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
Answer:
The net amount of energy change of the air in the room during a 10-min period is 120 KJ.
Explanation:
Given that
Heat loss from room (Q)= 60 KJ/min
Work supplied to the room(W) = 1.2 KW = 1.2 KJ/s
We know that 1 W = 1 J/s
Sign convention for heat and work:
1. If heat is added to the system then it is taken as positive and if heat is rejected from the system then it is taken as negative.
2. If work is done by the system then it is taken as positive and if work is done on the system then it is taken as negative.
So
Q = -60 KJ/min
In 10 min Q = -600 KJ
W = -1.2 KJ/s
We know that
1 min = 60 s
10 min = 600 s
So W = -1.2 x 600 KJ
W = -720 KJ
WE know that ,first law of thermodynamics
Q = W + ΔU
-600 = - 720 + ΔU
ΔU = 120 KJ
The net amount of energy change of the air in the room during a 10-min period is 120 KJ.
Answer:
The correct answer is A : Orientation dependence of normal and shear stresses at a point in mechanical members
Explanation:
Since we know that in a general element of any loaded object the normal and shearing stresses vary in the whole body which can be mathematically represented as

And 
Mohr's circle is the graphical representation of the variation represented by the above 2 formulae in the general oriented element of a body that is under stresses.
The Mohr circle is graphically displayed in the attached figure.
Answer:A. No one has ever beat Nancy.
Explanation:
The dormain of discourse in a simple language is the set of entities upon which our discussions are based when discussing about something.
The dormain of discourse is also known simply as universe, can also be said to be a set of entities o
upon which certain variables of interest in some formal treatment may range.
The dormain of discourse is generally attributed to Augustus De Morgan, it was also extensively used by George Boole in his Laws of Thought.
THE LOGICAL UNDERSTANDING OF THE THE QUESTION IS THAT NO ONE HAS EVER BEAT NANCY.