3/31/2008 ME 145 - Solution to Supplemental Problem 5
m 1000
p 15MPa
v f 15MPa 0.00165
f
Natural buoyancy pressure rise pB gH f R
1 606.1 vf
pdrop
9.812606.1 300 3.6 x104 Pa
3/8/2008 ME 145 - Solution to Supplemental Problem 4 (SI units with kJ) a) STEAM GEN 1 TURBINE 2 CONDENSER 3 PUMP 4
T1=300C x1=1.0 h1=2748.7 s1=5.7042 p1=8.58MPa T2=40C s2=s1 h2=1774.6 p2=0.
3/8/2008 ME 145 - Solution to Supplemental Problem 5
m 1000
p 15MPa
v f 15MPa 0.00165
f
Natural buoyancy pressure rise pB gH f R
1 606.1 vf
pdrop
9.815606.1 250 5.23 x104 Pa
1/26/2008 ME 145 - Solution to Supplemental Problem 2 Powerfc = Power rating/DC to AC efficiency = 2x106/0.80 = 2.5 MW(e) DC Assume 0.8 V output (unrealistic but the numbers will work to give a reason
2/6/2008 ME 145 - Solution to Tester Text Problem 15.2
3.2 x1012 x1000 Average power needed in Watts 3.65 x1011 24 x 365.25
Power provided in Watts CFxN T x500x1000 0.25 x500x1000xN T 125000N T
2/13/2008 ME 145 - Solution to Tester Text Problem 14.2
Part a
P COP
Part b PROS:
3 AcVavg
2
0.3
1025xx 170 2
x2 2
2
3
27.9 x106 W(e) 28 MW(e)
higher energy density than wind 40 uni
ME 145 POWERPLANT ENGINEERING ANALYSIS AND DESIGN In-Class Midterm Exam
SPRING 2004
NAME: 1. (3) Fuel cells are limited by the Carnot efficiency true or false? Explain your answer. False. The Carn
Department of Mechanical Engineering ME 145 Power Generation Systems - Spring 2008
In-Class Midterm Exam 20 points total
Allowed Materials: Tester et al. text and one 8 x 11 sheet. PLEASE WRITE ANS
2/17/2008 ME 145 - Solution to Problem 10.3 of Tester et al. Text 12x106 people x 2 lbm/day x 0.5 x 7500 BTU/lbm = 9x1010 BTU/day 9x1010 BTU/day x 365.25 days/year = 3.29x1013 BTU/yr 3.29x1013 BTU/yr