What rate does water flow through the pipes?

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a horizontal pipe 10.0 cm in diameter has a smooth reduction to a pipe 5cm in diameter. if the pressure of the water in the larger pipe is 8x10^4 Pa and the pressure in the smaller pipe is 6x10^4 Pa, at what rate does the water flow?

ive been able to get the answer of 6.53 m/s i dont know if this is right, but the answer is 12.8 kg/s, so i mite have the wrong answer, but im not sure how to convert the units??

What rate does water flow through the pipes?

This is the rate of water flow in the pipe 5 cm in diameter.

We can change unit

The horizontal section of pipe:

S = pi x (d/2)^2 = pi x (5/2)^2 = 19.6 cm^2 = 0.0019 m^2

Thus

the rate

6.53 m/s x S x Density of water

6.53 m/s x 0.0019 m^2 x 1000 kg/m^3 = 12.4 kg/s

What rate does water flow through the pipes?

d1 = 10 cm = 10*10^-2 m

d2 = 5 cm = 5*10^-2 m

p1 = 8*10^4 Pa =8*10^4 N/m2

p2 = 6*10^4 Pa =6*10^4 N/m2

Continuity equation:

V' = S*v = const. ; V'1=V'2=V'

V'=dV/dt=Volume flow

S=surface

v=velocity

(Pi/4)d1^2 *v1 = (Pi/4)d2^2 * v2

d=diameter

Pi=3.14...

---> v1 = [v2*(Pi/4)d2^2]\(Pi/4)d1^2 = v2*(d2^2/d1^2) (I)

Bernoulli equation:

(rho) g h + p + 1/2[(rho)*v^2]/2 = const.

(rho)1 g h1 + p1 + 1/2[(rho)1*v1^2]/2 = (rho)2 g h2 + p2 + 1/2[(rho)2*v2^2]/2

(rho)=density in kg/m^3 = const. --->(rho)1=(rho)2=(rho)=1000kg/m^3

g=9.81 m/s^2

h=height in m

p=pressure in N/m^2

v=velocity in m/s

horizontal pipe --->h1=h2=h

p1 + 1/2[(rho)1*v1^2]/2 = p2 + 1/2[(rho)2*v2^2]/2

Use the substitution (I) from the continuity equation for Bernoulli:

p1-p2 = delta p = v2^2[1/2(rho)*(d2/d1)^4 -1]

v2^2 = delta p/[1/2(rho)*(d2/d1)^4 -1]

v2 = Sqrt{delta p/[1/2(rho)*(d2/d1)^4 -1]} = 6.53 m/s

m = rho*V

m' = rho*V'=rho*S*v

m'= 1000kg/m^3 *(Pi/4)*(5*10^-2 m)^2 *6.53 m/s

m'=12.83 kg/s



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