Renewable Energy And Technology Assignment

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RENEWABLE ENERGY TECHNOLOGY
by Student’s Name
Code + Course Name
Professor’s Name
University Name
City, State
Date

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Q 1
Part a
Assumptions: the car accelerates from 0 to 60km/hr immediately and it also decelerates
immediately. The car is travelling during. The sun rises at 6 am and sets at 6 pm. The student
number ends with 6 so the month of June shall be used for calculations.
Time = distance
speed
=
3000 km
60 km
hr
= 50 hours
Number of days = number of hours
number of hours per day
= 50
12
= 2.0833 days
= 4 days and 2 hours
Part b
Solar constant = 1.366 kW/m2
For this part, the sun shall be assumed to shine maximally on the panels from 8.00 am to
4.00 pm. at that time, the car has its speed at 120 km/hr.

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The acceleration and deceleration will be assumed to be 10m/s2.
Time take to accelerate from 0 – 60 km/hr.
t = v−u
a
= 60−0
10
= 10 seconds
Distance covered in first ten seconds of acceleration and deceleration
S=ut+ 0.5at2
= 0.5 x 10 x 100
= 500 m
Time taken to accelerate between 60 km/hr. to 120 km/hr. = 10 seconds and so is the
distance covered = 500
The graph below shows the speed against time for the two days

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Part c
In the month June the sun is fur from cancer of Capricorn but on the equator. This means
that the length of the days is equal to length of nights. However, if it is in the month of
September the days are longer on the North Pole than the length nights. The opposite is in the
month of March. Other months makes the charging of the car to be unevenly distributed when
the car is charged for long hours and other month there are few hours that the car has few hours
of charging. The distance travelled is directly proportion to the time the sun is shining. The
longer the sun is shining the longer the car can travel.
Question 2
Part a

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The firm produces 130 tons per week. It shall be assumed that all the biogas produced is
consumed in the part of the question
It is assumed that for a total household consumption, the generator consumes 100
m3/hour of methane. Now converting the flow rate into m3/s. it is also assumed that the farmer
uses all the biogas produced every week in this section.
100m3/hr. = 100
3600 m3 /s
= 0.02778 m3/s
1m3 produces
→ 22 MJ
To get the power in J/s that is released gas the following is used.
=22,000,000 J m−3 x ( 100
3600 m3 /s)
= 611,111 Watts
= 611.11kW
611.11kW is the power that is produced by the generator in one second after combusting
0.02778 m3 in one second. However, the turbine is able to capture 250 kW only.
The turbine efficiency is obtained by getting the ratio between the energy (in joules)
captured by turbine to the energy released by methane gas per second.
Turbine efficiency ηturbine= energy extracted by turbine
the output power ¿ chicken manure ¿

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= 250000 watts
611111 watts
= 0.4091
= 40.91 %
When it comes to using the chicken manure, only dry mass is accounted for conversion
into biogas. Since only dry mass that is accounted when calculating the power output, then
Efficiency of the chicken manure ηchickenmanure= 100% - 75%
= 25%
The combined efficiency is the product of two efficiencies (the chicken manure
efficiency and the turbine efficiency)
ηcombined =ηchickenmanure x ηturbine
= 25% x 40.91%
= 10.2275 %
Part b
130 tons are produced every week. The amount of dry mass for biogas production
= 130,000 kg x 25%
= 32500 kg

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The amount of gas produced for 0.3 m3/k g−1
V gas= 32500 x 0.3 m3/k g−1
= 9750 m3 of the gas is produced
0.3 m3/k g−1 equivalent
−−−−−−−−−−−−−→ 3.33 k g−1/m3
50 kg/m3 equivalent
−−−−−−−−−−−−−→ 0.02 m3/k g−1
It is assumed that the same amount of gas produced now is equivalent to when the gas
was produced by lower density of dry mass.
The volume of gas V gas= 9750 m3
From the above volume, it is possible to get the mass of gas used during production
mwater= 9750 m3/ 0.02 m3/k g−1
= 487,500 kg
The remaining mass of the manure is equivalent to the mass of water.
Density of water = 1 kg /m3
Therefore, the volume of water
vwater= 487,500 kg / 1 kg /m3
= 487,500 m3

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