water is flowing through a horizontal pipe of non uniform cross section the speed of water is 30 cm/s where pressure is 10 . Calculate the speed of water at the speed of water at the place where pressure half the first .
Answers
Answer:
Correct Question:
Water is flowing through a horizontal pipe of non-uniform cross-section. the speed of water is 30 cm/s at a place where pressure is 10 cm (of water). calculate the speed of water at the other place where the pressure is half of that at the first place?
Answer:
Speed of the Water (v₂) = 1.04 m/s.
Given:
Velocity at one End (v₁) = 30 cm/s.
Height where pressure (h₁) = 10 cm.
Height half the Pressure (h₂) =5 cm
Note:-
We took h₂ as 5 cm as Question states to find the Velocity where the pressure is half of that at the first place.
Explanation:
This is an Application of Bernoullis theorem.
From Bernoullis theorem we Know,
But Question says that the Pipe is Horizontal. Therefore the Height (h) will be Zero.
Therefore, Equation becomes,
We Know, P = hρg
Now,
Substituting the values,
∵ [g = 10 m/s² = 1000 cm/s²]
We know, [1cm = 10⁻²m]
∴ Velocity will be 1.04 m/s.
Answer:
Correct question:-
Water is flowing through a horizontal pipe of non-uniform cross-section. the speed of water is 30 cm/s at a place where pressure is 10 cm (of water). calculate the speed of water at the other place where the pressure is half of that at the first place?
Answer:
Speed of the Water (v₂) = 1.04 m/s.
Given:
Velocity at one End (v₁) = 30 cm/s.
Height where pressure (h₁) = 10 cm.
Height half the Pressure (h₂) =5 cm
Note:-
We took h₂ as 5 cm as Question states to find the Velocity where the pressure is half of that at the first place.
Explanation:
This is an Application of Bernoullis theorem.
From Bernoullis theorem we Know,
But Question says that the Pipe is Horizontal. Therefore the Height (h) will be Zero.
Therefore, Equation becomes,
We Know, P = hρg
Now,
Substituting the values,
∵ [g = 10 m/s² = 1000 cm/s²]
We know, [1cm = 10⁻²m]
∴ Velocity will be 1.04 m/s.