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faust18 [17]
3 years ago
13

Prebions Now that you are about to complete this module, I'm sure you

Engineering
1 answer:
kvasek [131]3 years ago
7 0

Explanation:

These are probably the most used tool in any Plumber’s tool box. Pliers are not just another tool for a Plumber, they become an extension of their arms. Most people think that sounds odd, but pliers are more than just a tool to grab or turn things.

These are probably the most used tool in any Plumber’s tool box. Pliers are not just another tool for a Plumber, they become an extension of their arms. Most people think that sounds odd, but pliers are more than just a tool to grab or turn things.Sometimes a piece of copper pipe won’t quite go into a fitting. By using the handle end as a mallet you can gently force it in without damaging/denting the pipe or fittings. Or, when a brute force is needed the jaw end becomes a hammer. On an old pair of pliers I took a grinder to form one side of the handle into a flathead screwdriver/pry bar.

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Generally the primary source of all water supply is to be said as
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b) River

the primary source of all water supply is to be said as River

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A. Briefly describe the microstructural difference between spheroidite and tempered martensite. Explain why tempered martensite
masha68 [24]

Answer:

Answered below.

Explanation:

A) Both spheroidite & tempered martensite possess sphere - like cementite particles within their microstructure known as a ferrite matrix. However, the difference is that these particles are much larger for spheroidite than tempered.

B) Tempered martensite is much harder and stronger than spheroidite primarily because there is much more ferrite - cementite phase boundary area for its sphere - like cementite particles.

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4 0
3 years ago
A heat pump receives heat from a lake that has an average wintertime temperature of 6o C and supplies heat into a house having a
Dafna1 [17]

Answer:

a) \dot W = 1.062\,kW

Explanation:

a) Let consider that heat pump is reversible, so that the Coefficient of Performance is:

COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}

COP_{HP} = \frac{298.15\,K}{298.15\,K-279.15\,K}

COP_{HP} = 15.692

The minimum heat received by the house must be equal to the heat lost to keep the average temperature constant. Hence:

\dot Q_{H} = 60000\,\frac{kJ}{h}

The minimum power supplied to the heat pump is:

\dot W = \frac{\dot Q_{H}}{COP}

\dot W = \frac{\left(60000\,\frac{kJ}{h}  \right)\cdot \left(\frac{1\,h}{3600\,s}  \right)}{15.692}

\dot W = 1.062\,kW

5 0
3 years ago
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