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Question: Water is flowing through a horizontal tube \( 4 \) km in \( 4 \) cm in radius at a rate of \( 20{\te...

Water is flowing through a horizontal tube 44 km in 44 cm in radius at a rate of 20 litre s−120{\text{ litre }}{{\text{s}}^{ - 1}} . Calculate the pressure required to maintain the flow in terms of the height of mercury column?

Explanation

Solution

First of all note down all the given measures and convert all the units in the same system of units. Then find the correlation between the known and the unknown terms and place the values and simplify for the required value.

Complete step by step solution:
Convert the litre unit in meters.
Volume, v is 20litre=0.02m320litre = 0.02{m^3} …. (A)
Also, given that radius, r =4cm=0.04m= 4cm = 0.04m
Now, the area of the tube =πr2= \pi {r^2}
Place the given data in the above expression.
Area of the tube =(3.14)×(0.04)2= (3.14) \times {(0.04)^2}
Simplify the above equation-
Area of the tube =0.005024m2= 0.005024{m^2} …. (B)
Velocity of water can be defined as the ratio of volume per unit area.
Velocity of water =volumeArea= \dfrac{{volume}}{{Area}}
Place the values using equations (A) and (B)
Velocity of water =0.020.005024= \dfrac{{0.02}}{{0.005024}}
Simplify the above expression
Velocity of water =3.98m/s= 3.98m/s ….. (C)
Now, the Pressure P=12ρv2P = \dfrac{1}{2}\rho {v^2}
Where, the density of water is ρ=1000kg/m3\rho = 1000kg/{m^3}
Pressure, P=12(1000)(3.98)2P = \dfrac{1}{2}(1000){(3.98)^2}
Simplify the above expression-
Pressure, P=7920PaP = 7920Pa
Now, 1 atm = 101,300 Pa (760 mm of Hg)1{\text{ atm = 101,300 Pa (760 mm of Hg)}}
Pressure, P=7920101300×760P = \dfrac{{7920}}{{101300}} \times 760
Simplify the above expression –
Pressure, P=60mm HgP = 60mm{\text{ Hg}}
This is the required solution.

Note :
Be very careful while placing the values in the equation. All the units should be in the same format either MKS (meter kilogram system) or the CGS (cenit-meter gram second) system. Remember the correct formula or the conversion relation among the units to solve these types of solutions. Rest goes well, just substitution and simplification. Also, refer to the types of systems of units to know the correlation among the terms. Know the basic conversational relation and apply accordingly as per the requirement. Since one kilogram and one gram differs a lot.