Respuesta :
Answer:
Pressure drop across the contraction section = 133 kPa
The pressure difference due to frictional losses is = 39.7 kPa
The pressure difference due to kinetic energy changes = 93 kPa
Explanation:
If [tex]V = \frac{Q}{A}[/tex] -----------equation (1)
where ;
V = velocity
Q = flow rate
A = area of cross-section
As we know that Area (A) = [tex](\frac{\pi }{4}D^2)[/tex]
substituting [tex](\frac{\pi }{4}D^2)[/tex] for A in equation (1); we have:
[tex]V= \frac{Q}{\frac{\pi }{4}D^2 }[/tex]
[tex]V= \frac{4Q}{\pi D^2 }[/tex] ----------------- equation (2)
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!
