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Gala2k [10]
3 years ago
13

5. Switch a in the circuit has been open for a long time and switch b has been closed for a long time. Switch a is closed at t =

0. After remaining closed for 1s, switches a and b are opened simultaneously and remain open indefinitely. Determine the expression for the inductor current i that is valid when (a) 0 ≤ t ≤ 1s and (b) t ≥ 1s
Engineering
1 answer:
andriy [413]3 years ago
8 0

Answer:

(a) 1/L∫Vdt; integral t [0,1]

(b) 1/L∫Vdt; integral t [ 1, infinity]

Explanation:

An Inductor current I, flowing through an inductor depends on the voltage, V, across the inductor and the inductance, L, of the inductor. The switch 1, 2 timing varies the voltage V with time t

The expression for inductor current is given as:

I= 1/L∫Vdt,

where I is equal to the current flowing through the inductor, L is equal to the inductance of the inductor, and V is equal to the voltage across the inductor.

The formula can also be written as:

I= I0 + 1/L∫Vdt, where I is inductor current at time t, and io is inductor current at t = 0. Time can be varied by controlling the switch

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What is the answer of this question please tell me <br><br>dadada please please​
charle [14.2K]

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6 0
3 years ago
A tank with a volume of 8 m3 containing 4 m3 of 20% (by volume) NaOH solution is to be purged by adding pure water at a rate of
lawyer [7]

Answer:

The time necessary to purge 95% of the NaOH is 0.38 h

Explanation:

Given:

vfpure water(i) = 3 m³/h

vNaOH = 4 m³

xNaOH = 0.2

vfpure water(f) = 2 m³/h

pwater = 1000 kg/m³

pNaOH = 1220 kg/m³

The mass flow rate of the water is = 3 * 1000 = 3000 kg/h

The mass of NaOH in the solution is = 0.2 * 4 * 1220 = 976 kg

When the 95% of the NaOH is purged, thus the NaOH in outlet is = 0.95 * 976 = 927.2 kg

The volume of NaOH in outlet after time is = 927.2/1220 = 0.76 m³

The time required to purge the 95% of the NaOH is = 0.76/2 = 0.38 h

4 0
3 years ago
A gasoline engine has a piston/cylinder with 0.1 kg air at 4 MPa, 1527◦C after combustion, and this is expanded in a polytropic
Roman55 [17]

Answer:

The expansion work is 71.24 kJ and heat transfer is -16.89 kJ

Explanation:

From ideal gas law,

Initial volume (V1) = nRT/P

n is the number of moles of air in the cylinder = mass/MW = 0.1/29 = 0.00345 kgmol

R is gas constant = 8314.34 J/kgmol.K

T is initial temperature = 1527 °C = 1527+273 = 1800 K

P is initial pressure = 4 MPa = 4×10^6 Pa

V1 = 0.00345×8314.34×1800/(4×10^6) = 0.013 m^3

V2 = 10×V1 = 10×0.013 = 0.13 m^3

The process is a polytropic expansion process

polytropic exponent (n) = 1.5

P2 = P1(V1/V2)^n = 4×10^6(0.013/0.13)^1.5 = 1.26×10^5 Pa

Expansion work = (P1V1 - P2V2) ÷ (n - 1) = (4×10^6 × 0.013 - 1.26×10^5 × 0.13) ÷ (1.5 - 1) = 35620 ÷ 0.5 = 71240 J = 71240/1000 = 71.24 kJ

Heat transfer = change in internal energy + expansion work

change in internal energy (∆U) = Cv(T2 - T1)

T2 = PV/nR = 1.26×10^5 × 0.13/0.00345×8314.34 = 571 K

Cv = 20.785 kJ/kgmol.K

∆U = 20.785(571 - 1800) = -25544.765 kJ/kgmol × 0.00345 kgmol = -88.13 kJ

Heat transfer = -88.13 + 71.24 = -16.89 kJ

5 0
3 years ago
For a steel alloy it has been determined that a carburizing heat treatment of 15 h duration will raise the carbon concentration
Free_Kalibri [48]

Answer:

135 hour

Explanation:

It is given that a carburizing heat treatment of 15 hour will raise the carbon concentration by 0.35 wt% at a point of 2 mm from the surface.

We have to find the time necessary to achieve the same concentration at a 6 mm position.

we know that \frac{x_1^2}{Dt}=constant where x is distance and t is time .As the temperature is constant so D will be also constant

So \frac{x_1^2}{t}=constant

then \frac{x_1^2}{t_1}=\frac{x_2^2}{t_2} we have given x_1=2 mm\ ,t_1=15 hour\ ,x_2=6\ mm and we have to find t_2 putting all these value in equation

\frac{2^2}{15}=\frac{6^2}{t_2}

so t_2=135\ hour

5 0
3 years ago
A 200-mm-long strip of metal is stretched in two steps, first to 300 mm and then to 400 mm. Show that the total true strain is t
Neko [114]

Explanation:

For true Strain:

step 1:

E true = Ln(1 + 0.5 ) = 0.40

Step 2:

E true = Ln(1 + 0.33 ) = 0.29

By single step process:

E true = Ln(1 + 1 ) = 0.69

total strain of step process = 0.40 + 0.29 = 0.69 units

SO TRUE STRAIN IS ADDITIVE.

4 0
2 years ago
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