Answer:
The number of kilo- grams of hydrogen that pass per hour through this sheet of palladium is
Explanation:
Given
x1 = 0 mm
x2 = 6 mm = 6 * m
c1 = 2 kg/
c2 = 0.4 kg/
T = 600 °C
Area = 0.25
D = 1.7 *
First equation
J = - D
Second equation
J =
To find the J (flux) use the First equation
J = - 1.7 * *
To find M use the Second equation
=
M =
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
