Download Workshop 2 Energy-solution - energy-and-environment

Physics 107
Workshop #2 Solution
Energy and Efficiency
Spring, 2011
1.
. You are riding your bicycle at 10 m/s on a level road. You decide to coast up a hill without pedaling. How
high up the hill can you roll before you stop? Neglect the effects of friction.
a)
Just after you stop pedaling, your energy is all kinetic energy (assuming this is the height you are
considering to be “0”, so there is not potential energy). At your highest point on the hill, you have come to
a stop, so the kinetic energy is zero, and all of your initial energy has been converted to potential energy.
b)
Final
0 m/s
y=hfinal
Initial
10 m/s
y=0
c)
Conservation of energy is useful because we know something about our initial energy- our
speed, which relates to kinetic energy- and we want to calculate something about our
final energy- the final height, which relates to potential energy.
d) Write the equations (above) for kinetic and potential energy. Keep this in terms of variables like
“m” and “v” rather than trying to plug in numbers.
Total Initial energy = Kinetic Energy = KE = ½ mvinit2
Total final energy= Potential Energy = PE = mghfinal
e) Solve for your height symbolically first, then plug in numbers.
Using conservation of energy, we know the Total Initial energy = Total final energy
No energy is lost to heat (due to friction or wind resistance), so the energies just involve KE and
PE, as written in part (d)
Total final energy = Total initial energy
mghfinal = ½ mvinit2
(dividing both sides by m) ½ vinit2 = ghfinal
2
 m
2
10 
vinit
s
(dividing both sides by g)
 5m
h final 
 
2g
 m
210 2 
 s 
2. a)
If a power plant uses 1 ton of coal, how much electrical energy is produced (in kWh)? Assume the
power plant is 35% efficient in converting coal to electricity.
Knowns:
Efficiency=useful energy output/total energy input = 35% = 35/100 = 0.35
Total energy input=1 ton coal
Unknown: useful energy output
To solve for useful energy output, multiply both sides of the efficiency equation by total energy
input.
=
=useful energy output
1 ton coal * 0.35 = useful energy output = 0.35 ton coal
To convert to kWh, use the relationship 1 ton coal = 7800 kWh
Useful energy output = 0.35
= 2730 kWh
How much electrical energy reaches your home? Assume the transmission lines are 90% efficient.
The total energy input (to the transmission line) is the output of the coal power plant, 2730 kWh
The useful energy output is the energy that comes out at the other end of the transmission line, at
your house, to be used.
Useful energy output = total energy input * efficiency = 2730 kWh * 0.9 = 2457 kWh
p. 2
b)
You use 1000 kWh of energy in your home each month. How much coal must be burned to supply
your home with electricity?
There are two possible approaches to this problem. One way is to do it just like problem (a), but
now, the useful energy output is known (1000 kWh), but the total energy input—to the power
lines, and to the coal power plant, is unknown. We know the relationship
Useful energy output = total energy input * efficiency
So, to solve for total energy input, divide both sides by efficiency
So, the energy that must be input to the power lines, to supply a useful output of 1000 kWh, is
1000
0.9
1111
So, the coal power plant must be outputting 1111 kWh to the power lines.
This means the energy input to the coal power plant is
1111
0.35
3175
Now, we want to express this as an amount of coal, so again, we use the relationship that
1 ton coal = 7800 kWh
3175
3175
0.41
We could also convert the units to pounds of coal:
0.41
820
I said there was another possible approach to this problem. This would be to recognize, from (a),
that an input of 1 ton coal results in 2457 kWh of electricity reaching your home. Thus, to
receive 1000 kWh at your home, a proportionally smaller amount of coal is needed.
0.41
1000
p. 3
3. Suppose you left a 100W light bulb on continuously for one month. If the electricity generation and
transmission efficiency is 30%, how much chemical energy (in Joules) was wasted at the power plant for
this oversight? If the fuel consumption for one meal in China using a kerosene wick stove is 6 MJ, how
many equivalent meals could be prepared with this wasted energy?
Known:
Power=100W
Time=1 month
Efficiency = useful energy output/total energy input
We also clearly need to relate power and energy… from the definition of power, we
know
Power=Energy/time ; this can also be written as Energy = Power * time
Unknown: How much energy is wasted?
One way to read this problem is that ALL of the energy input to the power plant is
wasted, assuming the light bulb shouldn’t have been left on and wasn’t used.
Alternatively, you could say that the output of the bulb (light and heat) was used, but the
rest of the energy input to the power plant was wasted. (then, the energy wasted
would be the difference, total energy input to the power plant minus useful energy
output)
Taking the first way of reading the problem, we would find the energy wasted by solving
for the energy input in to the power plant.
Energy at home = Power * time = 100W  1 month 
30 days 24 hours 1 kW


 72 kWh
month
day
1000 W
Energy at home = Energy into power plant * efficiency of generation and transmission
Energy into power plant = Energy at home / efficiency of generation and transmission
Energy into power plant = 72 kWh / 0.3 = 240 kWh
240 kWh 
3.61x10 6 J 1 MJ 1 meal
 6 
 144 meals
kWh
10 J 6 MJ
p. 4