Suppose the tube in the previous problem is kept vertical with A upward but the other conditions remain the same. The separation between the cross sections at A and B is 15/16 cm. Repeat parts (a), (b) and (c) of the previous problem. Take g = 10 m/s².
Answers
See the Question given in that link, and then watch this solution.
https://brainly.in/question/8681778
Now, (a). We know volume rate of the flow does not depends upon the the height of the pipe, since pipe.
∴ Velocity of the fluid will remains unchanged.
∴ (a). At A,
Area × Velocity = 1 cm³/s.
4/100 cm² × Velocity = 1
∴ Velocity = 25 cm/seconds.
________________________
(b). At B,
Area × Velocity = 1 cm³/s.
2/100 × Velocity = 1
Velocity = 50 cm/seconds.
________________________
(c). Pressure difference, will be affected, this is because, there will be the gravitational work also which will be included.
Using the Bernoulli's equation,
Using the Bernoulli's Equation,
P₁ + h₁ρg + 1/2 ρv₁² = P₂ + h₂ρg + 1/2 ρv₂²
∴ P₁ - P₂ = 1/2 ρ(v₂² - v₁²) + ρg(h₂ - h₁)
∴ P₁ - P₂ = 1/2 × 1 × (50² - 25²) + 1 × 1000(-15/16)
[Note ⇒ Minus sign is used because h₁ - h₂ = 15/16 but h₂ - h₁ = -15/16.]
∴ P₁ - P₂ = 937.5 - 937.5
∴ P₁ - P₂ = 0
Hence, the pressure difference is zero.
Hope it helps.
The speed at A =25 cm/s
The speed at B =50 cm/s
The Bernoulli theorem = zero
Explanation:
As the discharge is the same through the tube
The speeds at A and B would be the same, i.e. 1 cc / s.
(a) The speed at A,
v =1/0.04 cm/s =100/4 cm/s =25 cm/s
(b) The speed at B,
v' =1/0.02 cm/s =100/2 cm/s =50 cm/s
(c) The Bernoulli theorem
Pₐ+ρgh+1/2 ρv² = Pᵦ+1/2 ρv'²
Taking point B for height as a reference level.
Given that
h=15/16 cm
Pₐ-Pᵦ =1/2 ρ(v'²-v²) -ρgh
=1/2×1000×(0.50²-0.25²) -1000×10×15/1600 N/m²
=1/2×1000×0.75×0.25 - 1500/16 N/m²
=93.75 - 93.75 N/m²
= zero