Water flowing at 0.040 m3/s in a 0.12 m dia pipe encounters asudden contraction to a 0.06 m diameter, as illustrated. Determinethe pressure drop across the contraction. How much of thepressure difference is due to frictional losses and how much is dueto changes in kinetic energy?

Now, having gotten that; lets find out the corresponding velocity [tex](V_1)[/tex] of the water at point (1) of the pipe and velocity [tex](V_2)[/tex] of the water at point 2 using the derived formula.

For velocity [tex](V_1)[/tex] :

[tex]V_1= \frac{4Q_1}{\pi D_1^2 }[/tex]

From, the question; we are given that:

water flow rate at point 1 [tex](Q_1)[/tex] = [tex]0.040m^3/s[/tex]

Diameter of the pipe at point 1 [tex](D_1)[/tex] = 0.12 m

∴ [tex]V_1= \frac{4(0.04m^3/s)}{\pi (0.12m/s)^2 }[/tex]

[tex]V_1=3.5367 m/s[/tex]

For velocity [tex](V_2)[/tex]:

[tex]V_2= \frac{4Q_2}{\pi D_2^2 }[/tex]

water flow rate at point [tex](Q_2)[/tex] = [tex]0.040m^3/s[/tex]

Diameter of the pipe at point 2 [tex](D_2)[/tex] = 0.06 m

[tex]V_2= \frac{4(0.04m^3/s)}{\pi (0.06m/s)^2 }[/tex]

[tex]V_2=14.1471 m/s[/tex]

Similarly, since we have found out our veocity; lets find the proportion of the area used in both points. So proportion of [tex](\frac{A_2}{A_1})[/tex] can be find by replacing [tex](\frac{\pi }{4}D_2^2)[/tex] for [tex]A_2[/tex] and [tex](\frac{\pi }{4}D_1^2)[/tex] for [tex]A_1[/tex].

So: [tex]\frac{A_2}{A_1} = \frac{\frac{\pi }{4}D^2_2 }{\frac{\pi }{4}D^2_1 }[/tex]

[tex]\frac{A_2}{A_1} = \frac{D_2^2}{D_1^2}[/tex]

[tex]\frac{A_2}{A_1} = \frac{(0.06m)^2}{(0.12m)^2}[/tex]

= 0.25

However, let's proceed to the phase where we determine the pressure drop across the contraction Δp by using the expression.

Δp = [tex][\frac{p_{water}}{2} (V_2^2-V_1^2)+ \frac{p_{water}}{2} K_LV_2^2][/tex]

where;

[tex]K_L[/tex] = standard frictional loss coefficient for a sudden contraction which is 0.4

[tex]p_{water[/tex] = density of the water = 999 kg/m³

[tex]\frac{p_{water}}{2} K_LV_2^2[/tex] = pressure difference due to frictional losses.

[tex]\frac{p_{water}}{2} (V_2^2-V_1^2)[/tex] = pressure difference due to the kinetic energy

So; we are calculating three terms here.

a) the pressure drop across the contraction = Δp

b) pressure difference due to frictional losses. = [tex]\frac{p_{water}}{2} K_LV_2^2[/tex]

c) pressure difference due to the kinetic energy = [tex]\frac{p_{water}}{2} (V_2^2-V_1^2)[/tex]

a) Δp = [tex][\frac{p_{water}}{2} (V_2^2-V_1^2)+ \frac{p_{water}}{2} K_LV_2^2][/tex]

Δp = [tex][\frac{999kg/m^3}{2} ((14.1m/s)^2-(3.53m/s)^2)+ \frac{999kg/m^3}{2} (0.4)(14.1m/s)^2][/tex]

Δp = [(93081.375) + (39722.238)]

Δp = (93 kPa) + (39.7 kPa)

Δp = 132.7 kPa

Δp ≅ 133 kPa

∴ the pressure drop across the contraction Δp = 133 kPa

the pressure difference due to frictional losses [tex]\frac{p_{water}}{2} K_LV_2^2[/tex] = 39.7 kPa

the pressure difference due to the kinetic energy [tex]\frac{p_{water}}{2} (V_2^2-V_1^2)[/tex] = 93 kpa

I hope that helps a lot!