Assignment Details
There are five questions/problems on this assignment. Attempt ALL problems
Please show your working so that any errors can be tracked as part of your feedback.
1. Thiazole (C3H3NS) occurs natrually as part of vitamin B1 (thiamin)
such as in pasta and bread.
Thiazoles have anti-tumor and anti-viral properties
but most thiazole compounds are flavourings. If it is combusted perfectly in
air the products are carbon dioxide, water, hydrogen sulphide (H2S) and nitrogen.
What is the balanced reaction?
[8 marks]
Ans : Chemical reaction :
(Chemistry) a process that involves changes in the structure and energy content of atoms, molecules, or ions
but not their nuclei.
Balance chemical reaction : a reaction that has the same # of atoms of each element before the reaction
as after it.
What is the RMM of thiazole?
[2 marks]
Ans: Mass of thiazole = 3*mass of carbon + 3* mass of hytrogen + mass of niterogen
+ mass of sulphar
= 3* 12 + 3* 1 + 14+ 32
= 85 gram per mole .
What is the fuel-air mass ratio?
[5 marks]
Ans : fuel-air mass ratio = air/fuel
= CO2/H2S
= 12*2(16) / 2 (1) * 32
= 384/64
= 6
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What is the oxygen depletion?
[4 marks]
Ans : OXYGEN DEPLATION:
Microorganisms that live in water feed on biodegradable substances. When too much biodegradable material is
added to water, the number of microorganisms increase and use up the available oxygen. This is called oxygen
depletion. When oxygen levels in the water are depleted, relatively harmless aerobic microorganisms die and
anaerobic microorganisms begin to thrive. Some anaerobic microorganisms are harmful to people, animals and
the environment, as they produce harmful toxins such as ammonia and sulfides.
What is the yield of CO2?
[2 marks]
Ans : C(s) + O2 ( g) → CO2 (g)
If you burn 12 gram of carbon to make CO2, then amount of carbon dioxide expected
Is one mol of CO2 or 44grams of CO2
Sadly the amount you will get will probably be less than 44 grams and more like 34
grams of CO2 . The problem is a competing reaction that happens . Some carbon reacts
To make CO.
2C(s) + O2 (g) → 2 CO (g)
The carbon participating in this “side” reaction will not be able to make CO2 . The
reaction will not yield 100 % of the expected 44 grams .
Percent yield = 100* (34 gram CO2 actual / 44 grams CO2 predicted )= 77 %
What is the yield of H2O?
[2 marks]
Ans: determine the theoretical yield of H2O. This is the maximum amount of H2O possible.
write a balanced equation
2H2 + O2 -------> 2H2O
moles H2 = mass / molar mass
moles H2 = 16 g / 2.016 g/mol
= 7.937 moles H2
The balanced equation shows that
2 moles H2 react to from 2 moles H2O
Therefore 7.937 moles H2 can form up to 7.937 moles H2O
mass H2O possible = moles x molar mass
= 7.937 mol x 18.016 g/mol
2
= 143 g of H2O is possible
% yield = actual yield / theoretical yield x 100/1
= 138 g / 143 g x 100
What is the yield of N2?
[1 mark]
Ans: 112g of nitrogen gas reacts with hydrogen gas to produce 40.8g of ammonia gas
according to the equation given below:
N2(g) + 3H2(g)
2NH3(g)
Calculate the percentage yield of ammonia.
a. Actual yield of ammonia (NH3) = 40.8g
b. Theoretical yield of ammonia (NH3) is calculated using the equation:
moles = mass ÷ molar mass
From the balanced chemical equation the mole ratio N2:NH3 is 1:2
moles NH3 = 2 x moles N2
moles NH3 = 2 x (mass N2 ÷ molar mass N2) = 2 x 112 ÷ 28 = 8 moles
theoretical yield NH3 = predicted mass NH3
predicted mass NH3 = moles NH3 x molar mass NH3 = 8 x 17 = 136g
c. Percentage yield = (actual yield ÷ theoretical yield) x 100
percentage yield NH3 = (40.8 ÷ 136) x 100 = 30%
For ideal combustion, what is the yield of carbon-monoxide?
[1 mark]
If 100g of CO reacts with an excess of H2 to form 75g CH3OH, what is the percent yield of the
reaction?
The balanced equation is CO + 2H2 = CH3OH.
First, determine the theoretical yield. You have mass of the reactant CO, so you need to change this to
moles of CO. Use the mole ratio and then determine the theoretical amount of CH3OH that will be
made.
Mass of CO (use the molar mass equivalent here) = moles CO (use the mole ratio here to change
moles to moles) = moles CH3OH (use the molar mass equivalent here) = mass CH3OH = percent yield.
100g CO x 1 mole CO/28g CO = 3.6 moles CO
3.6 moles CO x 1 mole CH3OH/1 mole CO = 3.6 moles CH3OH
3.6 moles CH3OH x 32g CH3OH/1 mole CH3OH = 115.2 g CH3OH. This is the theoretical yield.
Percent yield = actual yield/theoretical yield x 100%
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Percent yield = 75g CH3OH/115.2 g CH3OH x 100% = 65%
The percent yield for example one is determined by solving
this equation
So the percent yield from this reaction is 65%. They expected to get 115.2g of product but only got 75g.
You may assume the composition of air is 21% O2 and 79 % N2. A periodic table is
provided overleaf, should you need one.
[25 marks for this question]
