4. Water is flowing at 12m/s in a horizontal pipe under a pressure of 600kpa radius 2cm.
a. What is the speed of the water on the other side of 0.5cm radius?
b. What is the pressure of the other side?

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.

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s