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vodomira [7]
3 years ago
15

6. We want to make a temperature controller using the diode. Consider your results and suggest which parameter/characteristic of

the PN junction diode should be monitored and used for designing the controller. The chosen parameter/characteristic (could be more than one), that is most suitable as the controller input, is expected have most linear (or otherwise reliable) relationship with the temperature.
Engineering
1 answer:
ella [17]3 years ago
5 0

Answer:

The forward biased and reverse biased temperature coefficients of the diode can be used to design a temperature controller.

Explanation:

The forward and reverse biased temperature coefficients vary linearly with temperature. They almost linearly increase as the temperature increases hence, they can be used as the controller input when making the temperature controller using the diode.

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What is a heat engine? State its efficiency (a sketch would be appropriate).
dimulka [17.4K]

Answer:

\eta =\dfrac{Q_1-Q_2}{Q_1}

Explanation:

Heat engine

 Heat engine is a device which produce work while consuming some amount of energy .This engine works in cycle.In other words we can say that Heat engine is a device which covert thermal or chemical energy in to mechanical energy and after that this mechanical energy can be use for producing mechanical work.

Heat engine takes heat from high temperature and rejects heat to lower temperature and producing work.

We know that efficiency given as

\eta =\dfrac{out\ put}{in\ put}

From diagram we can say that

Q_1=Q_2+W

So efficiency

\eta =\dfrac{w}{Q_1}

\eta =\dfrac{Q_1-Q_2}{Q_1}

3 0
3 years ago
The following true stresses produce the corresponding true plastic strains for a brass alloy: True Stress (psi) True Strain 4840
Assoli18 [71]

Answer:

70,900

Explanation:

Given :

True stress (psi) _____ True strain (psi)

48400 ______________ 0.11

60400 ______________ 0.19

Using ratio simplification :

Let :

s = True stress ; t = true strain

s1 = 48400

s2 = 60400

t1 = 0.11

t2 = 0.19

True stress, s0 ; needed to produce a True plastic strain, tp = 0.26

(s0 - s1) / (s2 - s1) = (tp - t1) / (t2 - t1)

(s0 - 48400)/(60400 - 48400) = (0.26 - 0.11)/(0.19 - 0.11)

(s0 - 48400)/12000 = 0.15/0.08

Cross multiply :

0.08(s0 - 48400) = 0.15 * 12000

0.08s0 - 3872 = 1800

0.08s0 = 1800 + 3872

0.08s0 = 5672

s0 = 5762 / 0.08

s0 = 70,900

The true stress required to produce a true plastic strain of 0.26 is 70,900

5 0
3 years ago
Air is a....<br> O Solid<br> O Liquid<br> O Gas<br> O Plasma
Galina-37 [17]

Answer:

Air is a gas

Explanation:

i think. beavuse it cant be a liqued or a solid. i dont think a plasma. i would answer gas

6 0
3 years ago
Water at 60°F passes through 0.75-in-internal diameter copper tubes at a rate of 1.2 lbm/s. Determine the pumping power per ft
Lelu [443]

Answer:

The pumping power per ft of pipe length required to maintain this flow at the specified rate 0.370 Watts

Explanation:

See calculation attached.

- First obtain the properties of water at 60⁰F. Density of water, dynamic viscosity, roughness value of copper tubing.

- Calculate the cross-sectional flow area.

- Calculate the average velocity of water in the copper tubes.

- Calculate the frictional factor for the copper tubing for turbulent flow using Colebrook equation.

- Calculate the pressure drop in the copper tubes.

- Then finally calculate the power required for pumping.

8 0
3 years ago
Two blocks of rubber with a modulus of rigidity G =10 MPa are bonded to rigid supports and to a plate AB. Knowing that b = 200 m
g100num [7]

Answer:

Solution is given in the attachments.

5 0
4 years ago
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