2. Critically review and distinguish superficial, partial-thickness and full thickness burns
and explain the significance of ‘eschar’. How would eschar appear on a Lund-Browder
chart?
[15 marks for this question]
Ans: A burn is a type of injury to flesh or skin caused by heat, electricity, chemicals, friction,
or radiation.[1] Burns that affect only the superficial skin are known as superficial or first-degree burns.
When damage penetrates into some of the underlying layers, it is a partial-thickness or second-degree
burn.
In a full-thickness or third-degree burn, the injury extends to all layers of the skin. A fourth-degree burn
additionally involves injury to deeper tissues, such as muscle or bone.
The treatment required depends on the severity of the burn.
Superficial burns may be managed with little more than simple pain relievers, while major burns may
require prolonged treatment in specialized burn centers. Cooling with tap water may help relieve pain
and decrease damage; however, prolonged exposure may result in low body temperature.
Partial-thickness burns may require cleaning with soap and water, followed by dressings. It is not clear
how to manage blisters, but it is probably reasonable to leave them intact.
Full-thickness burns usually require surgical treatments, such as skin grafting. Extensive burns often
require large amounts of intravenous, because the subsequent inflammatory response causes
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significant capillary fluid leakage and edema. The most common complications of burns
involve infection.
An eschar (/ˈɛskɑr/; Greek: eschara, "scab") is a slough[1] or piece of dead tissue that is cast off from
the surface of the skin, particularly after a burn injury, but also seen ingangrene, ulcer, fungal
infections, necrotizing spider bite wounds, and exposure to cutaneous anthrax.
Black eschars are most commonly attributed to anthrax, which may be contracted through herd animal
exposure, but can also be obtained from Pasteurella multocida exposure in cats and rabbits. A newly
identified human rickettsial infection, R. parkeri rickettsiosis, can be differentiated from Rocky Mountain
spotted fever by the presence of an eschar at the site of inoculation.[2] Eschar is sometimes called
ablack wound because the wound is covered with thick, dry, black necrotic tissue.
Eschar may be allowed to slough off naturally, or it may require surgical removal (debridement) to
prevent infection, especially in immunocompromised patients (e.g. if a skin graft is to be conducted).
If eschar is on a limb, it is important to assess peripheral pulses of the affected limb to make sure blood
and lymphatic circulation is not compromised. If circulation is compromised, an escharotomy, or
surgical incision through the eschar, may be indicated.
An escharotic is a substance that causes tissue to die and slough off. Examples include acids, alkalis,
carbon dioxide, metallic salts andsanguinarine, as well as certain medicines like imiquimod. Escharotics
known as black salves, containing ingredients such as zinc chlorideand sanguinarine containing
bloodroot extracts, were traditionally used in herbal medicine as topical treatments for localised skin
cancers and can be effective in some cases, but often cause scarring and can potentially cause serious
injury and disfigurement. Consequently escharotic salves are very strictly regulated in most western
countries and while some prescription medicines are available with this effect, unauthorized sales are
illegal. Some prosecutions have been pursued over unlicensed sales of escharotic products such
as Cansema.
Lund and Browder chart—This chart, if used correctly, is the most accurate method. It compensates
for the variation in body shape with age and therefore can give an accurate assessment of burns area
in children.
It is important that all of the burn is exposed and assessed. During assessment, the environment
should be kept warm, and small segments of skin exposed sequentially to reduce heat loss.
Pigmented skin can be difficult to assess, and in such cases it may be necessary to remove all the
loose epidermal layers to calculate burn size.
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3. Critically review zone and field models used for compartment fire modelling. Provide
examples and critically analyse the main assumptions, limitations, advantages and
disadvantages of these models.
[15 marks for this part]
Ans : A fire model is a mathematical formulation intended to predict the nature of a fire, or the effects
of a fire on it’s surroundings. Fire modeling can range from a single handwritten equation, to a complex
finite element computer program requiring many days to calculate on high speed computers.
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Physical Fire Models
Mathematical Fire Models
Probabilistic Fire Models

