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Rzqust [24]
3 years ago
9

What is the major drawback in nanocrystalline alloys? a)- high brittleness b)-low hardness c)-rapid grain growth upon heating d)

-low thermal conductivity
Engineering
1 answer:
patriot [66]3 years ago
8 0

Answer: a) High brittleness

Explanation: Nano crystalline alloys are the alloys that have the packing of nano crystal that is the size of the crystal is very minute or small. The size of the nano crystalline alloys usually ranges around 100 nm. Due to such arrangement of the crystal the nano crystalline alloy becomes hard in nature and because of this property it becomes difficult to make them ductile. Therefore the major drawback in nano crystalline alloys are that it has high brittleness or hardness.

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Outdoors

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A 800-MW steam power plant, which is cooled by a nearby river, has a thermal efficiency of 40 percent. Determine the rate of hea
Arturiano [62]

Answer:

Rate of heat transfer to river=1200MW

So the actual amount of heat rejected ti the river will be less as there will some heat loss to surrounding and in pipes

Explanation:

In order to find the actual heat transfer rate is lower or higher than its value we will first find the rate of heat transfer to power plant:

Efficiency=\frac{work}{heat transfer to power plant}

Heat transfer=\frac{work}{Efficiency\\} \\\\Heat transfer=\frac{800}{0.40}\\\\Heat transfer=2000MW

From First law of thermodynamics:

Rate of heat transfer to river=heat transfer to power plant-work done

Rate of heat transfer to river=2000-800

Rate of heat transfer to river=1200MW

So the actual amount of heat rejected ti the river will be less as there will some heat loss to surrounding and in pipes.

4 0
3 years ago
Which task best fits the role of a planning engineer?
Nonamiya [84]

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D

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ensuring project end on time through carefully planning and organizing

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What i s the value of a capacitor with 250 V applied and has 500 pC of charge? (a) 200 uF (b) 0.5 pF (c) 500 uF (d) 2 pF
exis [7]

Answer:

(d) 2 pF

Explanation: the charge on capacitor is given by the expression

Q=CV

where Q=charge

           C=capacitance

           V=voltage across the plate of the capacitor

here we have given Q=500 pF, V=250 volt

using this formula C=\frac{Q}{V}

=500×10^{-12}×\frac{1}{250}

=2×10^{-12}

=2 pF

3 0
3 years ago
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