IDH2931/6988--”Energy”

Questions/Problems  (Q/P)










Brief answers to assigned problemsand to the multiple-choice (MC) questions in the R & K text

Occasional hints will be given in [brackets like these]

--->  If you have trouble working a problem or have any other questions,  ASK!!!!  Come by during office hours
       or email me.  Get help soon------don't delay until one or two days before the exam!
 
 

Chapter 1

Q/P:

3. 100 ft-lb; 136 J   [use conversion table at front (end papers) of text]

7.  (93.8  QBtu)/ (265 E6 people) = 3.5 E8 Btu/person.  [Use Table 1.1 &/or Fig. 1.5 & last par., p. 17]    Ans.  13.3 T coal/person

8. 5.7 E3 cal [Recall 1 W = 1J/s}

10. 3.1 E3 Btu/hr  {The area doesn't enter into the calculation]

11.  3.4 hr.  [1 gal water = 8 pints = 8 lb]

12. 6.8 E9 gal oil;   3.5 E7 T coal
 

  Multiple-Choice Answers

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Chapter 2

4.    About 5 yr.

7.    $ 9.89

11. 1.6 E-3  bbl, or about 0.07 gal!

12.  0.1 lb coal, or about 0.01 gal oil
 

Multiple-Choice Answers
 

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Chapter 3

3.  1 mole of C is 12g. = 0.0264 lb.  1 T of  C is then 76000 moles.
 76000 x 95 kcal =  7.2 E6 kcal = 2.9 E7 Btu  [close to the 2.7 E7 given in the tables]

5.  The heat delivered to the water by the electrical energy is [40% - 10%(40%)] = 36%,
 whereas that delivered directly by the gas heater is 60%.  So the gas water heater is
 60/36 = 1.7 times more efficient!

10. The chemical equation is:  CH4  + 2 O2  =  CO2  + 2 H2O  + heat .
      Summing  atomic weights:   (12 + 4) + 2 (2 x 16) = (12 + 2 x 16) + 2(2 + 16) + heat
or  16   +   64    =   44  +   36 ,  so 16 parts CH4  yields 44 parts CO2 .
 Thus the ratio is 44/16 = 2.75, or 2.75 tons CO2  for each ton of methane burned.

12.  Burning the fuel directly to produce heat can yield no more than 100% of its energy, assuming
100% heater/furnace efficiency [It's probably not!  Why??].
The heat pump delivers heat = [COP x  40%] = 4 x 40% = 160% of  the input energy.

14. The ideal [Carnot] efficiency is 39%.  The 20%  actual efficiency is 20/39 = 51% of ideal.

15.  The EER = (Btu/hr output)/ (watts input) --see text, p. 80.
 Since 1 kWh = 3413 Btu [tables] , 1 kW = 3413 Btu/hr and  1W = 3.4 Btu/hr .
 So EER = 3.4 'breaks even', pumping exactly as much heat as its input energy.

 EER = 10 thus moves 10/3.4 = 2.9 , thus 1 W removes 2.9 x 3.4 = 9.9 Btu/hr;
 Hence for EER=10,  1 kWh input energy removes 9900 Btu.

 Qh = Qc + W = 1 kWh + 9900 Btu = 13313 Btu

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Chapter 4

2. The glass is transparent to the visible wavelengths but is a poor transmitter of the infra-red.

5. Isolated locations where electric utilities aren't available; portable devices [hand-held calculators];  what else can you think
of?

7.  A recent catalog lists a 2' x 5' panel cell array that in full sun supplies 17 volts and output power of 120W [cost, $600].
    So 120V would require 120/17 = 7 of these panels
    2 kW would require 2000/120 = 17 panels.  17 x 10 sqft = 170 sq ft.

8.  The total heat required is 50 E6 Btu/(1000 sqft) x 2(1000sqft) = 100 E6 Btu.
    The daily heat requirement is 100 E6 Btu/(180 days) = 5.6 E6 Btu
    Area required = 5.6 E6/(1000 btu/sqft x 0.5) = 1120 sqft

9.  Heat required = c m (T2 - T1) = 1 Btu/lbF x 1200 gal x 8 lb/gal x 50F = 4.8 E5 Btu
    Collector area required = 4.8 E5 Btu/(1100 Btu/sqft x 0.5) = 9600 sqft.

10.  Concrete stores 22 Btu/ cuftF [Text, p. 103]
    We can use the formula used in problem 9, substituting volume for mass:  Q = c V (T2 - T1).
    So V = 2 E5 Btu/ (22 Btu/cuftF x 20F) = 455 cuft
    The thickness is thus 455 cuft/ 1000sqft = 0.45 ft or about 5.5 inches.

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Multiple-Choice Answers
 

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Chapter 5

4. Usingthe formula on p. 134 we see that power is proportional to  v3.  Therefore P1/P2  = v13/v2 3.

    10 mi/hr = 4.47 m/s, so P2 = P1[v2 /v1]3 = 23 [(1.5 x 4.47)/ 4.47]3 = 77.6 kW

5.  Theoretical max. = 59%, so the max. from this system is (0.59)(0.6) = 0.35

    The 10 mph wind produces [formula, p. 134]  0.17 kW which at 35% eff is 60 W
    A 20 mph wind then will produce 476 W and a 30 mph wind, 1.6 kW

6. The heat liberated is Q = c m (T2-T1) = 2.4 J/cm3 x 1 m3 x (100 cm/m)3 x (240 - 100) = 3.36 E8 J

9. 1000 gal/s produces 2.13 MW, so1000 MW will require 4.69 E5 gal/s .  If the flow rate is 4 m/s this means  a 4m length of the pipe must contain 4.69 E5 gallons.  This works out to 12 m diameter.  Quite a pipe!

