Answer:
a) Ql=33120000 kJ
b) COP = 5.6
c) COPreversible= 29.3
Explanation:
a) of the attached figure we have:
HP is heat pump, W is the work supplied, Th is the higher temperature, Tl is the low temperature, Ql is heat supplied and Qh is the heat rejected. The worj is:
W=Qh-Ql
Ql=Qh-W
where W=2000 kWh
Qh=120000 kJ/h
b) The coefficient of performance is:
c) The coefficient of performance of a reversible heat pump is:
Th=20+273=293 K
Tl=10+273=283K
Replacing:
The classical motion for an oscillator that starts from rest at location x₀ is
x(t) = x₀ cos(ωt)
The probability that the particle is at a particular x at a particular time t
is given by ρ(x, t) = δ(x − x(t)), and we can perform the temporal average
to get the spatial density. Our natural time scale for the averaging is a half
cycle, take t = 0 → π/
ω
Thus,
ρ =
Limit is 0 to π/ω
We perform the change of variables to allow access to the δ, let y = x₀ cos(ωt) so that
ρ(x) =
Limit is x₀ to -x₀
Limit is -x₀ to x₀
This has as expected. Here the limit is -x₀ to x₀
The expectation value is 0 when the ρ(x) is symmetric, x ρ(x) is asymmetric and the limits of integration are asymmetric.
Answer:
Diode equation for reverse saturation current
Voltage at which diode goes into Resistive region:V=-5 volts
Voltage at which high level injection occurs:Va=0.55 volt
Voltage at which avalanche multiplication occurs:V=5volts
Explanation:
we take here forward and reverse 0.7 volt and -5 volt
As Diode current equation is express as
....................1
here is total current through the diode and is reverse saturated current and is voltage drop across diode and is idealized factor and is thermal voltage
so here we know that when Bios is forward than
= .................2
ans Bios is Reverse than
= ..................3
so here
1. diode reverse saturation current is express as
and
2. Voltage at which diode go into Reverse behavior will be
= = -5 volt
and
3. voltage at which high level injection occur that is
Va = 0.55 volt
and
4. voltage at which avalanche multiplication occurs is
Va = 5 volt
Answer:
Fatigue lifetimes will be ranked as B>A>C
Explanation:
Start by calculating the mean stress for all samples
σa= (σamax + σamin)/2 = (450 - 350)/2 = 50MPa
σb= (σbmax + σbmin)/2 = (400 - 300)/2 = 50MPa
σa= (σcmax + σcmin)/2 = (340 - 340)/2 = 0MPa
Now calculate the stress amplitudes of all three samples
σa= (σamax - σamin)/2 = (450 + 350)/2 = 400MPa
σb= (σbmax - σbmin)/2 = (400 + 300)/2 = 350MPa
σc= (σcmax - σcmin)/2 = (340 + 340)/2 = 340MPa
The mean stress of samples A and B is the same which is higher than sample C. Hence samples A and B will have a higher fatigue life than sample C. However, the higher stress amplitude means lower fatigue life, hence, sample B will have a higher fatigue life than sample A. So the order will be B> A> C.
Attached picture shows the justification using and S- N plot.
Answer:50 , 20
Explanation:
Given
Diametrical Pitch
where T= no of teeths
D=diameter
module(m) of gears must be same
Let be the gears on two gears
Therefore Center distance is given by
thus
and Velocity ratio is given by
From 1 & 2 we get