Solar engineers work in any number of areas of engineering products that help harness energy from the sun. They may research, design, and develop new products, or they may work in testing, production, or maintenance. They may collect and manage data to help design solar systems.
Answer:
The volume percentage of graphite is 10.197 per cent.
Explanation:
The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:
Where:
- Volume occupied by the graphite phase, measured in cubic centimeters.
- Volume occupied by the graphite phase, measured in cubic centimeters.
The expression is expanded by using the definition of density and subsequently simplified:
Where:
, 
- Masses of the ferrite and graphite phases, measured in grams.
- Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.
If 
, 
, 
and 
, the volume percentage of graphite is:
The volume percentage of graphite is 10.197 per cent.
Answer:
(a) maximum positive reaction at A = 64.0 k
(b) maximum positive shear at A = 32.0 k
(c) maximum negative moment at C = -540 k·ft
Explanation:
Given;
dead load Gk = 400 lb/ft
live load Qk = 2 k/ft
concentrated live load Pk =8 k
(a) from the influence line for vertical reaction at A, the maximum positive reaction is
= 2*(8) +(1/2(20 - 0)* (2))*(2 + 0.4) = 64 k
See attachment for the calculations of (b) & (c) including the influence line
Answer:
See explanation
Explanation:
Given:
Initial pressure,
p
1
=
15
psia
Initial temperature,
T
1
=
80
∘
F
Final temperature,
T
2
=
200
∘
F
Find the gas constant and specific heat for carbon dioxide from the Properties Table of Ideal Gases.
R
=
0.04513
Btu/lbm.R
C
v
=
0.158
Btu/lbm.R
Find the work done during the isobaric process.
w
1
−
2
=
p
(
v
2
−
v
1
)
=
R
(
T
2
−
T
1
)
=
0.04513
(
200
−
80
)
w
1
−
2
=
5.4156
Btu/lbm
Find the change in internal energy during process.
Δ
u
1
−
2
=
C
v
(
T
2
−
T
1
)
=
0.158
(
200
−
80
)
=
18.96
Btu/lbm
Find the heat transfer during the process using the first law of thermodynamics.
q
1
−
2
=
w
1
−
2
+
Δ
u
1
−
2
=
5.4156
+
18.96
q
1
−
2
=
24.38
Btu/lbm
