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.
A1V1=A2V2
(10)(275)=(3.50)V2
2750=3 .50 V2
V2= 786 cm/s
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.
To calculate the smaller section's pressure (part b), we can use Bernoulli's equation. Bernoulli's equation relates the pressure, velocity, and height of a fluid. The equation is as follows:
P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2
Where:
P1 and P2 are the pressures at points 1 and 2 respectively.
v1 and v2 are the velocities at points 1 and 2 respectively.
ρ is the density of the fluid.
g is the acceleration due to gravity.
h1 and h2 are the heights at points 1 and 2 respectively.
Since the two sections of tubing are joined end to end and are horizontal, the heights h1 and h2 are at the same level. Therefore, we can cancel out ρgh1 and ρgh2 terms from the equation.
The equation simplifies to:
P1 + 1/2ρv1^2 = P2 + 1/2ρv2^2
We know the pressure (P1) and flow speed (v1) in the first section, so we can substitute those values into the equation:
1.20 × 10^5 Pa + 1/2(ρ)(275 cm/s)^2 = P2 + 1/2(ρ)(v2)^2
To find the smaller section's pressure (P2), we need to rearrange the equation and solve for P2.
P2 = 1.20 × 10^5 Pa + 1/2(ρ)(275 cm/s)^2 - 1/2(ρ)(v2)^2
Since we know the density of the liquid (ρ = 1.65 g/cm^3), we can substitute this value as well.
P2 = 1.20 × 10^5 Pa + 1/2(1.65 g/cm^3)(275 cm/s)^2 - 1/2(1.65 g/cm^3)(v2)^2
Now, all we need to do is calculate the value of P2. Feel free to substitute the value of v2 and calculate P2 yourself.
To solve part b of the problem, we can use Bernoulli's equation. Bernoulli's equation states that the total mechanical energy of a fluid flowing in a streamline is constant at all points along the streamline, neglecting friction. In equation form, it can be written as:
P1 + (½)ρv1^2 + ρgh1 = P2 + (½)ρv2^2 + ρgh2
Where:
P1 and P2 are the pressures at points 1 and 2 respectively,
ρ is the density of the fluid,
v1 and v2 are the velocities at points 1 and 2 respectively,
g is the acceleration due to gravity,
h1 and h2 are the heights at points 1 and 2 respectively.
In this problem, as the two sections of tubing are horizontal, the heights (h1 and h2) are the same and can be neglected. So the equation simplifies to:
P1 + (½)ρv1^2 = P2 + (½)ρv2^2
Since we are trying to find the pressure at the smaller section (P2), let's rearrange the equation:
P2 = P1 + (½)ρv1^2 - (½)ρv2^2
Now we can substitute the given values into the equation:
P1 = 1.20 × 10^5 Pa (given)
ρ = 1.65 g/cm^3 = 1650 kg/m^3 (since 1 g/cm^3 = 1000 kg/m^3)
v1 = 275 cm/s (given)
v2 = ? (to be found)
Before we substitute the values, let's convert the units so that they are consistent. In SI units:
P1 = 1.20 × 10^5 Pa
ρ = 1650 kg/m^3
v1 = 2.75 m/s (since 1 cm/s = 0.01 m/s)
Substituting the values into the equation:
P2 = 1.20 × 10^5 Pa + (½)(1650 kg/m^3)(2.75 m/s)^2 - (½)(1650 kg/m^3)(v2)^2
Now, we need to use the equation of continuity that you mentioned earlier: A1V1 = A2V2. We can solve for v2:
v2 = (A1/A2) * v1
Given that A1 = 10.0 cm^2 = 0.001 m^2 and A2 = 3.50 cm^2 = 0.00035 m^2, we can substitute the values to find v2:
v2 = (0.001 m^2 / 0.00035 m^2) * 2.75 m/s
Calculate this value to find v2. Once you have v2, substitute it into the equation for P2 that we derived earlier to find the pressure at the smaller section.