A Gaussian random voltage X volts is input to a half-wave rectifier and the output voltage is Y = Xu (X) Volts were u (x) is the
unit step function. Assume X has mean 0 V and variance σ^2 V^2. The output voltage Y is then applied across a (nonrandom) resistance of R ohms. The answers below should be expressed in terms of the Φ or Q functions or in closed form (no integrals).
(a) Find the probability that the current which flows through the resistor exceeds 1 Amp.
(b) Find the probability that the power which is dissipated in the resistor exceeds 1 watt.
(c) Find the mean and variance of the current which flows through the resistor.
(d) Find the mean and variance of the power which is dissipated in the resistor.
(a) Find the probability that the current which flows through the resistor exceeds 1 Amp.
(b) Find the probability that the power which is dissipated in the resistor exceeds 1 watt.
(c) Find the mean and variance of the current which flows through the resistor.
(d) Find the mean and variance of the power which is dissipated in the resistor.
1 answer:
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Answer:
a) 0.76
b) 0.80
c) 1964 kW
Explanation:
GIVEN DATA:
Assume Mechanical energy at exist is negligible
A) Take lake bottom as reference, and then kinetic and potential energy are taken as zero.
change in mechanical energy is givrn as
= 0.491 kJ/kg
B)
c)
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.