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Task 1
Contents
Task 1 .......................................................................................................................................................... 1
Part(a)..................................................................................................................................................... 1
Part b....................................................................................................................................................... 2
Task 2 .......................................................................................................................................................... 3
Task 3 .......................................................................................................................................................... 8
Task 4 ........................................................................................................................................................ 10
Part a..................................................................................................................................................... 10
Part b..................................................................................................................................................... 11
Task 5 ........................................................................................................................................................ 12
Part a..................................................................................................................................................... 12
Part b..................................................................................................................................................... 13
Part(a)
From Table 1,
W 1 =3.4 kN=3,400 N
w 2 =
820 N
m
L 1 =3.5 m.
Free body diagram
Since the system is at equilibrium then
∑
F= 0
R + R − 34 00 N −
820 N
× 3 m = 0
A B
(
m
)
R
A
+ R
B
=5,860 N.
Taking moments at A and considering the system is at equilibrium
∑
M
clockwise
∑
M
anticlockwise
( 3400 N × 3.5 m)
820 N
× 3 m× 6.5 m
=R ×9.5 m
(
m
)
B
11900 Nm+ 15990 Nm=9.5 m× R
B
R =
278990 N / m
=2,935.79 N
B
9 ⋅ 5 m
Therefore,
R
A
+2935.79 N=5,860 N
R
A
=5,860 N −2935.79 N =2,924.21 N.
Therefore, the reactions at A and B are 2,924N and 2,936N respectively
Part b
From Table 1,
Task 2
From the table
T 1 = 26 ℃=( 26 + 273 ) K = 299 K
T 3 = 1090 ℃=( 1090 + 273 )= 1363 K
The T-S diagram for an ideal Joule cycle and P-V diagram
Work input in the compressor=
h
2
−h
1
=C
p
(
T
2
−T
1
)
Work generated in the turbine=
h
3
h
4
C
p
(
T
3
T
4
)
Heat input in the heater=
h
3
−h
2
=C
p
(
T
3
−T
2
)
Heat rejected in the cooler=
h
4
−h
1
=C
p
(
T
4
−T
1
)
1 p
T
1
T
4
p
1
p
Thermal efficiency is therefore given by,
η=
∑
Q
C
p
(
T
3
−T
2
)
−C
p
(
T
4
−T
1
)
T
4
−T
1
……................. (1)
Q 1
C
p
(
T
3
−T
2
)
T 3 −T 2
Since processes 1-2 and 3-4 are isentropic compression and expansion respectively then,
T T p
(γ − 1 )
( γ− 1 )
2
3
2
γ
(
r
)
γ
Where,
r p = the compression
ratio Therefore,
T 2 =T 1
(
r
p)
( γ − 1
)
γ
T 3 =T 4
(
r
p )
( γ− 1
)
γ
Substituting in equation 1,
( γ− 1 )
(
r p )
γ
But γ =1.4,then
γ
γ 1.
Then thermal efficiency is equal to,
η= 1 −
(
r
p
)
Net work output is given by,
W =C
p
(
T
3
−T
4
)
−C
p
(
T
2
−T
1
)
¿ C
p
T
3
(
r
T
3
(
T
(
r
)
p
)
−T
1
)
η= 1 −
1 p
3
(
p
)
7
P=m´
C
(
T
(
)
−T
( (
r
)
0.285 7
)
)
But m´ =0.3 kg / s
For a compression ratio between 5 to 20,
Compression
ratios Thermal Efficiency Power Generated
kW
5 0.3686 98.
6 0.4006 104.
7 0.4265 108.
8 0.4479 110.
9 0.4662 112.
10 0.4820 114.
11 0.4959 115.
12 0.5083 115.
13 0.5194 116.
14 0.5295 116.
15 0.5387 116.
16 0.5471 115.
17 0.5549 115.
18 0.5621 115.
19 0.5688 114.
20 0.5751 114.
Plot of Net power against thermal efficiency
p
1 p
Power Generated [kW] againist Thermal efficiency
0.4000 0.4500 0.5000 0.
Thermal efficieny
Figure 1Plot of Net power output against Thermal efficiency
Fitting the data, we get
Net
power
From d’Alembert’s principle,
∑
F−ma= 0
From Table 2,
Hover acceleration , a=1.35 m s
− 2
hover acceletion at a path which is at angle ,
ϕ= 62 ° Surface Drag , D= 295 N
From the task,
Trust ¿ the propeller , T = 480
N Wind Force ,W = 32 N
The free body diagrams,
Thus,
∑
F= 480 N × cos ( 62 ° − 56 ° )+ 32 N ×cos ( 160 − 62 )−C ×cos (θ )− 295 N
¿ 177.912−C × cos (θ)
⸫
177.912−C× cos(θ) −ma= 0
177.912−C × cos (θ)− 240 × 1.35= 0
C × cos(θ)=146.08 N
Also,
480 × cos ( 56 )+ 240 ×1.35 ×cos ( 62 )− 295 ×cos ( 62 )− 32 × cos ( 180 − 160 )−Csin (θ )= 0
Csin (θ)=251.96 N
Therefore,
tan (θ )=
251.96 N
146.08 N
θ=tan
− 1
From that, the magnitude of the force exerted by the current by the current is given by;
C=
146.08 N
=292.72 N
cos ( 59.9)
And the direction of the force exerted by the current by the current is given by;
β= 160 +78.1=238.1 ° anticlockwise
Part b
Mass of the component ∈air =127.6 kg
Mass of the component ∈oil=11.8 kg
gravitional acceleration , g=9.81 m / s
2
Weight of the component ∈air =127.6 kg× 9.81=1251.8 N
Weight of the component ∈oil=11.8 kg× 9.81=115.8 N
From Archimedes’’ principle,
Weight of water displaced=Weight of thecomponent ∈ air−weight of the component ∈oil
¿ 1251.8 N −115.8 N = 1136 N
Weight of water displaced= volume of oil displace ×oil desity a× 9.
1136 N=9.81×
870 kg
×Volume of oil displaced
m
3
Volume of water dis placed=
1136 N
=0.1332 m
3
8526 N / m
3
Volume of oil displaced =Volume of the component
Therefore, the mean density of the component is given by,
ρ =
Mass of the component ∈air
kg
=957.7 kg m
− 3
component
Volume of the
component
0.1332m
3
Task 5
Part a
Since the valve stem is free to expand then the extension in length of the valve stem is given by,
dL=αTL
In this case,
Co−efficient of thermal expansion, α =12.
×
Original length , L= 125 mm
Rise∈Temperature ,T =T 4
From the table,
T
4
Therefore,
T = 857 ℃− 10 ℃= 847 ℃
− 6
Thus,
dL=12.
×
− 6
× 847 ℃ × 125 mm
dL=1.29 mm
Part b
Since the valve stem is free to expand then the extension in length of the valve stem is given by,
