P110 Unit 2 Assignment 2

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P110/120: Energy/Energy & Technology Fall 2015

Homework 02

1 What is the potential energy of the water in a lake of surface area 10 square kilometers, with an average depth of 15 m, and elevation above the electrical generator of 200 m? (Note that you can ignore 15 m depth of the water compared to the 200 m elevation, i.e., use the total mass of the lake at an elevation of 200 m. Also, a cubic meter, i.e., 1 m3, has a mass of 1000 kg. The wonders of the metric system!
PEG = weight x height= mgh g=9.8m/s2
m= 10km2 x 1000m x 1000m = 107m2 107m2 x 15= 1.5 x 1.8m3 = 1.5 x 1011kg
E= 1.5 x 1011kg (9.8m/s)2 (200m)= 1kg/s2 = 1 Joule

2. The per capita rural consumption of household energy in some Bangladesh villages is 7 x 109 J/year. …show more content…

A household furnace has an output of 100,000 Btu/hr. What size electrical heating unit (in kW) would be needed to replace this?
1Btu= 0.002928104kW (conversion table online)
100,000 Btu(0.0002928104kW)= ~29.281kW


4. An engine performs 5000 Joules of work in 20 seconds. What is its power output in kilowatts and in …show more content…

The lights in our Swain West 007 lecture room were accidentally left on overnight (5:00 p.m. to 8:00 a.m., i.e., 15 hours). If electricity costs \$0.10/kWh, how much does this oversight cost if the energy consumption of the lights is 2200 W?
kWh= 2200W/1000= 2.2kWh 2.2kWh x 15 hr = 33 kWh
cost= 33 kWh x 0.10kWh = $3.30

6. A small stream flowing at a rate of 8 liters per second has a vertical drop of 1.5 m. What is the maximum power that you can obtain from this stream? (1 liter of water has a mass of 1 kg).
1L water= 1kg water, so mass= 8kg h=1.5m P=E/t E=mgh
E= 8 (9.8m/s2) (1.5)= 107.6 J Watt= 1Joule/Sec
P=107.6 J/1s= 107.6 W

7. If a 60-kg sprinter running at 10 m/s could convert all of her kinetic energy into upward motion, how high could she jump?
KE= 1/2mv2 =mgh g=9.8m/s2 h=v2/2g = 102 (2 x 9.8) = 5.1m

8. If a coal-burning electrical generating plant burns 2 tons of coal to generate 6000 kWh of electricity, calculate the efficiency of the plant as the ratio of electricity output to fuel energy input (refer to Table 3.4).
1 ton coal= 25 x106Btu 1kWh= 3413

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