Event trees

Fault trees

Statistical distributions
Deterministic Fire Models
Simplified fire growth calculations
Simplified fire growth calculations are used to ...

Heat Release Rate (HRR)

Flame height

Plume centreline velocity

Plume centreline temperature

Virtual Origin

Radiant heat flux to a target

Pre-Flashover temperature estimates

Flashover prediction

Post-Flashover temperature estimates

Fire duration equivalency for ASTM E119

Smoke production rate

Enclosure smoke filling

Thermal fire detector response
Compartment fire models
A main subset of fire models is the compartment fire model, which is used to understand the effects of a
fire on one or more rooms or compartments inside a building. Typical effects studied are compartment
temperature, smoke production rates, gas species concentrations, visibility, fire duration, and ventilation
effects. There are two main types of compartment fire models: zone and field.
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Zone models
Zone fire models are based on the simplifying assumption that the compartment during fire
development can be split horizontally into two zones: an upper zone in which the products of
combustion have stratified against the ceiling due to the buoyancy of the hot gases and smoke; and the
lower zone in which the fire exists.

Multiple room fire and smoke spread

Forced ventilation

Post flashover fire resistance

Detection system activation

Suppression modelling
4. The underside of a smoky layer 14m x 9m is radiating like a flat, isotropic plate at
510°C to the floor of a compartment 2.15m below. The mean emissivity is 0.41 and the
floor is homogenous/flat plate at 35°C. What is the rate of heat transfer from the smoky
layer to the floor?
Useful figures & formulae:
Q  F12ATH4  TC4 
(1)



1  X 2 1  Y 2 
ln


2 

1 X 2 Y 2

F12 

XY 
X
Y
 Y 1  X 2 arctan
 X arctan X  Y arctan Y 
  X 1  Y 2 arctan
1Y 2
1 X 2


(2)
Where X  a / z and Y  b / z
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Figure 1. Parallel plates
[20 marks for this part]
Ans : Q= 2.14 £ 0.14 * 126 (
354
–
5104
)
F12 = 2/3.14*14*9 [ ln 1 + 126 arctan 0.04569 + 14 (9) – 9 arctan 9- 14 arctan 14 ]
=
9
2
3


atomic number
symbol
Hydrogen

name

4
3
2
1
Table 3. Periodic table of elements
0
2
He
Helium
1
3
Li
4
Be
relative atomic mass
5
B
6
C
7
N
8
O
9
F
4
10
Ne
Lithium
Beryllium
Boron
Carbon
Nitrogen
Oxygen
Fluorine
Neon
7
11
Na
9
12
Mg
11
13
Al
12
14
Si
14
15
P
16
16
S
19
17
Cl
20
18
Ar
Sodium
Magnesium
Transition Elements
Aluminum
Silicon
Sulphur
Chlorine
Argon
23
19
K
24
20
Ca

