Respuesta :
Answer:
The rate of flow of water is 71.28 kg/s
Solution:
As per the question:
Diameter, d = 18.0 cm
Diameter, d' = 9.0 cm
Pressure in larger pipe, P = [tex]9.40\times 10^{4}\ Pa[/tex]
Pressure in the smaller pipe, P' = [tex]2.80\times 10^{4}\ Pa[/tex]
Now,
To calculate the rate of flow of water:
We know that:
Av = A'v'
where
A = Cross sectional area of larger pipe
A' = Cross sectional area of larger pipe
v = velocity of water in larger pipe
v' = velocity of water in larger pipe
Thus
[tex]\pi \frac{d^{2}}{4}v = \pi \frac{d'^{2}}{4}v'[/tex]
[tex]18^{2}v = 9^v'[/tex]
v' = 4v
Now,
By using Bernoulli's eqn:
[tex]P + \frac{1}{2}\rho v^{2} + \rho gh = P' + \frac{1}{2}\rho v'^{2} + \rho gh'[/tex]
where
h = h'
[tex]\rho = 10^{3}\ kg/m^{3}[/tex]
[tex]9.40\times 10^{4} + \frac{1}{2}\rho v^{2} = 2.80\times 10^{4} + \frac{1}{2}\rho (4v)^{2}[/tex]
[tex]6.6\times 10^{4} = \frac{1}{2}\rho 15v^{2}[/tex]
[tex]v = \sqrt{\frac{2\times 6.6\times 10^{4}}{15\times 10^{3}}} = 8.8\ m/s[/tex]
Now, the rate of flow is given by:
[tex]\frac{dm}{dt} = \frac{d}{dt}\rho Al = \rho Av[/tex]
[tex]\frac{dm}{dt} = 10^{3}\times \pi (\frac{18}{2}\times 10^{- 2})^{2}\times 8.8 = 71.28\ kg/s[/tex]
Answer:
The rate of flow is 75.4 kg/s.
Explanation:
Given that,
Diameter of pipe = 18.0 cm
Diameter = 9.00 cm
Pressure [tex]P= 9.40 \times10^{4}\ Pa[/tex]
Pressure in the smaller pipe [tex]P'=2.80\times10^{4}\ Pa[/tex]
We need to calculate the velocity of longer pipe
Using formula of velocity
[tex]A_{1}v_{1}=A_{2}v_{2}[/tex]
[tex]\pi\times r_{1}^2\times v_{1}=\pi\times r_{2}^2\times v_{2}[/tex]
Put the value into the formula
[tex](9.00)^2\times v_{1}=(4.5)^2\times v_{2}^2[/tex]
[tex]v_{1}=\dfrac{20.25}{81}\times v_{2}[/tex]
[tex]v_{1}=0.25 v_{2}[/tex]....(I)
We need to calculate the velocity of smaller pipe
Using Bernoulli's equation
[tex]P_{1}+\dfrac{1}{2}\rho\times v^2+\rho g h=P_{2}+\dfrac{1}{2}\rho\times v^2+\rho g h[/tex]
[tex]P_{1}-P_{2}=\dfrac{1}{2}\rho(v_{2}^2-v_{1}^2)[/tex]
Put the value into the formula
[tex]9.40 \times10^{4}-2.80\times10^{4}=\dfrac{1}{2}\times1000(v_{2}^2-(0.25v_{2})^2)[/tex]
[tex]6.6\times10^{4}=\dfrac{1}{2}\times1000\times(0.9375)v_{2}^2[/tex]
[tex]v_{2}^2=\dfrac{2\times6.6\times10^{4}}{0.9375\times1000}[/tex]
[tex]v_{2}=\sqrt{\dfrac{2\times6.6\times10^{4}}{0.9375\times1000}}[/tex]
[tex]v_{2}=11.86\ m/s[/tex]
Put the value of v₂ in the equation (I)
[tex]v_{1}=0.25\times11.86[/tex]
[tex]v_{1}=2.965\ m/s[/tex]
We need to calculate the flow
Using formula of flow rate
[tex]Q=A_{2}\times v_{2}[/tex]
Where, Q = flow rate
A = area
v = velocity
Put the value into the formula
[tex]Q=\pi\times(4.5\times10^{-2})^2\times11.86[/tex]
[tex]Q=0.0754\ m^3/s[/tex]
We need to calculate the rate of flow
Using formula of rate
[tex]m=Q\times\rho[/tex]
Put the value into the formula
[tex]m=0.0754\times1000[/tex]
[tex]m=75.4\ kg/s[/tex]
Hence, The rate of flow is 75.4 kg/s.
