PLEASE SHOW ALL WORK TO RECEIVE 5 STARS Consider a small simple Rankine Power cy

Question

PLEASE SHOW ALL WORK TO RECEIVE 5 STARS

Consider a small simple Rankine Power cycle which produces 600kW power from an electric generator with an efficiency of 95%. The steam leaves the boiler at 1250psia as superheated steam and leaves the turbine at 2 psia with a quality of 90%. What is the mass flow rate of steam through the turbine, Ib/h? If the boiler has an efficiency of 85% and bums natural gas with a heating value of 100,000 btu/therm at a cost of $1.25/therm, what does it cost to provide natural gas to this plant in $/hr? What is the heat rejection for the condenser in btu/h? If the condenser is water cooled with cooling water entering at 85deg F and leaving at 100 deg F, what flow rate of cooling water is required in gpm?

Explanation / Answer

power output from electric generator = 600KW

efficiency of generator = 90%

therefore output from turbine = 12000/19 KW

a) State of steam leaving the turbine:

dryness fraction = x = 0.95 and pressure = P2 = 2 psia.

Now, from steam table

therefore

h2 = 2357.6 KJ/Kg and

s2 = 7.3056 KJ/Kg-K

State of steam entering the turbine:

s1 = s2 = 7.3056 KJ/Kg-K and P1 = 1250 psia

From steam table:

h1 = 3927 KJ/Kg

power output from turbine = h1 - h2 = (3927 - 2357.6)KJ/Kg = 1569.4 KJ/Kg

But, the net power output from turbine = 12000/19 KW

therefore m*1569.4 = 12000/19

mass flow rate = m = 0.4024 Kg/sec = 3193.7 lb/hr

b) Total cost = 1.25*3412*100*3600/100000 = 180.63

c) outlet state of steam from condenser:

  P3 = 2 psia (saturated).

h3 = 218.69 KJ/Kg

heat rejected = m*(h2 - h3) = 860.69 KJ/sec = 816 btu/hr

d) m'CdT (= heat absorbed by cooling water) = 860.69 (= heat rejected by condenser)

where m' = mass flow rate of cooling water

100F = 37.77 C

85F = 29.44 C

m'*4200*8.33 = 860.69 = 0.025 K/sec = 1476 g/min

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