Exam 1
General Comments: Score range: 50 - 78 . Approximate letter grades: A, 68 up; B, 55-67.
Problem Solutions:
7P. Furnace eff. = 0.85; Power plant
eff. = 0.34
Let Q = heat required for home. The furnace
burns Q/(0.85) = 1.18Q
Heat pump [HP] consumes 10% less energy than this
from power plant, so the plant requires (1.18Q - 0.12Q) = 1.06Q
HP input energy (W) is thus 1.06Q (0.34)
= 0.36Q
HP Coefficient of Performance [COP] =
Q/W = Q/(0.36Q) = 2.78
[Alternate method: choose a number for Q, say 50,000
Btu and work through with it]
8P. 128 births/10s X 86400 s/d
= 1.1 x 106 births/d
X 365 = 4.04 x 108 births/ yr
If population doubling time = 35 yr, then
annual % growth rate = 70/35 = 2%.
The annual & birth rate = (4.04 x 108
)/(6 x 109) = 0.067 = 6.7%
The growth rate must = [birth rate - death rate],
So death rate = [growth rate - birth rate] = 6.7%
- 2% = 4.7%
[Note: calculating the 35-year total of births using (12.8 b/s) x (# of s in 35 yr) is equivalent to assuming linear population growth & is thus incorrect!]
9P. (a)[5 points] Eff. = (Th -
Tc)/Th = [(273 + 20) - (273 + 5)]/ (273 + 20)
= 5.1 x 10-2 = 5.1%
(b) [5 points]10kW output = 10000 J/s ; (10000J/s)
/ (4.184 J/cal) = 2390 cal/s
1L water = 1 kg; 1 kg x 1000cal/kg C x 15C
= 15000 cal
(2390cal/s) / [0.051 x 15000cal/L] = 3.1 L/s
10P. (a) [2 points] Transportation
is about 30% of total energy expenditure = 28 Quads = 28 x 1015
Btu
(b) [4 points] 28 x 1015 Btu x 1055J/Btu
x 1 yr/(86400 x 365) = 9.4 x 1011 W = 940000 MW
(c) [4 points] 28 x 1015 Btu /yr
/[5.8 x 106 Btu/Bbl x 365 d/yr] = 13.2 x 106
Bbl/day