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3 years ago

Initially, a pump pressure of __________ pounds per square inch should be used to maintain a sprinkler or standpipe system.

Engineering

1 answer:

jasenka [17]3 years ago

6 0

Answer:

of 150 pounds per square inch

Explanation:

Note that the unit for measuring water pressure is called pounds per square inch (psi)

In the case of sprinklers and standpipe systems, a pressure of 150 pounds per square inch was used initially.

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A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and gauge pressure of 300 kPa. The gas is h

Sergio [31]

Answer:

the final temperature is 77.1 °C

Explanation:

Given the data in the question;

Initial temperature; T₁ = 27°C = ( 27 + 273)K = 300 K

Initial absolute pressure P₁ = 300 kPa = ( 300 + 101.325 )kPa = 401.325 kPa

Final absolute pressure P₂ = 367 kPa = ( 367 + 101.325 )kPa = 468.325 kPa

Now, to calculate the final temperature, we use the ideal gas equation;

P₁V/T₁ = P₂V/T₂

but it is mentioned that the rigid tank is closed,

so the volume is the same both before and after.

Change in volume = 0

hence;

P₁/T₁ = P₂/T₂

we substitute

401.325 kPa / 300 K = 468.325 kPa / T₂

T₂ × 401.325 kPa = 300 K × 468.325 kPa

T₂ = [ 300 K × 468.325 kPa ] / 401.325 kPa

T₂ = 140497.5 K / 401.325

T₂ = 350.08 K

T₂ = ( 350.08 - 273 ) °C

T₂ = 77.1 °C

Therefore, the final temperature is 77.1 °C

3 0

3 years ago

Workers who work with what kind of chemicals chemicals may require regular medical checkups on a more frequent basis as a result

vazorg [7]

Workers who work with TOXIC chemicals may require regular medical checkups on a more frequent basis as a result of contact.

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2 years ago

The volume at a section of a 2-lane highway is 1800 vph in each direction and the density is approximately 30 bpm. A slow moving

katrin [286]

Answer:

Idk

Explanation:

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3 years ago

A geothermal heat pump absorbs 15 KJ/s of heat from the Earth 15 m below a house. This heat pump uses a 7.45 kJ/s compressor.

Oxana [17]

Answer:

COP of heat pump=3.013

COP of cycle=1.124

Explanation

W = Q2 - Q1 ----- equation 1

W = work done

Q2 = final energy

Q1 = initial energy

A) calculate the COP of the heat pump

COP =Q2/W

from equation 1

Q2 = Q1 + W = 15 + 7.45 = 22.45 KW

therefore COP =22.45/7.45 = 3.013

B) COP when cycle is reversed

COP = Q1/W

from equation 1

Q1 + W = Q2 ------ equation 2

Q2 = 15 Btu/s = 15 * 1.055 = 15.825 KW therefore from equation 2

Q1 = 8.375 KW

COP =8.375/7.45 = 1.124

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Answer: skip what

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