Consider two pipes with r1 = 2r2. Join them together and let water flow. The pressure difference between the two parts is P1−P2 = 100Pa. What is the velocity in the larger half?

 Consider two pipes with r1 = 2r2. Join them together and let water flow. The pressure difference between the two parts is P1−P2 = 100Pa. What is the velocity in the larger half?

Solution

A1V1=A2V2 where A1=pi(r1)^2=3.14r1^2 and A2=pi(r2)^2=3.14(r2)^2 but r1=2r2 so r2=r1/2; A2=3.14(r1/2)^2=3.14(r1^2/4)=0.7854r1

3.14r1(V1)=0.7854r1(V2); V1=0.7854/3.14V2=0.25V2 or V2/4. It is 1/4 of V2 because the diameter of pipe 1 is bigger so the velocity is smaller

P1+1/2rhoV1^2=P2+1/2rhoV2^2

P1-P2=1/2rhoV2^2–1/2rhoV1^2 but P1-P2=100Pa and V1=V2/4. So substitute the values of P1-P2=100 and V1=V2/4 to the following:

100=1/2(1000)(V2^2)-1/2(1000)(V2/4)^2

100=500V2^2–500(V2^2/16)=500V2^2–31.25V2^2

100=468.75V2^2; V2^2=468.75/100=4.6875

V2=(4.6875)^1/2=2.165 meters/second

V1= 2.165/4= 0.54 meter/second velocity of the larger half(answer)