21
22
23
24
2526
27
28
29
30
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Phosphoru
s
27
31
Ga
28
32
Ge
31
33
As
32
34
Se
35.5
35
Br
40
36
Kr
Potassiu
m
Calcium
Scandium
Titanium
Vanadium
39
37
Rb
40
38
Sr
45
39
Y
48
40
Zr
51
41
Nb
Rubidium
Strontium
Yttrium
Zirconium
85
55
Cs
88
56
Ba
89
Cesium
Iron
Cobalt
Nickel
Copper
Zinc
Gallium
Germanium
Arsenic
Selenium
Bromine
Krypton
55
43
Tc
56
44
Ru
59
45
Rh
59
46
Pd
64
47
Ag
65
48
Cd
70
49
In
73
50
Sn
75
51
Sb
79
52
Te
80
53
I
84
54
Xe
Molybdenum
Technetium
Rhodium
Palladium
Silver
Cadmium
Indium
Tin
Antimony
Tellurium
Iodine
Xenon
93
73
Ta
96
74
W
(98)
75
Re
Rutheniu
m
101
76
Os
103
77
Ir
106
78
Pt
108
79
Au
112
80
Hg
115
81
Tl
119
82
Pb
122
83
Bi
128
84
Po
127
85
At
131
86
Rn
Hafnium
Tantalum
Tungsten
Rhenium
Osmium
Iridium
Platinum
Gold
Mercury
Thallium
Lead
Bismuth
Polonium
Astatine
Radon
178
104
Rf
181
105
Db
184
106
Sg
186
107
Bh
190
108
Hs
192
109
Mt
195
110
Uun
197
111
Uuu
201
112
Uub
204
113
Uut
207
114
Uuq
209
115
Uup
(209)
116
Uuh
(210)
117
|Uuh
(222)
118
Uuo
Rutherfordiu
m
Dubnium
Seaborgium
Bohrium
Hassium
Meitnerium
(262)
(263)
(262)
(265)
(266)
Chromiu
m
Mangane
se
Niobium
91
72
Hf
Barium
133
87
Fr
137
88
Ra
Francium
Radium
(223)
(226)
57-71
89-103
(261)
52
42
Mo
Ununnilium Unununium
(269)
Ununbium
(272)
(277)
Ununtritium Ununquadiu Ununpentiu Ununhexium Ununhexiu Ununoctium
m
m
m
-
(289)
-
(289)
-

57
La
58
Ce
59
Pr
60
Nd
61
Pm
62
Sm
63
Eu
64
Gd
65
Tb
66
Dy
67
Ho
68
Er
69
Tm
70
Yb
71
Lu
Lanthanu
m
Cerium
Praseodymiu
m
Neodymiu
m
Promethiu
m
Europium
Gadolinium
Terbium
Dysprosiu
m
Holmium
Erbium
Thulium
Ytterbium
Lutetium
139
89
Ac
140
90
Th
141
91
Pa
144
92
U
(147)
93
Np
Samariu
m
150
94
Pu
152
95
Am
157
96
Cm
159
97
Bk
163
98
Cf
165
99
Es
167
100
Fm
169
101
Md
173
102
No
175
103
Lr
Actinium
Thorium
Protactiniu
m
Uranium
Plutonium
Berkelium
Californium
Einsteinium
Fermium
Mendelevium
Nobelium
Lawrenciu
m
232
231
238
Americiu
m
Curium
(227)
Neptuniu
m
(247)
(247)
(249)
(254)
(253)
(256)
(254)
(257)
(237)
(244)
(243)
10
(293)
6
1
1
H
5. A new type of insulating board has been developed by that esteemed
construction company Kaput Ltd. They warn that at extremely high heat fluxes it
could be ignited, but they don’t think it’s very likely and it would take hours, so
there’s no real risk!
As an expert on the ignitability of materials you are asked to perform a thick/thin
calaculation given the following data on the material:
Density
Thermal conductivity
Specific heat capacity
Thickness of board
Initial/ambient laboratory temperature
Ignition temperature
2300 kg m-3
0.82 W m-1 K-1
824 J kg-1 K-1
7mm
18°C
410°C
It is considered that if 20 kW m-2 would be enough to ignite most materials (i.e.
indicative of flashover fires). Perhaps something easy to ignite would only require
10 kW m-2. Would this material ignite within ten minutes if exposed to 10 kW m -2?
Useful formulae:
  k / c
  k c
  t
 T 
tTHICK  4 kc  
 Q 
T
tTHIN  cL 
Q
2

[25 marks]
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work.
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
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