# Homework Help: One more

1. Apr 1, 2005

### Jayhawk1

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?

2. Apr 1, 2005

### xanthym

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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3. Apr 1, 2005

### Jayhawk1

...quick question

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

4. Apr 1, 2005

### Staff: Mentor

What's the area of a circle in terms of its diameter?