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:
Following are the responses to the given question:
Explanation:
Answer:
Rate of Entropy =210.14 J/K-s
Explanation:
given data:
power delivered to input = 350 hp
power delivered to output = 250 hp
temperature of surface = 180°F
rate of entropy is given as
T = 180°F = 82°C = 355 K
Rate of heat = (350 - 250) hp = 100 hp = 74600 W
Rate of Entropy
Answer:
(a) Relative Humidity = 48%,
Specific humidity = 0.0095
(b) Enthalpy = 65 KJ/Kg of dry sir
Specific volume = 0.86 m^3/Kg of dry air
(c/d) 12.78 degree C
(e) Specific volume = 0.86 m^3/Kg of dry air
Answer:
Multiplying impulse response by t ( option D )
Explanation:
We can obtain The impulse response of strength 1 considering a unit step response by Multiplying impulse response by t .
When we consider the Laplace Domain, and the relationship between unit step and impulse, we can deduce that the Impulse response will take the inverse Laplace transform of the function ( transfer ) . Hence Multiplying impulse response by t will be used .
