Answer:
the pressure after contraction is 2×10^5 Pa
the speed after contraction is 15m/s
Explanation:
We were given Pressure P to be 3.5 x 10^5 that is Flowing with speed of 5.0 m/s,
For us to calculate pressure we need to calculate the area first as;
Let initial Area = A₁
And Final area A₂
We were told that in a horizontal pipe it contracts to 1/3 its former area. Which means
A₂= A₁/3.................
V₁ is the speed
the pressure and speed of the water after the contraction can be calculated using equation of continuity below
A₂V₂ = A₁V₁
But
If we substitute given value in the expresion we have
V₂ = (3A *5)/A
V₂ = 15m/s
Therefore, the speed after contraction is 15m/s
Now we can calculate the pressure using
Bernoulli's equation
p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂
But we know that the pipe is horizontal, then "h" terms cancel out then
p₁ + ½ρv₁² = p₂ + ½ρv₂²
Making P₂ subject of formula we have
p₂ = 0.5ρ( V ₁² - v₂² ) + P₁
P₂=. 0.5 × 1000 (5² -15² ) + 3*10^5
=2×10^5 Pa
Therefore, the pressure after contraction is 2×10^5 Pa
(a) the final speed of the water after contraction is 15 m/s.
(b) The final pressure of the water after contraction is 2.5 x 10⁵ Pa.
The given parameters;
Let the initial area of the pipe = A₁
Apply the continuity equation to determine the final speed of the water after contraction as follows;
[tex]A_1 V_1 = A_2 V_2\\\\V_2 = \frac{A_1V_1}{A_2} \\\\V_2 = \frac{A_1 \times 5}{\frac{1}{3} A_1 } \\\\V_2 = 15 \ m/s[/tex]
The final pressure of the water after contraction is determined by applying Bernoulli's equation for horizontal pipe;
[tex]P_1 + \frac{1}{2} \rho V_1^2= P_2 + \frac{1}{2} \rho V_2^2\\\\P_2 = \frac{1}{2} \rho (V_1^2 - V_2^2) + P_1\\\\P_2 = \frac{1}{2} \times 1000(5^2 - 15^2) + 3.5 \times 10^5\\\\P_2 = 2.5 \times 10^5 \ Pa[/tex]
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