Solutions for Exercises – Renewable Energy Resources (4th Edition) by John Twidell

This document presents Solutions for Exercises from Renewable Energy Resources, 4th Edition by John Twidell. It provides insights into solar, wind, hydro, and biomass systems with detailed problem-solving support for renewable energy engineering students.

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INSTRUCTORS' SOLUTIONS TO Renewable Energy Resources, 4th edition, (2022) by John Twidell

All references to ‘RER Figures and equation numbers, e.g. (9.12), are to this fourth edition

These solutions are maintained and updated by Dr Julie Alexander of Camosun College,
Victoria, Canada ( ) The original material was produced by Dr John Twidell
and Dr Tony Weir for the earlier editions.

SOLUTIONS TO PROBLEMS FOR CHAPTER 1 PRINCIPLES OF RENEWABLE ENERGY

Solution to Problem 1.1

1(a) Average absorbed irradiance = (absorbed flux) / (total surface area of Earth)
From RER Fig. 1.2, Absorbed flux = 120,000 x 10
12
W =

1.2 X 10
17
W .
From RER (Appendix B) (or numerous other data sources, e.g. Kaye & Laby
www.kayelaby.npl.co.uk/ ) ,
Radius of Earth, RE = 6390 km = 6.39 X 10
6
m, so
Surface area of Earth = 4π RE
2 = 5.1 X 10
14
m
2

Hence average absorbed irradiance = 230 Wm
-2
Comment: the question as asked takes no consideration for latitude or climate, so the answer is of
‘ballpark’ relevance. See RER Fig 2.12 for more precise data.
==============
1(b) Solar radiation is intercepted by the Earth as a disc of area πR E
2 , so the average irradiance absorbed
on this disc is 4x 230 Wm
-2
= 920 Wm
-2
. If a device points at the Sun during daylight (~50% of time),
then the incident irradiance is 920/2= 460 Wm
-2
≈ 500 Wm
-2
. This estimate is accurate to only to 1
significant figure at best (e.g. atmospheric absorption varies with latitude, as discussed in RER sec. 2.6,
also local cloud, aerosol, pollution etc affect irradiance at ground level ). Such ‘back of the envelope’
estimations are most important to check complex solutions and measurements.
=========================================================================
Solution to Problem 1.2
Consider costs after 10,000 h of usage, with an electricity price of 0.10 Euro/ kWh.

(a) Direct cost = (10 x 0.5 €) + (10,000h x 0.100 kW x 0.10 €/kWh) = (5 + 100) € = 105€.
Note: the dominant cost is for the electricity
==========
(b) The CFL is 22/5=4.4 times more efficient than an incandescent light of equivalent illumination,
therefore uses 100W/4.4 = 23W of electricity.
Direct cost during 10,000 h of usage = (1x3.0 €) + (10,000h x 0.023 kW x 0.10 €/kWh) = (3+23) €=26 €.
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Again, the dominant cost is for the electricity, which nevertheless is much less than for the incandescent
lights.

Savings over the lifetime of the CFL (10,000 h), 105€ - 26€ = 79€
This lifetime expenditure is 24% of that in (a); so the direct saving is 76% of (a) , but with a larger initial
expenditure. .

(c) The phrase ‘payback time’ or ‘payback period’ is not uniquely defined. Several definitions are used
below.
* For (a), payback time is the time for the return on the investment to ‘repay’ the original investment.
The return in this case may be considered the reduced electricity costs and reduced replacement light bulb
costs.
Electricity cost reduction =(100–23)W x T x 0.1€/kWh = [(77x0.1)/1000]T €/h = 0.0077T€/h
Net cost of CFL = cost of CFL less cost of incandescent= 3.0€ - (T/1000) x 0.5€
Equating these 0.0077T/h = 3.0 – 0.0005T/h
So T = 3.0h/0.0072 = 300h/0.72 = 420 h

* For (b) payback time T is when the total cost of buying and using the CFL to that time equals that of
buying and using the series of incandescent lamps to the same time.
Thus [3.0€+0.023 kW x T x 0.1€/kWh] = [(T/1000h) x 0.5€] + 0.10 kW x T x 0.1 €/kWh
3.0 = [T/h] x [0.5/1000 + 0.01 – 0.002) = [T/h] x [0.0005 + 0.01 – 0.0023] = 0.0082 T/h
T= 3.0 h /0.0082 = 350 h
Note: RER chapter 17 discusses ‘payback time’ and related concepts more fully


Solution to Problem 1.3

Students to find and use their own information, and work as Problem 1.2. Team work may be best.

=========================================================================1.4

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Solution to Problem 1.4
Guidance and references relating to the lifetime of oil ‘reserves’:

• A web search on ‘peak oil’ will lead to both informed and uninformed discussions of proven and
unproven reserves and the increasingly complex relationship between supply and demand that are
constrained by commercial and governmental interests.
• Wikipedia https://en.wikipedia.org/wiki/Peak_oil provides a brief summary of the current
situation.
• Discussion should include the rate of increase of alternate energy sources at the present time and
the possibility of new technologies that could replace petroleum. How will these affect demand
and consequent pricing of petroleum products that may not behave as predicted by traditional
price-demand models.


========
Solution to Problem 1.5

For users of this book, discussion of ‘sustainable ‘ will probably focus on what forms of energy are being
used and how efficiently.

Becoming more sustainable in energy obviously involves minimising use of fossil fuels (including those
used for electricity generation and transport). Drawing on renewable energy is clearly more sustainable in
terms of resources than depleting fossil resources. RER chapter 16 gives many examples of ways in
which end-use efficiency can be improved.

Note that the question can be taken to include not just personal energy use but use by your household or
wider community.

