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Answer:
Hello there, please check explanations for step by step procedures to get answers.
Explanation:
Given that;
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College Engineering 10+5 pts
The switch has been in position a for a long time. At t = 0, the switch moves from position a to position b. The switch is a make-before-break type so there is no interruption of the inductor current. a. Find the expression for i(t) for t ≥ 0. b. What is the initial voltage across the inductor after the switch has been moved to position b? c. Does this initial voltage make sense in terms of circuit behavior? d. How many milliseconds after the switch has been put in position b does the inductor voltage equal 24V? e. Plot both i(t) and v(t) versus t
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Answer:
the rate of entropy change of the air is -0.1342 kW/K
the assumptions made in solving this problem
- Air is an ideal gas.
- the process is isothermal ( internally reversible process ). the change in internal energy is 0.
- It is a steady flow process
- Potential and Kinetic energy changes are negligible.
Explanation:
Given the data in the question;
From the first law of thermodynamics;
dQ = dU + dW ------ let this be equation 1
where dQ is the heat transfer, dU is internal energy and dW is the work done.
from the question, the process is isothermal ( internally reversible process )
Thus, the change in internal energy is 0
dU = 0
given that; Air is compressed by a 40-kW compressor from P1 to P2
since it is compressed, dW = -40 kW
we substitute into equation 1
dQ = 0 + ( -40 kW )
dQ = -40 kW
Now, change in entropy of air is;
ΔS = dQ / T
given that T = 25 °C = ( 25 + 273.15 ) K = 298.15 K
so we substitute
ΔS = -40 kW / 298.15 K
ΔS = -0.13416 ≈ -0.1342 kW/K
Therefore, the rate of entropy change of the air is -0.1342 kW/K
the assumptions made in solving this problem
- Air is an ideal gas.
- the process is isothermal ( internally reversible process ). the change in internal energy is 0.
- It is a steady flow process
- Potential and Kinetic energy changes are negligible.