Question

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?

Answer 1

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

Step-by-step explanation:

If
`V = (Q)/(A)` -----------equation (1)

where ;

V = velocity

Q = flow rate

A = area of cross-section

As we know that Area (A) =
`((\pi )/(4)D^2)`

substituting
`((\pi )/(4)D^2)` for A in equation (1); we have:

`V= (Q)/((\pi )/(4)D^2 )`

`V= (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= (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= (4(0.04m^3/s))/(\pi (0.12m/s)^2 )`

`V_1=3.5367 m/s`

For velocity
`(V_2)`:

`V_2= (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= (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
`((A_2)/(A_1))` can be find by replacing
`((\pi )/(4)D_2^2)` for
`A_2` and
`((\pi )/(4)D_1^2)` for
`A_1`.

So:
`(A_2)/(A_1) = ((\pi )/(4)D^2_2 )/((\pi )/(4)D^2_1 )`

`(A_2)/(A_1) = (D_2^2)/(D_1^2)`

`(A_2)/(A_1) = ((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 =
`[(p_(water))/(2) (V_2^2-V_1^2)+ (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

`(p_(water))/(2) K_LV_2^2` = pressure difference due to frictional losses.

`(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. =
`(p_(water))/(2) K_LV_2^2`

c) pressure difference due to the kinetic energy =
`(p_(water))/(2) (V_2^2-V_1^2)`

a) Δp =
`[(p_(water))/(2) (V_2^2-V_1^2)+ (p_(water))/(2) K_LV_2^2]`

Δp =
`[(999kg/m^3)/(2) ((14.1m/s)^2-(3.53m/s)^2)+ (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
`(p_(water))/(2) K_LV_2^2` = 39.7 kPa

the pressure difference due to the kinetic energy
`(p_(water))/(2) (V_2^2-V_1^2)` = 93 kpa

I hope that helps a lot!