Question 1. A simple Rankine cycle works between pressures 28 bar and 0.06 bar, the initial condition of steam being Dry saturated. Calculate the cycle efficiency, work ratio and specific steam consumtion.

Solution.

From steam tables,

At 28 bar: h1 = 2802 KJ/Kg, s1 = 6.2104 KJ/Kg K h2

At 0.06 bar: hf2 = hf3 = 151.5 KJ/Kg, hfg2 = 2415.9 KJ/Kg

Sf2 = 0.521 KJ/Kg K, sfg2 = 7.809 KJ/kg K

Vf = 0.001 m3/kg

Considering turbine process 1-2, we have:

S1= s2

6.2104 = sf2 + x2 sfg2 = 0.521 + x2 7.809

X2 = (6.2104 – 0.521)/ 7.809 = 0.728

h2 = hf2 + x2 hfg2

= 151.5 + 0.728 x 2415.9 = 1910.27 kJ/ Kg

Turbine work, WTurbine = h1 – h2 = 2802 – 1910.27 = 891.73 KJ/Kg

Pump work, WPump = hf4 – hf3 = vf (p1 – p2 )

= 0.001 (28 – 0.06) x 105 / 1000 = 2.79 KJ/kg

[hf4 = hf3 + 2.79 = 151.5 +2.79 = 154.29 KJ/kg]

Net Work, Wnet = wturbine - wpump

= 891.73 – 2.79 = 888.94 KJ/kg

Cycle Efficiency = Wnet / Q1 = 888.94 / h1 – hf4 = 888.94/ 2802-154.29 = 0.3357 or 33.57%

Work Ratio = Wnet / Wturbine = 888.94 / 891.73 = 0.997

Specific steam consumption = 3600 / Wnet = 3600 / 888.94 = 4.049 KJ/kWh

Question 2. A 35 KW (I.P) system engines consumes 284 Kg per hour at 15 bar and 2500 C. If condenser pressure is 0.14 bar, determine:

1. Final condition of steam 2. Rankine efficiency 3. Relative efficiency

Solution. Power developed by the engine = 35 KW (I.P)

Steam consumption = 284 kg/hour

Condenser pressure = 0.14 bar

Steam inlet pressure = 15 bar, 2500 C

From Steam Tables, At 15 bar, 2500 c: h = 2923.3 KJ/kg, S = 6.709 KJ/Kg K

hf = 220 KJ/Kg, hfg = 2376.6 KJ/Kg

Sf = 0.737 KJ/Kg K, Sfg = 7.296 KJ/Kg K

1. Final Condition of Steam: Since Steam expands isentropically S1 = S2 = Sf2 + x2 Sfg2

6.709 = 0.737 + x2 7.296

X2 = (6.709 – 0.737) / 7.296 = 0.818

h2 = hf2 + x2 hfg2 = 220 + 0.82 x 2376.6 = 2168.8 KJ/Kg

2. Rankine efficiency = h1 – h2 / h1 – hf2 = (2923.3 – 2168.8) / (2923.3 – 220) = 0.279 or 27.9%

3. Relative efficiency: Thermal efficiency = (I.P) / [ ( h1 – hf2 ) ] = 35 / [ 284/3600(2923.3 – 220)] = 0.1641 or 16.41%

Relative efficiency = Thermal efficiency / Rankine Efficiency = 0.1641 / 0.279 = 0.588 or 58.8%

Submitted By:

Jaswant Singh Sandhu