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2 years ago

Illustrate the crowbar protection for silicon controlled rectifier

Engineering

1 answer:

julsineya [31]2 years ago

4 0

Answer:

The SCR over voltage crowbar or protection circuit is connected between the output of the power supply and ground. ... It also clamps the gate voltage at ground potential until the Zener turns on. The capacitor C1 is present to ensure that short spikes to not trigger the circuit.

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A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric c

Paraphin [41]

Answer:

$10.8\ \text{lb/ft^2}$

$101.96\ \text{lb/ft}^2$

Explanation:

$v_1$ = Velocity of car = 65 mph = $65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}$

$\rho$ = Density of air = $0.00237\ \text{slug/ft}^3$

$v_2=0$

$P_1=0$

$h_1=h_2$

From Bernoulli's law we have

$P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}$

The maximum pressure on the girl's hand is $10.8\ \text{lb/ft^2}$

Now $v_1$ = 200 mph = $200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}$

$P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2$

The maximum pressure on the girl's hand is $101.96\ \text{lb/ft}^2$

5 0

3 years ago

2. A counter flow tube-shell heat exchanger is used to heat a cold water stream from 18 to 78oC at a flow rate of 1 kg/s. Heatin

Anastaziya [24]

Answer:

a) $L = 220\,m$ , b) $U_{o} \approx 0.63\,\frac{kW}{m^{2}\cdot ^{\textdegree}C}$

Explanation:

a) The counterflow heat exchanger is presented in the attachment. Given that cold water is an uncompressible fluid, specific heat does not vary significantly with changes on temperature. Let assume that cold water has the following specific heat:

$c_{p,c} = 4.186\,\frac{kJ}{kg\cdot ^{\textdegree}C}$

The effectiveness of the counterflow heat exchanger as a function of the capacity ratio and NTU is:

$\epsilon = \frac{1-e^{-NTU\cdot(1-c)}}{1-c\cdot e^{-NTU\cdot (1-c)}}$

The capacity ratio is:

$c = \frac{C_{min}}{C_{max}}$

$c = \frac{(1\,\frac{kg}{s} )\cdot(4.186\,\frac{kW}{kg^{\textdegree}C} )}{(1.8\,\frac{kg}{s} )\cdot(4.30\,\frac{kW}{kg^{\textdegree}C} )}$

$c = 0.541$

Heat exchangers with NTU greater than 3 have enormous heat transfer surfaces and are not justified economically. Let consider that $NTU = 2.5$ . The efectiveness of the heat exchanger is:

$\epsilon = \frac{1-e^{-(2.5)\cdot(1-0.541)}}{1-(2.5)\cdot e^{-(2.5)\cdot (1-0.541)}}$

$\epsilon \approx 0.824$

The real heat transfer rate is:

$\dot Q = \epsilon \cdot \dot Q_{max}$

$\dot Q = \epsilon \cdot C_{min}\cdot (T_{h,in}-T_{c,in})$

$\dot Q = (0.824)\cdot (4.186\,\frac{kW}{^{\textdegree}C} )\cdot (160^{\textdegree}C-18^{\textdegree}C)$

$\dot Q = 489.795\,kW$

The exit temperature of the hot fluid is:

$\dot Q = \dot m_{h}\cdot c_{p,h}\cdot (T_{h,in}-T_{h,out})$

$T_{h,out} = T_{h,in} - \frac{\dot Q}{\dot m_{h}\cdot c_{p,h}}$

$T_{h,out} = 160^{\textdegree}C + \frac{489.795\,kW}{(7.74\,\frac{kW}{^{\textdegree}C} )}$

$T_{h,out} = 96.719^{\textdegree}C$

The log mean temperature difference is determined herein:

$\Delta T_{lm} = \frac{(T_{h,in}-T_{c, out})-(T_{h,out}-T_{c,in})}{\ln\frac{T_{h,in}-T_{c, out}}{T_{h,out}-T_{c,in}} }$

$\Delta T_{lm} = \frac{(160^{\textdegree}C-78^{\textdegree}C)-(96.719^{\textdegree}C-18^{\textdegree}C)}{\ln\frac{160^{\textdegree}C-78^{\textdegree}C}{96.719^{\textdegree}C-18^{\textdegree}C} }$

$\Delta T_{lm} \approx 80.348^{\textdegree}C$

The heat transfer surface area is:

$A_{i} = \frac{\dot Q}{U_{i}\cdot \Delta T_{lm}}$

$A_{i} = \frac{489.795\,kW}{(0.63\,\frac{kW}{m^{2}\cdot ^{\textdegree}C} )\cdot(80.348^{\textdegree}C) }$

$A_{i} = 9.676\,m^{2}$

Length of a single pass counter flow heat exchanger is:

$L =\frac{A_{i}}{\pi\cdot D_{i}}$

$L = \frac{9.676\,m^{2}}{\pi\cdot (0.014\,m)}$

$L = 220\,m$

b) Given that tube wall is very thin, inner and outer heat transfer areas are similar and, consequently, the cold side heat transfer coefficient is approximately equal to the hot side heat transfer coefficient.

$U_{o} \approx 0.63\,\frac{kW}{m^{2}\cdot ^{\textdegree}C}$

5 0

3 years ago

LINKS GET BODIED ON SITE! RAWR

bulgar [2K]

Answer:

False

Explanation:

MRK ME BRAINLIEST PLZZZZZZZZZZZZZZZZZZ

7 0

3 years ago

Type the correct answer in the box. Spell the word correctly. What is the best option possible for Gary to create an effective p

tresset_1 [31]

Answer:

Computer Aided Design (CAD)

Explanation:

The computer-aided design, CAD, tools are design and technical documentation technology, that makes use of an automated drafting process of drafting, rather than a manual drafting process.

Computer-aided design programs include CAD programs with 2D and/or 3D capability such as AutoCAD LT, AutoCAD, and SmartDraw which are common construction documentation drafting application with which concepts and ideas can be created and visualized

Therefore, we have;

He would use Computer-Aided-Design (CAD) tools to create accurate and uniform technical drawings for his device.

8 0

3 years ago

If, for a particular junction, the acceptor concentration is 1017/cm3 and the donor concentration is 1016/cm3 , find the junctio

natima [27]

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

Download docx

8 0

3 years ago

Illustrate the crowbar protection for silicon controlled rectifier​

Illustrate the crowbar protection for silicon controlled rectifier