A 8.0 cm-diameter horizontal pipe gradually narrows to 4.5 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 33.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow? in m^3/s

7.11x10^-3

Explanation:

We are to get the volume rate of flows

1/2pv1² + pl = 1/2pv2²

Such that A1V1 = A2V2

V1 = A2V2/A1

From the attachment I uploaded, we have a formula named equation 1 from which I have plugged in these values

P2 = 33000

P2 = 24000

P = 1000

r2 = 2.25

r1 = 4

When we put these values into the equation,

V2 = 4.47

A2V2 = pi(0.0225)²x4.47

= 7.7x19^-3m³/s

volts times amps 6 x 0.5 3w