Answer:
[tex]2.772 m^{3}/s[/tex]
Explanation:
Given information
Pressure difference [tex]\triangle P= 5 kPa[/tex]
Diameter 1, [tex]d_1=1 m[/tex]
Diameter 2, [tex]d_2= 1.5 m[/tex]
Assumption
Density of water, [tex]\rho= 1000 kg/m^{3}[/tex]
Calculations
Using the principle of conservation of mass
[tex]Q_1=Q_2[/tex] where Q is the volume flow rate and subscripts 1 and 2 represent first and second nozzles
[tex]A_1 V_1= A_2 V_2[/tex] where v is the velocity
[tex]\frac {\pi}{4} d_1^{2} V_1=\frac {\pi}{4} d_2^{2} V_2[/tex]
[tex]1^{2}V_1=1.5^{2}V_2[/tex]
[tex]V_1=2.25V_2[/tex]
Using Bernoulli’s equation
[tex]\frac {P_1}{\rho g}+\frac {V_1^{2}}{2g}=\frac {P_2}{\rho g}+\frac {V_2^{2}}{2g}[/tex]
[tex]\frac {V_1^{2}-V_2^{2}}{2g}=\frac {P_2 – P_1}{\rho g}=\frac {\triangle P}{\rho g}[/tex]
[tex]\frac {V_1^{2}-V_2^{2}}{2}=\ frac {\triangle P}{\rho}[/tex]
[tex]\frac {(2.25V_2)^{2}-V_2^{2}}{2}=\frac {5000 Pa}{100}[/tex]
[tex]2.03125V_2^{2}= 5[/tex]
[tex]V_2=\sqrt{\frac {5}{2.03125}}= 1.568929 m/s[/tex]
Since [tex]V_1=2.25V_2[/tex]
[tex]V_1=2.25(1.568929 m/s)= 3.53009\approx 3.53 m/s[/tex]
Now, the volume flow rate, [tex]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[/tex]
Keywords: Bernoulli's equation, volumetric flow
