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3 years ago

Estimate the weight of 25 km of gas pipeline, DN = 610 mm and wall thickness = 12.7 mm.

Physics

1 answer:

balu736 [363]3 years ago

6 0

Answer:

Weight = 45880 kN

Explanation:

Gas pipeline material is steel.

Density of steel = 7850 kg/m³

Mass = Volume x Density

Volume = Area x Length

Outer diameter = 610 mm

Internal diameter = 610 - 2 x 12.7 = 584.6 mm

$\texttt{Area}=\frac{\pi}{4}(610^2-584.6^2)=23831.21mm^2=0.0238m^2$

Length = 25 km = 25000 m

Volume = 25000 x 0.0238 = 595.78 m³

Mass = 595.78 x 7850 = 4676875.12 kg

Weight = Mass x Acceleration due to gravity = 4676875.12 x 9.81 = 45880 kN

Weight = 45880 kN

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Answer:

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Explanation:

The expression for the calculation of work done is shown below as:

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Also,

Considering the ideal gas equation as:-

$PV=nRT$

where,

P is the pressure

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R is Gas constant having value = 8.314 J/ K mol

So,

$V=\frac{nRT}{P}$

Also, for change in volume at constant pressure, the above equation can be written as;-

$\Delta V=\frac{nR\times \Delta T}{P}$

So, putting in the expression of the work done, we get that:-

$w=P\times \frac{nR\times \Delta T}{P}=nR\times \Delta T$

Given, initial temperature = 28.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (28.0 + 273.15) K = 301.15 K

W=1770 J

n = 6 moles

So,

$1770\ J=6 moles\times 8.314\ J/ Kmol \times (T_2-301.15\ K)$

Thus,

$T_2=301.15\ K+\frac{1770}{6\times 8.314}\ K$

$T_2=336.63\ K$

The temperature in Celsius = 336.63-273.15 °C = 63.5 °C

The final temperature is:- 63.5 °C

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