Answer:
a. 192 m/s
b. -17,760 kPa
Explanation:
First let's write the flow rate of the liquid, using the following equation:
Q = A*v
Where Q is the flow rate, A is the cross section area of the pipe (A = pi * radius^2) and v is the speed of the liquid. The flow rate in both parts of the pipe (larger radius and smaller radius) needs to be the same, so we have:
a.
A1*v1 = A2*v2
pi * 0.02^2 * 12 = pi * 0.005^2 * v2
v2 = 0.02^2 * 12 / 0.005^2
v2 = 192 m/s
b.
To find the pressure of the other side, we need to use the Bernoulli equation: (600 kPa = 600000 N/m2)
P1 + d1*v1^2/2 = P2 + d1*v2^2/2
Where d1 is the density of the liquid (for water, we have d1 = 1000 kg/m3)
600000 + 1000*12^2/2 = P2 + 1000*192^2/2
P2 = 600000 + 72000 - 1000*192^2/2
P2 = -17760000 N/m2 = -17,760 kPa
The speed in the smaller part of the pipe is too high, the negative pressure in the second part means that the inicial pressure is not enough to maintain this output speed.
