Water flowing out of a horizontal pipe emerges through a nozzle. the radius of the pipe is 1.7 cm, and the radius of the nozzle is 0.46 cm. the speed of the water in the pipe is 0.45 m/s. treat the water as an ideal fluid, and determine the absolute pressure of the water in the pipe.

By equation of continuity we can find the speed of flow through nozzle

[tex]A_1v_1 = A_2v_2[\tex]

[tex]\pi*1.7^2*0.45 = \pi*.46^2*v_2[\tex]

[tex]\pi*1.7^2*0.45 = \pi*.46^2*v_2[\tex]

[tex]v_2 = 6.15 m/s[\tex]

Now we can use equation of continuity to find the pressure

[tex]P_1 + 0.5 \rho v_1^2 = P_2 + 0.5 \rho v_2^2[\tex]

At exit through nozzle pressure will be same as atmospheric pressure

[tex]P_1 + 0.5 *1000*0.45^2 = 1.01*10^5 + 0.5*1000*6.15^2[\tex]

[tex]P_1 = 1.2* 10^5 Pa[\tex]

i don't know, you need to give me some answers.

explanation: