A rigid tank contains 3 kg of water initially at 43.97% quality and at a temperature of 120°C. The water is heated until it reac
hes a pressure of 7 bar. Neglect changes in kinetic and potential energy.
a. At what temperature, in °C, will the water become saturated vapor?
b. Determine the heat transfer, in kJ.
c. The work of the cycle, in kJ
a. At what temperature, in °C, will the water become saturated vapor?
b. Determine the heat transfer, in kJ.
c. The work of the cycle, in kJ
1 answer:
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Answer:
a)
b) attached below
c) type zero system
d) k >
e) The gain K increases above % error as the steady state speed increases
Explanation:
Given data:
Motor voltage = 12 v
steady state speed = 200 rad/s
time taken to reach 63.2% = 1.2 seconds
<u>a) The transfer function of the motor from voltage to speed</u>
let ; be the transfer function of a motor
when i/p = 12v then steady state speed ( k1 ) = 200 rad/s , St ( time constant ) = 1.2 sec
hence the transfer function of the motor from voltage to speed
=
<u>b) draw the block diagram of the system with plant controller and the feedback path </u>
attached below is the remaining part of the detailed solution
c) The system is a type-zero system because the pole at the origin is zero
d) ) k >
Answer:
a) at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
b) daylight (d) = 0.50 μm
Incandescent ( i ) = 1 μm
Explanation:
To Calculate the band emission fractions we will apply the Wien's displacement Law
The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as
F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )
<em>Values are gotten from the table named: blackbody radiati</em>on functions
<u>a) Calculate the band emission fractions for the visible region</u>
at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
attached below is a detailed solution to the problem
<u>b)calculate wavelength corresponding to the maximum spectral intensity</u>
For daylight ( d ) = 2898 μm *k / 5800 k = 0.50 μm
For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm
