0) Practice exercises

Question 1

energy formula using mass, gravity factor, and height

Answer 1

E = mgh
m = mass in kg
g = 9.81 on earth
h = height in m

Question 2

connection between J, Ws and kWh

Answer 2

1J = 1Ws (important, remember! rest can be calculated)
1kWh = 1 * 1000 * 60 * 60 Ws
1kWh = 3’600’000 J

Question 3

Assume you must satisfy your daily calorie needs (2500kcal) with crude oil. How much would it cost if a barrel of crude oil is $80? (assume 1l of oil = 0.9kg)

Answer 3

find out how many OE 2500 kcal are
1 OE is about 1l=0.9kg of oil, 1 barrel of oil has 159l

Question 4

In Europe, 255 GW of wind capacity are installed. Wind mills can be used 2’300h per year. How much energy in BTU is produced yearly? How much would this energy cost in natural gas prices if they are at $3/MMBtu?

Question 5

John consumes 0.0000072 peta-joules of electricity per year, Sarah consumes 51’159’618 micro-therm. At an electricity price of $0.14/kWh, how much do they each pay for electricity yearls?

Question 6

The first-ever oil tanker had a capacity of 3’000 tons of crude oil. Today’s biggest carriers can load 320’000 tons. How many times would the old and new tankers have to land in Switzerland’s port to deliver Switzerland’s entire yearly energy consumption (57 billion kWh)?

Question 7

If the population is 5.86 million, GDP per capita is $6’950, and each $ of GDP produces 0.586kg of CO2 emissions, what is the impact (total emissions) in millions of tons?

Answer 7

I = PxAxT
impact = 5.86mio6’9500.586 = 2.3866*10^10 kg of CO2
convert to millions of tons: divide by 1000 (because 1000kg in 1 ton), then divide by 1 million
-> 23.866 million tons of CO2

Question 8

The city of Basel has a population of 200’000 people, with an average per capita income of 190’000Fr. If the city’s total energy usage per year is 4 TWh, what is T (technology) in kWh/Fr.?

Answer 8

I = PxAxT
I = 4 TWh = 4*10^9 kWh
P = 200’000 people
A = 190’000 Fr./person
T = I/P/A = 0.105 kWh/Fr.

Question 9

If Basel’s population grows by 15’000 and its total (!) GDP grows by 38% until 2050, while technology remains the same, how much energy will it consume in 2050 in TWh?

Answer 9

new P = 215’000
GDP before: GDP per capita * population = 3.810^10
GDP now: old GDP1.38 = 5.24410^10
A is GDP per capita, new A:
5.24410^10 / 215’000 = 243’907 Fr./person
T still 0.105 kWh/Fr.
I = 215’000243’9070.105 = 5’506’200’525kWh=5.506TWh

Question 10

Assume extraction costs are $50 per unit, the prices per unit are $100 in the current period (t), and the interest rate is 10%. What unit price will result in the next period t+1?

Answer 10

p(t) = price per unit in year t
R(t) = units extracted in year t
c = cost per unit (constant)
profit (t) = (p(t)-c)R(t)
profit (t) = 50R(t)
profit(t+1) = (p(t+1)-50)R(t+1)
we assume R(t) = R(t+1) = R, either firm extracts everything now or next year
profit(t)=50R
profit(t+1)=(p(t+1)-50)R
we also know
profit (t+1) = profit (t) * (1+i)
so 50R1.1=55R= profit (t+1)
55R = (p(t+1)-50)R
get rid of R by dividing out of both sides, solve for
p(t+1) = 105

Question 11

Demand curve: P=10-Q
Dominant firm MC = 1
Competitive fringe MC = Q
What is the market outcome?