11.  Q = 265 E6 x 1000 lb x 4300 Btu/lb = 1.14 E15 Btu, or about 1 Quad. This is about1 % of the USA's 94 Quad consumption.

12.  From Table 5.3 [p. 151], a forest converts about 0.5 % of the sun's energy into fuel (wood).  Burned at 33% conversion efficiency the overall efficiency = (0.5 x 0.33) = 0.165% .  At 600w/m2 insolation (p.94) the forest's output is 600 x 0.00165 = 1 W/m2.
Thus to obtain 1000MW requires 1000E6 = E9 sq. m of growing area.  This amounts to about 386 sq. mi., or a square that is about 20 mi along one edge.

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Multiple-Choice Answers

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Chapter 6

1. a) At 198 Mev/fission, 1 g of U-235 contains 1/235 [Avogadro's No.] atoms, which is

1/235 x 6.02 E23 = 2.56 E21 atoms.

From 1 kg of U235 we will obtain: 1000 x 2.56 E21 x 198 MeV x 1.6 E-13 J/MeV = 8.1 E13 J

b) 1 kg natural U is 0.7% U25, so the energy content is 8.1 E13 x 0.007 = 5.7 E11 J

9. a) At 33% eff., we will require input [thermal] power = 3 x E12 W.

3 E12 J/s x 86400 s/day x 365 d/yr = 9.46 E19 J annual energy needed.

From table, 8.28 E10 J/gram x E6 gm/tn = 8.28 E16 J/tn

9.46 E19 J/(8.28 E16 J/tn) = 1143 tn of U235 annually

b) At 0.07% U235 per tonne of U ore, 1143/0.007 = 1.63 E5 tonnes would have to mined

c) Coal required for the same power output:

9.46 E19 J/(2.81 E10 J/T) = 3.37 E9 T = 3.05 E9 tonne

d) 3.05 E9 tn /(1.7 E5 tn/A x 640 A/sqmi) = 28 sqmi

10. Probability of significant release of radioactivity:

WASH-1400: "1 in 2 million, per reactor year" = 0.5 E-6

NUREG-1150: 4 E-6 (average for all five reactors analyzed)

12. See pp.196-197

13. Ptotal = P1 P2 = (0.1)(0.5) = 0.005 or 0.5 %

14. 48days/8 = 6 half-lives; 26 = 64, so 20 Ci/64 = 0.3 Ci

16. See pages 180-181 and 189-190.
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Multiple-Choice Answers

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Chapter 7

1.  Q = A/R (T2 - T1) t  = (8 ft x 24 ft)/ 16.2  x (68F - 16F) (8hr) = 4930 Btu

2.   DD = 60 d (65 - 55) + 50 d (65 - 45) + 30 d (65 - 30) = 2650 deg days

5. There’s some drudgery in this one---we have to compute the heat flow through each of  the four components of the house!
Using Q = [(24 hr x Area)/R] x DD :

a) windows:  Qa = 24 x 544/0.88 x 6040  = 8.96 E7 Btu   [the R-value is in the text of the paragraph below Table 7.3 in the textbook]

Similarly, Qb = 4.83 E6;
Qc = 3.68 E7;
Qd = 2.96 E7;  and the total Q = 16 E7 Btu

8.  Per sq ft, for one year {DD= 14160 from Table 7.2] we have

 Qsingle = 24/0.88 x 14160 = 3.86 E5 Btu  and similarly,
 Qdouble = 1.5 E5 Btu.  The difference or heat saved is 2.36 E5 Btu.

But since this heat is supplied by a furnace witn eff. = 75%,  the Btu’s consumed would be
2.36 E5/ 0.75 = 3.15 E5 Btu.

At $5.50 per million Btu this energy would cost $1.73.  Since the double-pane windows cost $2.50 /sq ft, 2.50/1.73  =  1.4;  thus
they pay for themselves in savings in about 1.4 years.

9.  For each day:  without setback, Q = 24 A/R (68 - 15) = 1272 A/R
 With setback, Q = 12 A/R [(68 - 15) + (50 - 15)] = 1176 A/R

The difference is 96 A/R, or 96/1272 = 7.5 %

14.  135 W x 0.5 x 8 hr x 160 d = 86.4 E3 Wh = 86.4 kWh.   At, say, $0.10 per kWh, the  operating cost for the year is $8.64.
An ordinary blanket on top of the electric blanket would provide some insulation and thus retard heat flow into the room, causing the electric blanket to operate for less time per day.  Estimated saving:  a few percent.

17.  Fluorescent:  10000 hr x 15 W = 150 kWh; 150 kWh x $0.08/kWh = $12 operating cost
 Total cost = 14 + 12 = $26

Incandescent: 10000 hr/1000hr/bulb = 10 bulbs.  Cost = 10 x $0.50 = $5
Operating cost = 10000 hr x 75 W x $0.08 /1000Whr = $60; Total cost = $65
Saving is $65 - $26 = $39

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Multiple-Choice Answers

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Chapter 8

The ‘problems’ mostly involve discussion, which requires that the text be read with reasonable care.  “Answer” given here is only
a bare outline!

1.  The air that’s pushed out of the way by the moving vehicle becomes very turbulent (think of how a fast boat’s wake looks).
The turbulence eventually is dissipated as heat.

2. Using the aerodynamic drag formula, we get
 v [in ft/s] = Sq.rt.(370 x F/CA) = 19.2 ft/s
 which converts to 13 mph

4. See recent news articles on the now-available Honda and Toyota hybrids.

5. See Section 8.5.  A bus costs about $2 per passenger-mile; a train, between $35 and $40.
Based on your own experience(s), what is your car costing you, per mile, to drive?  [Besides initial cost and fuel, don’t forget
tires, insurance, maintainance, etc.]
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Multiple-Choice Answers

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