Respuesta :
Answer:
(a).The speed of the water in the nozzle is 3.014 m/s.
(b). The pressure in the nozzle is 1.86 atm.
Explanation:
Given that,
Nozzle diameter = 0.25 in = 0.00635 m
Hose pipe diameter = 0.64 in = 0.016256 m
Pressure = 1.9 atm =192518 Pa
(a). We need to calculate the speed of the water in the nozzle
Flow Speed at the inlet pipe will be given by using Continuity Equation
[tex]Q_{1}=Q_{2}[/tex]
[tex]v_{1}A_{1}=v_{2}A_{2}[/tex]
[tex]v_{1}=v_{2}\times(\dfrac{A_{2}}{A_{1}})[/tex]
Where, A = area of pipe
[tex]A=\pi\times \dfrac{d^2}{4}[/tex]
[tex]v_{1}=v_{2}\times(\dfrac{d_{2}^2}{d_{1}^2})[/tex]
Put the value into the formula
[tex]v_{1}=0.46\times\dfrac{(0.016256)^2}{(0.00635)^2}[/tex]
[tex]v_{1}=3.014\ m/s[/tex]
The speed of the water in the nozzle is 3.014 m/s.
(b). We need to calculate the pressure in the nozzle
Using Bernoulli's Theorem,
[tex]P_{1}+\dfrac{1}{2}\rho\times v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho\times v_{2}^2+\rho gh_{2}[/tex]
Where, [tex]h_{1}=h_{2}[/tex]
[tex]P_{1}+\dfrac{1}{2}\rho\times v_{1}^2=P_{2}+\dfrac{1}{2}\rho\times v_{2}^2[/tex]
[tex]P_{1}=P_{2}+\dfrac{1}{2}\rho(v_{2}^2-v_{1}^2)[/tex]
Put the value into the formula
[tex]P_{1}=192518 +\dfrac{1}{2}\times1000\times((0.46)^2-(3.014)^2)[/tex]
[tex]P_{1}=188081.702\ Pa[/tex]
[tex]P=1.86\ atm[/tex]
Hence, (a).The speed of the water in the nozzle is 3.014 m/s.
(b). The pressure in the nozzle is 1.86 atm.
