Fluid Mechanics: Solving 8 Min Water Flow Problem

AI Thread Summary
Water flows through a horizontal pipe with varying diameters, delivering water at a speed of 15.9 m/s. The volume of water delivered in an 8-minute period is calculated to be approximately 3.746 m³. The flow speed in the left section of the pipe is determined to be about 1.3754 m/s. The gauge pressure in the left section is calculated to be 125,459 N/m². The area of a circle is derived using the formula A = π/4 * d², where d is the diameter.
Jayhawk1
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This is the last problem with fluid mechanics and I have no clue about this one... any ideas?

Water flows through a horizontal pipe and is delivered into the atmosphere at a speed of v1=15.9 m/s. The diameters of the left and right sections of the pipe are 8.5 cm and 2.5 cm, respectively. (a) What volume of water is delivered into the atmosphere during a 8 min period? (b) What is the flow speed of the water in the left section of the pipe? (c) What is the gauge pressure in the left section of the pipe?
 
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Jayhawk1 said:
This is the last problem with fluid mechanics and I have no clue about this one... any ideas?

Water flows through a horizontal pipe and is delivered into the atmosphere at a speed of v1=15.9 m/s. The diameters of the left and right sections of the pipe are 8.5 cm and 2.5 cm, respectively. (a) What volume of water is delivered into the atmosphere during a 8 min period? (b) What is the flow speed of the water in the left section of the pipe? (c) What is the gauge pressure in the left section of the pipe?
From problem statement:
(Water flows left to right.)
{Diameter Left Section} = (8.5 cm) = (8.5e(-2) m)
{Area Left Section} = AL = (π/4)*(8.5e(-2) m)^2 = (5.6745e(-3) m^2)
{Velocity Left Section} = vL
{Diameter Right Section} = (2.5 cm) = (2.5e(-2) m)
{Area Right Section} = AR = (π/4)*(2.5e(-2) m)^2 = (4.9087e(-4) m^2)
{Velocity Right Section} = vR = (15.9 m/s)

(a):
{Volume in (8 Min)} = {Volume in (480 sec)} =
= (480 sec)*vR*AR =
= (480 sec)*(15.9 m/s)*(4.9087e(-4) m^2) =
= (3.746 m^3)

(b):
ρ*AL*vL = ρ*AR*vR
::: ⇒ vL = ρ*AR*vR/{ρ*AL} =
= AR*vR/{AL} =
= (4.9087e(-4) m^2)*(15.9 m/s)/{(5.6745e(-3) m^2)} =
= (1.3754 m/sec)

(c):
PL + (1/2)*ρ(vL)2 + ρ*g*h = PR + (1/2)*ρ(vR)2 + ρ*g*h
::: ⇒ {Gauge Pressure Left Section} = PL - PR =
= (1/2)*ρ(vR)2 - (1/2)*ρ(vL)2 =
= (1/2)*ρ*{(vR)2 - (vL)2} =
= (1/2)*(1000 kg/m^3)*{(15.9 m/s)^2 - (1.3754 m/sec)^2}
= (125,459 N/m^2)


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...quick question

why is pi divided by four to get the area of the pipe?
 
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