Answer:
second-law efficiency = 62.42 %
Explanation:
given data
temperature T1 = 1200°C = 1473 K
temperature T2 = 20°C = 293 K
thermal efficiency η = 50 percent
solution
as we know that thermal efficiency of reversible heat engine between same temp reservoir
so here
efficiency ( reversible ) η1 = 1 -
............1
efficiency ( reversible ) η1 = 1 -
so efficiency ( reversible ) η1 = 0.801
so here second-law efficiency of this power plant is
second-law efficiency =
second-law efficiency =
second-law efficiency = 62.42 %
Answer:a
a) Vo/Vi = - 3.4
b) Vo/Vi = - 14.8
c) Vo/Vi = - 1000
Explanation:
a)
R1 = 17kΩ
for ideal op-amp
Va≈Vb=0 so Va=0
(Va - Vi)/5kΩ + (Va -Vo)/17kΩ = 0
sin we know Va≈Vb=0
so
-Vi/5kΩ + -Vo/17kΩ = 0
Vo/Vi = - 17k/5k
Vo/Vi = -3.4
║Vo/Vi ║ = 3.4 ( negative sign phase inversion)
b)
R2 = 74kΩ
for ideal op-amp
Va≈Vb=0 so Va=0
so
(Va-Vi)/5kΩ + (Va-Vo)74kΩ = 0
-Vi/5kΩ + -Vo/74kΩ = 0
Vo/Vi = - 74kΩ/5kΩ
Vo/Vi = - 14.8
║Vo/Vi ║ = 14.8 ( negative sign phase inversion)
c)
Also for ideal op-amp
Va≈Vb=0 so Va=0
Now for position 3 we apply nodal analysis we got at position 1
(Va - Vi)/5kΩ + (Va - Vo)/5000kΩ = 0 ( 5MΩ = 5000kΩ )
so
-Vi/5kΩ + -Vo/5000kΩ = 0
Vo/Vi = - 5000kΩ/5kΩ
Vo/Vi = - 1000
║Vo/Vi ║ = 1000 ( negative sign phase inversion)
Answer:
1200KJ
Explanation:
The heat dissipated in the rotor while coming down from its running speed to zero, is equal to three times its running kinetic energy.
P (rotor-loss) = 3 x K.E
P = 3 x 300 = 900 KJ
After coming to zero, the motor again goes back to running speed of 1175 rpm but in opposite direction. The KE in this case would be;
KE = 300 KJ
Since it is in opposite direction, it will also add up to rotor loss
P ( rotor loss ) = 900 + 300 = 1200 KJ
Answer:
c = 18.0569 mm
Explanation:
Strategy
We will find required diameter based on angle of twist and based on shearing stress. The larger value will govern.
Given Data
Applied Torque
T = 750 N.m
Length of shaft
L = 1.2 m
Modulus of Rigidity
G = 77.2 GPa
Allowable Stress
г = 90 MPa
Maximum Angle of twist
∅=4°
∅=4*
/180
∅=69.813 *10^-3 rad
Required Diameter based on angle of twist
∅=TL/GJ
∅=TL/G*
/2*c^4
∅=2TL/G*
*c^4
c=
∅
c=18.0869 *10^-3 rad
Required Diameter based on shearing stress
г = T/J*c
г = [T/(J*
/2*c^4)]*c
г =[2T/(J*
*c^4)]*c
c=17.441*10^-3 rad
Minimum Radius Required
We will use larger of the two values
c= 18.0569 x 10^-3 m
c = 18.0569 mm
A chemical engineer can clearly see from this kind of test if a substance stays in a system and builds up or if it just passes through.
<h3>What is a chemical engineer?</h3>
- Processes for manufacturing chemicals are created and designed by chemical engineers.
- To solve issues involving the manufacture or usage of chemicals, fuel, medications, food, and many other goods, chemical engineers use the concepts of chemistry, biology, physics, and math.
- A wide range of sectors, including petrochemicals and energy in general, polymers, sophisticated materials, microelectronics, pharmaceuticals, biotechnology, foods, paper, dyes, and fertilizers, have a significant demand for chemical engineers.
- Chemical engineering is undoubtedly difficult because it requires a lot of physics and math, as well as a significant number of exams at the degree level.
To learn more about chemical engineer, refer to:
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