Answer:
160 W
Explanation:
The relation is ...
Real Power = (pf)(V)(A)
= 0.8(100)(2) = 160 . . . . watts
Answer:
Therefore, the horizontal displacement of end F of rod EF is 0.4797 mm
Explanation:
solution is mentioned in steps.
Answer:
While resistance welding you should wear clear grinding glasses, unbreakable plastic face shields or clear unbreakable plastic goggles. When resistance welding a #10, or more, shade lens should be worn. All hand and portable tools should be inspected for loose parts, cleanliness, or worn power cords.
Answer:
Explanation:
given data:
Diameter =
from continuity equation
![v_2 = [\frac{d_1}{d_2}]^2 v_1](https://tex.z-dn.net/?f=v_2%20%3D%20%5B%5Cfrac%7Bd_1%7D%7Bd_2%7D%5D%5E2%20v_1)
![= [\frac{0.200}{0.158}]^2 \times 100](https://tex.z-dn.net/?f=%3D%20%5B%5Cfrac%7B0.200%7D%7B0.158%7D%5D%5E2%20%5Ctimes%20100)
by energy flow equation
and q =0, w =0 for nozzle
therefore we have
but we know dh = Cp dt
hence our equation become
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
