Question
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 size schedule 40 steel pipe would your recommend for this flow? And what would the actual loss factor be for this selection if ft/100ft?(hint: you use 2.5 ft/100 ft, in the range of 1 to 4 ft/100 ft for selection but once you make a selection, there is an actual head loss factor then for that selection). The cooling water is circulated through the Rankine cycle condenser and then to a cooling tower where heat is rejected to the atmosphere by evaporating some of the water, this lowers the temperature from the 100 deg F to the 85 deg F. There are several functions that take place in a cooling tower:1)some water is evaporated 2)some water is drained off to the sewer to control the build up of dissolved solids and some water is blown out of the tower due to wind. Let's assume for our tower that the water loss due to windage and blowdown is negligible and that the major loss of water is through evaporation. Assume that for every 969btu of heat rejection by the Rankine cycle condenser, that 1 lbm of water must be evaporated (approximately hfg for water in this temperature range). Assume the density of water to be 8.34 lbm/gallon. City water is piped to the cooling tower to make up the water due to evaporation. The water originates in a water main at 55psig pressure. The float value at the cooling tower that introduces the make-up water to the tower o replace the water being evaporated requires a pressure of 10psig to work properly. There is a water meter (assume 5/8 inch water meter) and a backflow preventor (pressure drop of 10psi) in the supply line to the cooling tower from the water main. The pipe developed length is 150 ft. What size steel pipe (fairly rough) would you recommend the supply this make-up water? Pick larger pipe size if you bracket two pipe sizes.
Explanation / Answer
power output from electric generator = 600KW
efficiency of generator = 95%
therefore output from turbine = 12000/19 KW
a) State of steam leaving the turbine:
dryness fraction = x = 0.90 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
