Question

7. Consider a Carnot-cycle heat engine with water as the working fluid. The heat transfer to the water occurs at 150°C, during which process the water changes from saturated liquid to saturated vapor. The heat is rejected from the water at 25°C. (a) Show the Carnot-cycle on a T-s diagram (b) Find the properties (P, T, v, h, s) water at each state. Here v is the specific volume. (c) Determine the cycle thermal efficiency, net heat added and work output per kilogram of water. (30 points)

Answer 1

" Carnot cycle saturated liquid curve k saturated vapour curve. Grada .) Ti=150% T3=25c to 3 or reje ->s 2- 3 - 1-2 -Isothermal neat addition i sentropic expansion. 3-4 Isothermal heat rejection 4-1 - isentropic comprenion given during heat addition, water changed from Saturated liquid (1) to saturated vapour. (2) From steam table of water at 150°c saturad steam Psat specific volume = 0.4759 MPa De 0.001090 miling 2 h = 632.18 kJ 0.39278 m2log ng = Se = 1.8417 kollak specific entropy 6.8378 kJ kJ /kg = 2746.4 KJ long og Sg = 168378 kJ kJ ligh = Y, h 3 Sf = s, Sg=s, h, hy=h₂ Vg P, = P₂ = Psatte T = T2 = 150°c heat is rejected at 25°C 25°c water, at From steam table. forsaturated Sg= 8.5579 kJ IES. Psot = 3.1691 kPa Sf 0.3673 kJ /kg.k 2-3 is isentropic process fraction at 3 잘 doynen S₂ = Sg+ X₂ (Sg-Sg 53 = 0.3673+2, (8.5579 -0.3673) 52=53) 6-8378 - 0-36734ng (88.1905) x=0.79 from steam table = 104.87 kJ lieg; hg = 2547.2 kJ/kg hz = hq + ²3 Chg-he) = 104.877 0.79( 2547-2 - 104.87) 2034: 28 kc) kJ/kg ho 3 similarly = 0.001003 m3 ac lig, =: 43.359 miling 3 = 1 Of + 23 (Vg - Up) 0.00100340-79(43.359 -0.001003) 83 = 34.254 miling Similarly 4-1 is isentropic process. S4=5 &q=drynen fraction at a Sa Stap (Sg - Sg) 0.8673+ tg (8.5538 – 0.3673) 1.8417 =0:18 ha= hetaq (ing-by) = 104.87 + 0.18 ( 2547.2 104.87) h = 544.48 kJ /kg & +24 VA .(Alg-U) 0.001003 + 0.18 (43.359-0.001003) V4 = 7.805 m3 Blikega n = thermal 001) Tint Dladdes Whet T-T₂ efficiency for carmot T, cycle T,,Ty are in Kelvin. 2 = of ays 493-298 - 0.295 423 = 29.5% is thermal efficienly Radded h₂-hi = 274 6.4 -632.18 = 2114.3 kJ lkg Wuet = (h₂ ha) = (hi-ha) = 624.42 kJ /ke Scanned with CamScanner

Thus we can find thermal efficiency of carnot cycle using above mentioned formula and it's value is 29.5% and heat added is 2114.3 kj/kg and net work output is 624.42 kJ/kg. And T3=T4= 25 degrees

P3 =P4= 3.1691 kPa

Thanks