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A non-viscous liquid of constant density 1000 kg /m 3 flows in a streamline motion along a tube of variable cross section. The tube is kept inclined in the vertical plane as shown in Figure. The area of cross section of the tube two points P and Q at height of 2 metres and 5 metres are respectively 4 x 10 -3 m 2 and 8 x 10 -3 m 2 . The velocity of the liquid at point P is 1 m/s. Find the work done per unit volume by the pressure and the gravity forces as the fluid flows from point P to Q

A non-viscous liquid of constant density 1000 kg /m3 flows in a streamline motion along a tube of variable cross section. The tube is kept inclined in the vertical plane as shown in Figure. The area of cross section of the tube two points P and Q at height of 2 metres and 5 metres are respectively 4 x 10-3 m2 and 8 x 10-3 m2. The velocity of the liquid at point P is 1 m/s. Find the work done per unit volume by the pressure and the gravity forces as the fluid flows from point P to Q

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Grade:11

1 Answers

Kevin Nash
askIITians Faculty 332 Points
9 years ago
Hello Student,
Please find the answer to your question
Given that
ρ = 1000 kg/m3, h1 = 2m, h2 = 5 m
A­1 = 4 x 10-3 m2, A2 = 8 x 10-3 m2 , v1 = 1 m/s
Equation of continuity
A1 v1 = A2v2 ∴ v2 = A1 v1 / A2 = 0.5 m/s
According to Bernoulli’s theorem,
(p1 – ­p1) = ρg (h2 – h­1) – ½ ρ (v22) – v21
Where (p1­ – p2) = work done /vol. [by the pressure ]
ρg (h2 – h1) = work don/vol. [by gravity forces]
Now, work done / vol. by gravity forces
= ρg (h2 – h1) = 103 x 9.8 x 3 = 29.4 x 103 J/m3
And ½ ρ (v22 – v21 ) = ½ x 103 [1/4 – 1 ] = - 3/8 x 103 J/m3
= - 0.375 x 103 J/m3
∴ Work done / vol. by pressure
= 29.4 x 103 – 0.375 x 103 J / m3 = 29.025 x 103 j/m3
Thanks
Kevin Nash
askIITians Faculty

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