Ask question

Ask question

All categories

3 years ago

9

A voltage regulator is to provide a constant DC voltage Vl=10V to a load Rl from a nominal Vcc=15V supply voltage. The load can

vary from 20Ω to 1KΩ. The supply voltage Vcc can vary from 13V to 16V. The op-amp can provide a maximum output current of 20mA. a)Find the βnecessary for the transistor to provide the needed current. b)Find the maximum power the transistor must dissipate.

1 answer:

Luden [163]3 years ago

7 0

Answer:

Beta values can be from the equation=change in Vcc/nominal Vcc

Beta=16-3/15=3/15=1/5=0.20

Maximum power=I^2*R=40 W

You might be interested in

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

How do you use the brakes in an airplane?

Paraphin [41]

Answer:

When a pilot pushes the top of the right pedal, it activates the brakes on the right main wheel/wheels, and when the pilot pushes the top of the left rudder pedal, it activates the brake on the left main wheel/wheels. The brakes work in a rather simple way: they convert the kinetic energy of motion into heat energy.

3 0

3 years ago

Consider the velocity boundary layer profile for flow over u flat plate to be of the form u = C_1 + C_2 y. Applying appropriate

ra1l [238]

Answer:

The result in terms of the local Reynolds number ⇒ Re = [μ_∞ · x] / v

Explanation:

See below my full workings so you can compare the results with those obtained from the exact solution.

4 0

3 years ago

A well-insulated rigid vessel contains 3 kg of saturated liquid water at 40oC. The vessel also contains an electrical resistor t

user100 [1]

Answer:

The final temperature is 111.66°C

Explanation:

The given conditions :-

i) Well insulated means no heat loss.

ii) Rigid vessels means volume remains same.

iii) Initial temperature ( T₁ ) = 40°C. = 273 + 40 = 313 K

iv ) Mass of water in vessel = 3 kg.

v) current drawn by resistor ( i ) = 10 ampere.

vi) Voltage applied ( V ) = 50 volts.

vii) The time for which resistor operating ( t ) = 30 minute = 30 * 60 = 1800 seconds.

Now we have to calculate heat developed by resistor in vessel.

Q = V * i * t = 50 * 10 * 1800 = 900,000 J = 900 KJ.

Since it is a rigid container so the work done is zero.

Q = du ( du - change in internal energy)

Q = m * C * dT ( C = 4.186 KJ/KgK )

Q = 3 * 4.186 * (T₂ - T₁ )

900 = 12.558 * ( T₂ - 313 )

T₂ - 313 = 71.6674

T₂ = 384.6674 K

T = 384.6674 - 273 = 111.66°C

So the final temperature is 111.66°C.

3 0

3 years ago

By law, who is responsible for providing Safety Data Sheets?

nalin [4]

Answer:

OSHA Hazard Communication Standard is responsible.

Explanation:

5 0

3 years ago

Read 2 more answers