Helium gas expands in a piston-cylinder in a polytropic process with n=1.67. Is the work positive, negative or zero?
1 answer:
Answer:
work will be positive when it is under polytropic expansion process
Explanation:
It states a polytropic process with n equal to 1.67. there is a polytropic expansion that mean work is positive and if it was polytropic compression then it would be negative
Also work during the process of polytropic is given as
the work will be positive when it is under the polytropic expansion process
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Answer:
a)31%
b)34MW
Explanation:
A rankine cycle is a generation cycle using water as a working fluid, when heat enters the boiler the water undergoes a series of changes in state and energy until generating power through the turbine.
This cycle is composed of four main components, the boiler, the pump, the turbine and the condenser as shown in the attached image
To solve any problem regarding the rankine cycle, enthalpies in all states must be calculated using the thermodynamic tables and taking into account the following.
• The pressure of state 1 and 4 are equal
• The pressure of state 2 and 3 are equal
• State 1 is superheated steam
• State 2 is in saturation state
• State 3 is saturated liquid at the lowest pressure
• State 4 is equal to state 3 because the work of the pump is negligible.
Once all enthalpies are found, the following equations are used using the first law of thermodynamics
Wout = m (h1-h2)
Qin = m (h1-h4)
Win = m (h4-h3)
Qout = m (h2-h1)
The efficiency is calculated as the power obtained on the heat that enters
Efficiency = Wout / Qin
Efficiency = (h1-h2) / (h1-h4)
For this problem, we will first find the enthalpies in all states
h1=3231kJ/Kg
h2=2310kJ/Kg
h3=h4=272kJ/Kg
A) using the eficiency ecuation
Efficiency = (h1-h2) / (h1-h4)
Efficiency =(3231-2310)/(3231-272)=0.31=<u>31%</u>
b)using ecuation for Wout
Wout = m (h1-h2)
Wout=37(3231-2310)=34077KW=<u>34.077MW</u>
Answer:
Attached below are the sketches
answer :
c) G(s) = 100 / ( s + 100 )
d) y'(t) + 100Y(s) = 100 X(s)
e) g(t) = e^-100t u(t)
Explanation:
a) Sketch the bode plot
The filter here is a low pass filter
b) Sketch the s-plane
attached below. pole ( s ) is at 100
c) write the transfer function of the filter
Transfer function ; G(s) = 100 / ( s + 100 )
d) write the differential equation
Y(s) / X(s) = 100 / s + 100
Y(s) [ s + 100 ] = 100 X(s)
= sY(s) + 100Y = 100 X(s)
∴ differential equation = y'(t) + 100Y(s) = 100 X(s)
e) write out the unforced transient response
g(t) = e^-100t u(t)
f) write out the frequency response
attached below
