x (htg at (P = Gok pa hi = 3.9 +0.95% ( 223.9) (the = 216.605 kJ/kg] Now but this value in above equation of work Win = 10.2308) ( 278-93 -2 16.605) Win = 14-38461 IKI like Sec I win - 863. 0766 |KJ min Ang