Chemistry, 26.02.2020 02:51 chasadyyy

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?

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Answer from: jake2920

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 $V = \frac{Q}{A}$ -----------equation (1)

where ;

V = velocity

Q = flow rate

A = area of cross-section

As we know that Area (A) = $(\frac{\pi }{4}D^2)$

substituting $(\frac{\pi }{4}D^2)$ for A in equation (1); we have:

$V= \frac{Q}{\frac{\pi }{4}D^2 }$

$V= \frac{4Q}{\pi D^2 }$ ----------------- equation (2)

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

For velocity (V_1) :

$V_1= \frac{4Q_1}{\pi D_1^2 }$

From, the question; we are given that:

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

Diameter of the pipe at point 1 (D_1) = 0.12 m

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

V_1=3.5367 m/s

For velocity (V_2) :

$V_2= \frac{4Q_2}{\pi D_2^2 }$

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

Diameter of the pipe at point 2 (D_2) = 0.06 m

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

V_2=14.1471 m/s

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

So: $\frac{A_2}{A_1} = \frac{\frac{\pi }{4}D^2_2 }{\frac{\pi }{4}D^2_1 }$

$\frac{A_2}{A_1} = \frac{D_2^2}{D_1^2}$

$\frac{A_2}{A_1} = \frac{(0.06m)^2}{(0.12m)^2}$

= 0.25

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

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

where;

K_L = standard frictional loss coefficient for a sudden contraction which is 0.4

$p_{water$ = density of the water = 999 kg/m³

$\frac{p_{water}}{2} K_LV_2^2$ = pressure difference due to frictional losses.

$\frac{p_{water}}{2} (V_2^2-V_1^2)$ = 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. = $\frac{p_{water}}{2} K_LV_2^2$

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

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

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

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 $\frac{p_{water}}{2} K_LV_2^2$ = 39.7 kPa

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

I hope that helps a lot!

Answer from: Quest

D)can be increased in a closed system

Answer from: Quest

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