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tensa zangetsu [6.8K]
4 years ago
10

A zener diode exhibits a constant voltage of 5.6 V for currents greater than five times the knee current. IZK is specified to be

1 mA. The zener is to be used in the design of a shunt regulator fed from a 15-V supply. The load current varies over the range of 0 mA to 15 mA. Find a suitable value for the resistor R. What is the maximum power dissipation of the zener diode?
Engineering
1 answer:
8090 [49]4 years ago
5 0

Answer:

The maximum power dissipation of the zener diode 112mV.

Explanation:

The minimum zener current should be:

5 * Iza= 5 * 1=  5 mA.

Since the load current can be at maximum 15 mA, we should select R so that, IL= 15 mA.

A zener current of 5 mA is available, Thus the current should be 20 mA, which leads to,

R = \frac{15 - 5.6}{20 mA} = 470 Ω.

Maximum power dissipated in the diode occours when, IL=0 is

Pmax = 20 * 10^{3} * 5.6 = 112mV.

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Determine the carburizing time necessary to achieve a carbon concentration of 0.30 wt% at a position 4 mm into an iron–carbon al
Ahat [919]

Answer:

the carburizing time necessary to achieve a carbon concentration is 31.657 hours

Explanation:

Given the data in the question;

To determine the carburizing time necessary to achieve the given carbon concentration, we will be using the following equation:

(Cs - Cx) / (Cs - C0) = ERF( x / 2√Dt)

where Cs is Concentration of carbon at surface = 0.90

Cx is Concentration of carbon at distance x = 0.30 ; x in this case is 4 mm = ( 0.004 m )

C0 is Initial concentration of carbon = 0.10

ERF() = Error function at the given value

D = Diffusion of Carbon into steel

t = Time necessary to achieve given carbon concentration ,

so

(Cs - Cx) / (Cs - C0) = (0.9 - 0.3) / (0.9 - 0.1)

= 0.6 / 0.8

= 0.75

now, ERF(z) = 0.75; using ERF table, we can say;

Z ~ 0.81; which means ( x / 2√Dt) = 0.81

Now, Using the table of diffusion data

D = 5.35 × 10⁻¹¹ m²/sec at (1100°C) or 1373 K

now we calculate the carbonizing time by using the following equation;

z = (x/2√Dt)

t is carbonizing time

so we we substitute in our values

0.81 = ( 0.004 / 2 × √5.35 × 10⁻¹¹ × √t)

0.81 = 0.004 / 1.4628 × 10⁻⁵ × √t

0.81 × 1.4628 × 10⁻⁵ × √t = 0.004

1.184868 × 10⁻⁵ × √t = 0.004  

√t = 0.004 / 1.184868 × 10⁻⁵

√t = 337.5903

t = ( 337.5903)²  

t = 113967.21 seconds

we convert to hours

t = 113967.21 / 3600

t = 31.657 hours

Therefore, the carburizing time necessary to achieve a carbon concentration is 31.657 hours

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At steady state, a reversible heat pump cycle discharges energy at the rate to a hot reservoir at temperature TH, while receivin
Doss [256]

Answer:

A) β_max = 20.64

B) TH = 68.25°C

C) TC = 54.27°C

Explanation:

A) We are given;

TH = 16°C = 16 + 273K = 289K

TC = 2°C = 2 + 273K = 275K

Formula for maximum cycle coefficient of performance is given as;

β_max = TH/(TH - TC)

β_max = 289/(288 - 275)

β_max = 20.64

B) We are given;

Heat rejected to system at hot reservoir; Q_H = 10.5 KW

Heat provided to system at cold reservoir; Q_C = 8.75 KW

Cold reservoir temperature; TC = 0°C = 0 + 273K = 273K

Formula for actual cycle COP is given as;

β_actual = Q_C/W_cycle

Where W_cycle is the work done and is given by;

W_cycle = Q_H - Q_C

W_cycle = 10.5 - 8.75 = 1.75 KW

Thus,

β_actual = 8.75/1.75

β_actual = 5

Actual cycle COP is defined as;

β_actual = TH/(TH - TC)

And we are looking for TH.

Thus,

TH = TC/(1 - (1/β_actual))

TH = 273/(1 - 1/5)

TH = 273/(4/5)

TH = 341.25K = 341.25 - 273°C = 68.25°C

C) We are given;

TH = 27°C = 27 + 273 = 300°C

β_max = 12

Thus, from,

β_max = TH/(TH - TC)

TC = TH(1 - (1/β_max))

TC = 300/(1 - 1/12)

TC = 327.27K = 327.27 - 273 °C = 54.27°C

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