Water moves through a constricted pipe in steady, ideal flow. At the lower point

Question

Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is P_1 = 2.00 Times 10^4 Pa, and the pipe diameter is 4.0 cm. At another point y = 0.35 m higher, the pressure is P_2 = 1.25 Times 10^4 Pa and the pipe diameter is 2.00 cm. Find the speed of flow in the lower section. Your response differs from the correct answer by more than 10%. Double check your calculations, m/s Find the speed of flow in the upper section. m/s Find the volume flow rate through the pipe. m^3/s

Explanation / Answer

Bernouli's equation   

  
  

   P1+ rho*g*h1 + 1/2 rho v1^2 =P2+ rho*g*h2 + 1/2 rho v2^2 --------------(1)

here given that the mass flow rate is same at each point so we can equate the mass flow rate at two point that is lower and upper regions

   mass flow rate = rho*A*v , A area of cross section= pi*r^2, v is speed

   R1²v1 = R2²v2 ==> V2 = (R1/R2)^2*V1 substituting v2 in eq (1)

       p1/ + gh1 + (1/2)v1^2 = p2/ + gh2 + (1/2)((R1/R2)^2*v1)^2

       (p1 - p2)/ + g(y1 - y2) = (1/2)(R1/R2)^4*v1^2 - v1^2)

                   = 1/2(v1^2)((R1/R2)^4 -1)
a)
       v1 = sqrt(2*(p1 - p2)/ + g(y1 - y2)/(R1/R2)^4 -1)
= sqrt(2((2*10^4-1.25*10^4)/1000 + 9.8(0-0.35))/((2*10^-2/1*10^-2)^4-1))
b)       V1= 0.7366 m/s--------------------> lower section

now v2 = (R1/R2)^2*V1 = (2*10^-2/1*10^-2)^2*0.7366 = 2.9464 m/s-----------------> upper section

c)

volume flow rate is = Av
A1v1 = A2v2
pi*r1^2*v1 = 3.14(2*10^-2)^2*0.7366 = 0.0009251696 m3 /s

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