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3 years ago

A semiconductor is a material that can carry an electric current between a ______and an insulator

Engineering

1 answer:

Solnce55 [7]3 years ago

4 0

Answer:

semiconductor, any of a class of crystalline solids intermediate in electrical conductivity between a conductor and an insulator. Semiconductors are employed in the manufacture of various kinds of electronic devices, including diodes, transistors, and integrated circuits.

Explanation:

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The following is a list of metals and alloys:

viktelen [127]

Answer:

A) Gray cast iron

B) Aluminum

C) Titanium alloy

D) Tool steel

E) Titanium alloy

F) magnesium

G) Tungsten

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3 years ago

List two common units of measurement to describe height

ladessa [460]

Answer:

Explanation:

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3 years ago

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A gas turbine has a thermal efficiency of 20% and develops a power output of 8 MW. Determine the fuel consumption rate if heatin

AlexFokin [52]

Answer:

0.8 kilograms of fuel are consumed each second.

Explanation:

As turbines are steady-state devices, the thermal efficiency of a turbine is equal to the percentage of the ratio of the output power to fluid power, that is:

$\eta_{th} = \frac{\dot W}{\dot E} \times 100\,\%$

The fluid power is:

$\dot E = \frac{\dot W}{\eta} \times 100\,\%$

$\dot E = \frac{8\,MW}{20\,\%}\times 100\,\%$

$\dot E = 40\,MW$

Which means that gas turbine consumes 40 megajoules of fluid energy each second, which is heated and pressurized with help of the fuel, whose amount of consumption per second is:

$50\,\frac{MJ}{kg} = \frac{40\,MJ}{m_{fuel}}$

$m_{fuel} = 0.8\,kg$

0.8 kilograms of fuel are consumed each second.

3 0

3 years ago

A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequa

taurus [48]

Answer:

Explanation:

Given

$T_h=250^{\circ}C\approx 523\ K$

$T_L=30^{\circ}C\approx 303\ K$

$Q_1=6 kW$

From Clausius inequality

$\oint \frac{dQ}{T}=0$ =Reversible cycle

$\oint \frac{dQ}{T}$ =Irreversible cycle

$\oint \frac{dQ}{T}>0$ =Impossible

(a)For $P_{out}=3 kW$

Rejected heat $Q_2=6-3=3\ kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{3}{303}=1.57\times 10^{-3} kW/K$

thus it is Impossible cycle

(b) $P_{out}=2 kW$

$Q_2=6-2=4 kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{4}{303}=-1.73\times 10^{-3} kW/K$

Possible

(c)Carnot cycle

$\frac{Q_2}{Q_1}=\frac{T_1}{T_2}$

$Q_2=3.47\ kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{3.47}{303}=0$

and maximum Work is obtained for reversible cycle when operate between same temperature limits

$P_{out}=Q_1-Q_2=6-3.47=2.53\ kW$

Thus it is possible

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4 years ago

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