IDH2931 Fall 2001

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