A 6.4

cm
diameter horizontal pipe gradually narrows to 5.2
cm
. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 30.0
kPa
and 20.2
kPa
, respectively

To determine the difference in velocity between the two sections of the pipe, we can use the principle of continuity, which states that the mass flow rate of a fluid is constant at any point in a closed system. In a horizontal pipe, this can be expressed as:

A1 * v1 = A2 * v2

Where:
A1 = cross-sectional area of the larger section (6.4 cm diameter)
A2 = cross-sectional area of the narrower section (5.2 cm diameter)
v1 = velocity of water in the larger section
v2 = velocity of water in the narrower section

First, let's calculate the cross-sectional areas of the two sections:

A1 = π * (d1/2)^2
= π * (6.4 cm/2)^2
= 32.17 cm^2

A2 = π * (d2/2)^2
= π * (5.2 cm/2)^2
= 21.24 cm^2

Next, we can rearrange the equation and solve for v2:

v2 = (A1 * v1) / A2

Now, we need to convert the diameter and gauge pressures into SI units:

d1 = 6.4 cm = 0.064 m
d2 = 5.2 cm = 0.052 m
P1 = 30.0 kPa = 30,000 Pa
P2 = 20.2 kPa = 20,200 Pa

Now we can substitute the values into the equation and solve for v2:

v2 = (A1 * v1) / A2
= [(π * (0.064 m/2)^2) * v1] / (π * (0.052 m/2)^2)
= [(0.064 m)^2 * v1] / (0.052 m)^2
= (0.004096 m^2 * v1) / 0.002704 m^2
= 1.518 v1

So, the velocity of water in the narrower section is 1.518 times the velocity of water in the larger section.

Note: This calculation assumes that the flow is incompressible and there are no losses due to friction or other factors.