A1V1=A2V2
(10)(275)=(3.50)V2
2750=3 .50 V2
V2= 786 cm/s
A liquid ( = 1.65 g/cm3) flows through two horizontal sections of tubing joined end to end. In the first section the cross-sectional area is 10.0 cm2, the flow speed is 275 cm/s, and the pressure is 1.20 105 Pa. In the second section the cross-sectional area is 3.50 cm2.
a.Calculate the smaller section's flow speed.
b.Calculate the smaller section's pressure.
I found part a but not b.
a. Use the equation of continuity A1V1=A2V2
b. Use bernoullis equation. If you post your work, perhaps we can help.
2 answers
To calculate the smaller section's pressure, you can use Bernoulli's equation:
P1 + 1/2pV1^2 + pgh1 = P2 + 1/2pV2^2 + pgh2
Assuming the tubing is horizontal and at the same height, we can simplify the equation to:
P1 + 1/2pV1^2 = P2 + 1/2pV2^2
Plugging in the given values, we get:
1.20 × 10^5 Pa + 1/2 × 1.65 g/cm^3 × (275 cm/s)^2 = P2 + 1/2 × 1.65 g/cm^3 × (786 cm/s)^2
Simplifying, we get:
P2 = 6.84 × 10^4 Pa
Therefore, the smaller section's pressure is 6.84 × 10^4 Pa.
P1 + 1/2pV1^2 + pgh1 = P2 + 1/2pV2^2 + pgh2
Assuming the tubing is horizontal and at the same height, we can simplify the equation to:
P1 + 1/2pV1^2 = P2 + 1/2pV2^2
Plugging in the given values, we get:
1.20 × 10^5 Pa + 1/2 × 1.65 g/cm^3 × (275 cm/s)^2 = P2 + 1/2 × 1.65 g/cm^3 × (786 cm/s)^2
Simplifying, we get:
P2 = 6.84 × 10^4 Pa
Therefore, the smaller section's pressure is 6.84 × 10^4 Pa.
0 answers