Other resources that may be an issue for sustainability include water, food (which brings in many other
issues, not least its sourcing) , money (spending more than you have or can repay) and materials
(recycling and/or re-use).

==========
Solution to Problem 1.6

Hydro-power requires some hills and wave power requires access to the ocean, so clearly not all forms
of RE are readily available everywhere. On the other hand, photovoltaics can provide a useful output
even in cloudy England – though of course less than the same panels would yield in sunny California.
The Introduction to each of the ‘technology’ chapters of RER indicates what are the key resource factors
for that technology, which should be a useful guide in your local context.

======

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Solution to Problem 1.7

This discussion can build on the answers to Problem 1.6.

The question asks about 30 years time, because over a shorter time frame (say, 5 years) change in energy
supply will be relatively small because large change requires utilities to replace their facilities and
infrastructure , which represents a substantial investment. And utilities are reluctant to have to replace
facilities until near the end of their ‘economic life’ [i.e. the time taken to recoup their initial investment,
which is usually taken as ~20 to 30 years] .

RER chapter 17 discusses such institutional factors in more depth. See especially section 17.7 on ‘the
way ahead’.


=======

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All references to ‘RER Figures and equation numbers, e.g. (9.12), are to this fourth edition

These solutions are maintained and updated by Dr Julie Alexander of Camosun College,
Victoria, Canada ( ) The original material was produced by Dr John Twidell
and Dr Tony Weir for the earlier editions.

SOLUTIONS TO PROBLEMS FOR CHAPTER 2 Solar radiation characteristics and impacts,
including the greenhouse effect

Solution to Problem 2.1

(a) The Solar Constant,
*
0
G
, is the irradiance perpendicular to the solar beam just outside our Earth’s
atmosphere (see RER sec.2.2).
Assume that the Earth maintains a constant distance L from the Sun (i.e. neglect the slightly elliptical
orbit of the Earth).

The power emitted by the Sun (as a black body) from its surface area is:


P
S
=(σT
S
4
)(4πR
S
2
)
=power passing through the surface of a pseudo sphere of radius equal to the Earth's distance, L
= (G
0
*
)(4πL
2
)


Hence

4242 1
* 2
0 22
2
91
8 -1 -4 4 2
11
-2
(4 )( )(4 )
44
1.392 10 m
(5.67 10 Wm K ) (5780K)
1.498 10 m
1365 Wm
SSSS
TDTR
G
LLσπσπ
ππ
−
= = =
××
=× ×× 
×
=


Comment: this answer equals the observed solar constant of 1366 W m
-2
within 0.15% (i.e. within
rounding errors) ; indeed working this sum backwards from satellite measured data is one way to
determine T
S .

(b) The area intercepting solar radiation at the Earth is
2
E
Rπ
, but the area of the Earth emitting
infrared radiation to maintain thermal equilibrium is almost exactly a spherical surface of area
2
4
E
Rπ. The depth of the Earth’s atmosphere is negligible compared with its radius for this
calculation. We initially assume the Earth is a black body of uniform absolute temperature T. and
zero reflectance. Hence at equilibrium:


4 2 *2
0
( )(4 )
EE
T R GRσπ π =


Hence this approximate analysis gives the radiative temperature of the Earth as

*
o0
1
4
278K 5 C
4
G
T
σ

= = =


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However there is significant reflection of incoming solar radiation from white cloud and the Earth’s non-
black surfaces. Therefore, reflection should be included, as in the related analysis of RER sec.2.9.1.
Then:

4 2 *2
00
( )(4 ) (1 )
EE
T R GRσπ π ρ= −


where ρ 0 is the effective short-wave reflectance of the Earth from Space. Approximately ρ 0 ≈ 0.3, hence


1
* -2 4
1
82400
8 -2 -41
4
(1 ) (1367Wm ) 0.7
42.0 10 2.55 10 255K 18 C
4 4(5.67 10 Wm K
G
TKρ
σ
°
− − ×
= = = × = ×= =−  
×


These calculations demonstrate the fundamental physical processes by giving reasonable answers to an
accuracy of about +/- 5%. The same principles, but with more detail and, are used for computer-
simulations, e.g. of weather.


Solution to Problem 2.2

Sketch diagrams should be drawn for the southern hemisphere, to replicate Figs 2.8 and 2.9. As
compared with the diagrams for the northern hemisphere, latitude φ is now negative and solar device
orientation γ shifts by 180
o
. Working through the equations in Sect 2.5.2, angle of incidence θ from Equ
2.8 is seen to be unchanged.

=========================================================================

Solution to Problem 2.3

This problem is a numerical example of eqns (2.8)-(2.11) of RER section 2.5.2. So, as in RER Worked
Example 2.1, the first step is to use the data to identify (determine) the various ‘input’ angles required to
calculate θ
z and θ. Here, we have:

latitude φ : given for Suva, capital of Fiji, in the Southern Hemisphere φ = -18
o

declination δ: calculated from RER eqn (2.5) thus:
day number = 139 for 20 May ( counted on a calendar)n
Hence, from RER (2.5):


δ=δ
0
sin[360×(284+139) / 365)]
=19.7deg with
δ
0
=23.5deg

Hour angle ω: at 9 a.m., using RER eqn (2.4):

o1 o
(15 h )( 3h) 45ω
−
= −=−
Neglecting ω eq and assuming ψ = ψzone (which is in fact true for Suva).

(a) With the above angles as input, RER eqn (2.11) gives the angle θz between the beam radiation and
the vertical:


cos
θ
z
=sinφsinδ+cosφcosωcosδ
=sin(−18
o
)sin19.7
o
+cos(−18
o
)cos45
o
cos19.7
o
=(−sin18
o
)sin19.7
o
+(cos18
o
)cos45
o
cos19.7
o
=0.528, and so
θ
z
= 58
o