At what forward voltage does a diode conduct a current equal to 10,000 Is ? In terms of Is , what current flows in the same diod

Answer:

Volume change percentage is 2.60%

Water level increase is 4.138 mm

Explanation:

Provided data

Water volume V = 500 L

Initial temperature T1 = 20°C

Final temperature T2 = 80°C

Diameter of the vat = 2 m

Objective

We aim to determine percentage change in volume and the rise in water level.

Solution

We will apply the bulk modulus equation, which relates the change in pressure to the change in volume.

It can similarly relate to density changes.

Thus,

E = ................1

And ............2

Here, ρ denotes density. The density at 20°C = 998 kg/m³.

The density at 80°C = 972 kg/m³.

Plugging in these values into equation 2 gives

dV = 0.0130 m³

Therefore, the percentage change in volume will be

dV % =  × 100

dV % =  × 100

dV % = 2.60 %

Hence, the percentage change in volume is 2.60%

Initial volume v1 = ................3

Final volume v2 = ................4

From equations 3 and 4, subtract v1 from v2.

v2 - v1 =  

dV =

Substituting all values yields

0.0130 =

Thus, dl = 0.004138 m.

Consequently, the water level rises by 4.138 mm.

Answer:

Explanation:

A liquid food containing 12% total solids is heated via steam injection at a pressure of 232.1 kPa (see Fig. E3.3). The product starts at a temperature of 50°C and has a flow rate of 100 kg/min, being elevated to a temperature of 120°C. The specific heat of the product varies with its composition as follows:

and the

specific heat of the product at 12% total solids is 3.936 kJ/(kg°C). The goal is to calculate the quantity and minimum quality of steam required to ensure that the leaving product has 10% total solids.

Given

Product total solids in () = 0.12

Product mass flow rate () = 100 kg/min

Product total solids out () = 0.1

Product temperature in () = 50°C

Product temperature out () = 120°C

Steam pressure = 232.1 kPa at () = 125°C

Product specific heat in () = 3.936 kJ/(kg°C)

The mass equation is:

Also

Therefore:

The energy balance equation is:

 kJ/(kg°C)

By substituting values into the energy equation:

From the properties of saturated steam at 232.1 kPa,

= 524.99 kJ/kg

= 2713.5 kJ/kg

% quality =

Any steam quality above 63.5% will result in higher total solids in the heated product.

Answer:

Time constant = 15.34 seconds

The thermometer indicates an error margin of 0.838°

Explanation:

Given

t = 1 minute = 60 seconds

c(t) = 98% = 0.98

According to the details provided, the thermometer functions as a first-order system.

The transfer function of such a system is expressed as;

C(s)/R(s) = 1/(sT + 1).

To determine the time constant, the step response must be evaluated.

This is defined as

r(t) = u(t) --- Applying Laplace Transformation

R(s) = 1/s

Replacing 1/s back into C(s)/R(s) = 1/(sT + 1).

What we have

C(s)/1/s = 1/(sT + 1)

C(s) = 1/(sT + 1) * 1/s

C(s) = 1/s - 1/(s + 1/T) --- Taking Inverse Laplace Transformation

L^-1(C(s)) = L^-1(1/s - 1/(s + 1/T))

Given that e^-t <–> 1/(s + 1) --- {L}

1 <–> 1/s {L}

Thus, the unit response c(t) = 1 - e^-(t/T)

Substituting 0.98 for c(t) and 60 for t

0.98 = 1 - e^-(60/T)

0.98 - 1 = - e^-(60/T)

-0.02 = - e^-(60/T)

e^-(60/T) = 0.02

ln(e^-(60/T)) = ln(0.02)

-60/T = -3.912

T = -60/-3.912

T = 15.34 seconds

It follows that the time constant = 15.34 seconds

The error signal is characterized by

E(s) = R(s) - C(s)

Where the temperature shifts at 10°/min; which equals 10°/60 s = 1/6

Thus,

E(s) = R(s) - 1/6 C(s)

Calculating C(s)

C(s) = 1/s - 1/(s + 1/T)

C(s) = 1/s - 1/(s + 1/15.34)

Note that R(s) = 1/s

Thus, E(s) turns into

E(s) = 1/s - 1/6(1/s - 1/(s + 1/15.34))

E(s) = 1/s - 1/6(1/s - 1/(s + 0.0652)

E(s) = 1/s - 1/6s + 1/(6(s+0.0652))

E(s) = 5/6s + 1/(6(s+0.0652))

E(s) = 0.833/s + 1/(6(s+0.0652)) ---- Taking Inverse Laplace Transformation

e(t) = 1/6e^-0.652t + 0.833

In a first-order system, a steady state condition is reached when the time is four times the time constant.

Thus,

Time = 4 * 15.34

Time = 61.36 seconds

Consequently, e(t) becomes

e(t) = 1/6e^-0.652t + 0.833

e(t) = 1/(6e^-0.652(61.36)) + 0.833

e(t) = 0.83821342824942664566211

e(t) = 0.838 --- Rounded off

Thus, the thermometer reveals an error of 0.838°

In the scenario of a metal ingot cooling slowly, the microstructure tends to be coarse. The surface, exposed to higher temperatures for extended periods during cooling, features smaller grain sizes as they have less time to form. However, as we delve deeper into the ingot, the grains gradually extend, leading to equiaxed grain formation at the center.

Response:

The solution to this question is 1273885.3 ∅

Clarification:

The first step is to ascertain the required hydraulic flow rate liquid based on the working pressure if a cylinder with a piston diameter of 100 mm is utilized.

Given that,

The distance = 50mm

The time t =10 seconds

The force F = 10kN

The piston diameter = 100mm

The pressure = F/A

10 * 10^3/Δ/Δ

P = 1273885.3503 pa

Subsequently

Power = work/time = Force * distance /time

= 10 * 1000 * 0.050/10

which amounts to =50 watt

Power =∅ΔP

50 = 1273885.3 ∅