Answer:
a) , b)
Explanation:
a) The thermal efficiency is:
b) The coefficient of performance is:
A spacecraft is fueled using hydrazine (N2H4; molecular weight of 32 grams per mole [g/mol]) and carries 1 comma 630 kilogram
s [kg] of fuel. On a mission to orbit a planet, the fuel will first be warmed from negative 199 degrees Fahrenheit [°F] to 100 degrees Fahrenheit [°F] before being used after the long space flight to reach the planet. The specific heat capacity of hydrazine is 0.099 kilojoules per mole kelvin [kJ/(mol K)]. If there is 250 watts [W] of power available to heat the fuel, how long will the heating process take in units of hours [h]?
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Answer:
a) the output voltage is 50 V
b)
- the average inductor current is 10 A
- the maximum inductor current is 13 A
- the maximum inductor current is 7 A
c) the output voltage ripple is 0.006 or 0.6%V₀
d) the average current in the diode under ideal components is 4 A
Explanation:
Given the data in the question;
a) the output voltage
V₀ = V/( 1 - D )
given that; V = 20 V, D = 0.6
we substitute
V₀ = 20 / ( 1 - 0.6 )
V₀ = 20 / 0.4
V₀ = 50 V
Therefore, the output voltage is 50 V
b)
- the average inductor current
= V
/ ( 1 - D )²R
given that R = 12.5 Ω, V = 20 V, D = 0.6
we substitute
= 20 / (( 1 - 0.6 )² × 12.5)
= 20 / (( 0.4)² × 12.5)
= 20 / ( 0.16 × 12.5 )
= 20 / 2
= 10 A
Therefore, the average inductor current is 10 A
- the maximum inductor current
= [V
/ ( 1 - D )²R] + [ V
given that, R = 12.5 Ω, V = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz
we substitute
= [20 / (( 1 - 0.6 )² × 12.5)] + [ (20 × 0.6 × 5) / (2 × 10) ]
= [20 / 2 ] + [ 60 / 20 ]
= 10 + 3
= 13 A
Therefore, the maximum inductor current is 13 A
- The minimum inductor current
= [V
/ ( 1 - D )²R] - [ V
given that, R = 12.5 Ω, V = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz
we substitute
= [20 / (( 1 - 0.6 )² × 12.5)] - [ (20 × 0.6 × 5) / (2 × 10) ]
= [20 / 2 ] -[ 60 / 20 ]
= 10 - 3
= 7 A
Therefore, the maximum inductor current is 7 A
c) the output voltage ripple
ΔV₀/V₀ = D/RCf
given that; R = 12.5 Ω, C = 40 μF = 40 × 10⁻⁶ F, D = 0.6, f = 200 Khz = 2 × 10⁵ Hz
we substitute
ΔV₀/V₀ = 0.6 / (12.5 × (40 × 10⁻⁶) × (2 × 10⁵) )
ΔV₀/V₀ = 0.6 / 100
ΔV₀/V₀ = 0.006 or 0.6%V₀
Therefore, the output voltage ripple is 0.006 or 0.6%V₀
d) the average current in the diode under ideal components;
under ideal components; diode current = output current
hence the diode current will be;
= V₀/R
as V₀ = 50 V and R = 12.5 Ω
we substitute
= 50 / 12.5
= 4 A
Therefore, the average current in the diode under ideal components is 4 A
Answer with Explanation:
The relation between power and energy is
Since the nuclear reactor operates at 1200 MW throughout the year thus the energy produced in 1 year equals
Now from the energy mass equivalence we have
where
'c' is the speed of light in free space
Thus equating both the above values we get
Since it is given that 1 kg of mass is 34% effective thus the mass reuired for the reactor is
Thus 1.235 kg of nuclear fuel is reuired for operation.
Answer:
a. Heat removal rate will increase
b. Heat removal rate will decrease
Explanation:
Given that
One end of rod is connected to the furnace and rod is long.So this rod can be treated as infinite long fin.
We know that heat transfer in fin given as follows
We know that area
Now when diameter will triples then :
So the new heat transfer will increase by 3 times.
Now when copper rod will replace by aluminium rod :
As we know that thermal conductivity(K) of Aluminium is low as compare to Copper .It means that heat transfer will decreases.