Question

The diameters in a converging-diverging nozzle at two different lo cations are 1 m and 1.5 m, respectively. A pressure difference of 5 kPa is measured between the two locations. Calculate the steady volumetric flow rate of water in the nozzle.

Answer 1

`2.772 m^(3)/s`

Step-by-step explanation:

Given information

Pressure difference
`\triangle P= 5 kPa`

Diameter 1,
`d_1=1 m`

Diameter 2,
`d_2= 1.5 m`

Assumption

Density of water,
`\rho= 1000 kg/m^(3)`

Calculations

Using the principle of conservation of mass

`Q_1=Q_2` where Q is the volume flow rate and subscripts 1 and 2 represent first and second nozzles

`A_1 V_1= A_2 V_2` where v is the velocity

`\frac {\pi}{4} d_1^(2) V_1=\frac {\pi}{4} d_2^(2) V_2`

`1^(2)V_1=1.5^(2)V_2`

`V_1=2.25V_2`

Using Bernoulli’s equation

`\frac {P_1}{\rho g}+\frac {V_1^(2)}{2g}=\frac {P_2}{\rho g}+\frac {V_2^(2)}{2g}`

`\frac {V_1^(2)-V_2^(2)}{2g}=\frac {P_2 – P_1}{\rho g}=\frac {\triangle P}{\rho g}`

`\frac {V_1^(2)-V_2^(2)}{2}=\ frac {\triangle P}{\rho}`

`\frac {(2.25V_2)^(2)-V_2^(2)}{2}=\frac {5000 Pa}{100}`

`2.03125V_2^(2)= 5`

`V_2=\sqrt{\frac {5}{2.03125}}= 1.568929 m/s`

Since
`V_1=2.25V_2`

`V_1=2.25(1.568929 m/s)= 3.53009\approx 3.53 m/s`

Now, the volume flow rate,
`Q= A_1V_1=\frac {\pi}{4} d_1^(2) V_1=\frac {\pi}{4} 1^(2) 3.53=2.772456 m^(3)/s\approx 2.772 m^(3)/s`

Keywords: Bernoulli's equation, volumetric flow