What rate does water flow through the pipes?

  • #1
138
0

Homework Statement



A horizontal pipe 10.9 cm in diameter has a smooth reduction to a pipe 4.50 cm in diameter. If the pressure of the water (density = 1000 kg/m3) in the larger pipe is 7.50×104 Pa and the pressure in the smaller pipe is 5.70×104 Pa, at what rate (as mass per unit time) does water flow through the pipes?

Homework Equations



Bernoulli's equation. Po+1/2(rho)v^2+(rho)gh= pfinal +1/2(rho)vfinal^2+(rho)(g)(h final)

The Attempt at a Solution


if i sub in the necessary parts we know that h will be the same so that becomes a constant. Po is 7.50×104 Pa and Pfinal is 5.70×104 Pa so if we simploify the above equation we get P1-P2 = ½[tex]\rho[/tex]v22-½[tex]\rho[/tex]v12.so i solved for [tex]\Delta[/tex]v by [tex]\sqrt{\frac{P1-P2}{0.5*\rho}}[/tex] i got an answer of 6 but am unsure of the units to put in the answer..if this is actually the right working..pls. help
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
8
The rate of flow of water can be written as
m = Avρ
m^2 = A^2*v^2*ρ^2
1/2*m^2/A^2*ρ = 1/2*ρ*v^2
Substitute this value in the equation to find m.
 
  • #3
138
0
i'm a little confused what value am i substituting in the equation to find m??and if i use the equation 1/2*m^2/A^2*ρ = 1/2*ρ*v^2

(m2/A2)*(1000)= (1000)v2

would my A= sum of both pipes diameter ....and what would v equal???l
 
  • #4
rl.bhat
Homework Helper
4,433
8
P1 - P2 =1/2*ρ*v2^2 - 1/2*ρ*v1^2 = 1/2*m^2/ρ*(1/A2^2 - 1/A1^2)
 
  • #5
138
0
P1 - P2 = 1/2*m^2/ρ*(1/A2^2 - 1/A1^2)
delta p =18000
A1= 9.331x10^-3
A2= 1.5904 x10^-3

i sub it in and get a value of 137030.15 ..does that look right??
 
  • #6
rl.bhat
Homework Helper
4,433
8
I am getting 68644 m^3/s
 

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