The Venturi tube shown in the figure below may be used as a fluid flowmeter. Suppose the device is used at a service station to measure the flow rate of gasoline (ñ = 7.00 102 kg/m3) through a hose having an outlet radius of 2.72 cm. The difference in pressure is measured to be P1 − P2 = 2.00 kPa and the radius of the inlet tube to the meter is 1.36 cm.

(a) Find the speed of the gasoline as it leaves the hose.
m/s

(b) Find the fluid flow rate in cubic meters per second.
m3/s

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To find the speed of the gasoline as it leaves the hose, we can use Bernoulli's equation:

P1 + 1/2 ρv1^2 + ρgh1 = P2 + 1/2 ρv2^2 + 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 gasoline, and
h1 and h2 are the heights at points 1 and 2 respectively.

In this case, point 1 is at the outlet of the hose and point 2 is at the Venturi tube.

(a) To find the speed of the gasoline as it leaves the hose (v1), we can assume that h1 = 0 (since they are at the same height) and h2 = 0 (since there is no height difference between the two points). We are given P1 − P2 = 2.00 kPa and the radii of the inlet tube (r1) = 1.36 cm and the outlet of the hose (r2)= 2.72 cm.

Using the equation: P1 + 1/2 ρv1^2 = P2 + 1/2 ρv2^2

Since h1 = h2 = 0, the equation becomes:

P1 + 1/2 ρv1^2 = P2 + 1/2 ρv2^2

Since there is no change in height, using the equation: 1/2 ρv1^2 = 1/2 ρv2^2

P1 = P2 + ρv1^2 - ρv2^2

Using the given values: ρ = 7.00 x 10^2 kg/m^3, P1 - P2 = 2.00 x 10^3 Pa, r1 = 1.36 cm, and r2 = 2.72 cm, we can calculate the speed of gasoline as it leaves the hose (v1):

P1 - P2 = ρ(v1^2 - v2^2)

2.00 x 10^3 Pa = (7.00 x 10^2 kg/m^3)((v1^2) - (v2^2))

v1^2 - v2^2 = (2.00 x 10^3 Pa) / (7.00 x 10^2 kg/m^3)

v1^2 - v2^2 = 2.857 m^2/s^2

(v1 + v2)(v1 - v2) = 2.857 m^2/s^2

v1 + v2 = v1 - v2 = √(2.857 m^2/s^2)

2v1 = √(2.857 m^2/s^2)

v1 = √(2.857 m^2/s^2) / 2

v1 ≈ 0.951 m/s

Therefore, the speed of the gasoline as it leaves the hose is approximately 0.951 m/s.

(b) To find the fluid flow rate in cubic meters per second, we can use the equation:

Q = A1v1

Where Q is the flow rate, v1 is the speed of the gasoline as it leaves the hose, and A1 is the cross-sectional area of the outlet of the hose.

Using the given value of r2 = 2.72 cm, we can calculate the flow rate (Q):

A1 = πr1^2

= π(1.36 cm)^2

= π(0.0136 m)^2

≈ 0.582 x 10^-3 m^2

Q = (0.582 x 10^-3 m^2)(0.951 m/s)

≈ 0.553 x 10^-3 m^3/s

Therefore, the fluid flow rate is approximately 0.553 x 10^-3 m^3/s.

To find the speed of the gasoline as it leaves the hose, we can use Bernoulli's equation:

P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2

Where:
P1 is the pressure at the inlet of the hose,
v1 is the speed of the gasoline at the inlet of the hose,
P2 is the pressure at the outlet of the hose,
v2 is the speed of the gasoline at the outlet of the hose,
ρ is the density of the gasoline,
g is the acceleration due to gravity,
h1 is the height at the inlet of the hose,
h2 is the height at the outlet of the hose.

Since the Venturi tube is horizontal, the height differences h1 and h2 are equal to zero. Also, since the tube is submerged in the gasoline, we can assume the height of the fluid columns cancel out. Therefore, we can simplify Bernoulli's equation to:

P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2

We are given the pressure difference P1 - P2 = 2.00 kPa = 2000 Pa, and the radii of the inlet tube and outlet hose, which are r1 = 1.36 cm = 0.0136 m, and r2 = 2.72 cm = 0.0272 m, respectively.

Let's find the speed of the gasoline as it leaves the hose (v2):

P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2

P2 - P1 = (1/2)ρv1^2 - (1/2)ρv2^2

v2^2 - v1^2 = (2(P2 - P1))/ρ

v2^2 = (2(P2 - P1))/ρ + v1^2

v2 = sqrt((2(P2 - P1))/ρ + v1^2)

Substituting ρ = 7.00 x 10^2 kg/m^3 and P2 - P1 = 2000 Pa:

v2 = sqrt((2(2000))/(7.00 x 10^2) + v1^2)

Now, we need to find the speed of the gasoline at the inlet of the hose (v1). We can use the principle of continuity for incompressible flow, which states that the mass flow rate remains constant throughout the tube.

The mass flow rate (ṁ) is given by:

ṁ = A1v1 = A2v2

Where:
A1 is the cross-sectional area of the inlet tube,
A2 is the cross-sectional area of the outlet hose.

The cross-sectional area of a circle is given by A = πr^2. Substituting the radii r1 and r2, the equation becomes:

πr1^2v1 = πr2^2v2

v1 = (πr2^2v2)/(πr1^2)

v1 = (r2^2v2)/(r1^2)

Now, substituting the values of r1, r2, and solving for v1:

v1 = (0.0272^2v2)/(0.0136^2)

With v1 and v2, substituting into the previously found formula for v2 to calculate the speed of the gasoline as it leaves the hose:

v2 = sqrt((2(2000))/(7.00 x 10^2) + ((0.0272^2 * v2)/(0.0136^2))^2)

Now, we can solve this equation for v2 using numerical methods or iteration techniques to obtain the speed of the gasoline as it leaves the hose.

To find the fluid flow rate in cubic meters per second (Q), we can use the formula:

Q = A2v2

Where:
A2 is the cross-sectional area of the outlet hose (radius r2).

Substituting the value of r2 and the previously calculated value of v2 into the formula:

Q = πr2^2v2

Q = π(0.0272)^2v2

Calculate this expression to find the fluid flow rate in cubic meters per second.

7.6

Try to use P2-P1 = ρ/2(v1^2-v2^2)

Derived from P1 + 1/2ρv1^2 = P2 + 1/2ρv2^2

Then A1V1 = A2V2 is from continuity and can be expressed V1 = A2V2/A1

You get the equation P2 - P1 = ρ/2((A2V2/A1)^2 - V2^2) via substitution

And then simplify to 2(P2-P1)/ρ = A2V2/A1 - V2^2

Plug in and solve