A horizontal pipe 17.00 cm in diameter has a smooth reduction to a pipe 8.50 cm in diameter. If the pressure of the water in the larger pipe is 8.00 104 Pa and the pressure in the smaller pipe is 6.00 104 Pa, at what rate does water flow through the pipes? The velocity is the same for both, how do i convert the answer to kg/s

Question

A horizontal pipe 17.00 cm in diameter has a smooth reduction to a pipe 8.50 cm in diameter. If the pressure of the water in the larger pipe is 8.00 104 Pa and the pressure in the smaller pipe is 6.00 104 Pa, at what rate does water flow through the pipes? The velocity is the same for both, how do i convert the answer to kg/s

Answer 1

Sorry, I did not see your last response/question.

You'll find it here:
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# physics - flame, Sunday, September 26, 2010 at 8:57pm

how do i convert the answer to kg/s

# physics - MathMate, Monday, September 27, 2010 at 9:44am

Flow is given by the product of the water velocity (m/s) and cross sectional area (m²), multiplied by the density ρ (1000 kg/m³).
or,
Q=A1*V1*ρ (in kg/s)

Answer 2

for the answers, which velocity do i chose to convert to kg/s

Answer 3

To find the rate at which water flows through the pipes, we can use the principle of continuity, which states that the mass flow rate of an incompressible fluid (such as water) remains constant along a pipe.

The formula for mass flow rate is given by:

𝑚̇ = 𝜌𝐴𝑣

Where:
𝑚̇ is the mass flow rate
𝜌 is the density of water
𝐴 is the cross-sectional area of the pipe
𝑣 is the velocity of water

Since the velocities in both pipes are the same, we can write:

𝑚̇1 = 𝑚̇2

𝜌1𝐴1𝑣 = 𝜌2𝐴2𝑣

Since the area of a circle is given by 𝐴 = 𝜋𝑟^2, we can substitute the radii of the pipes into the equation.

Let 𝑟1 be the radius of the larger pipe (17.00 cm in diameter = 8.50 cm in radius) and 𝑟2 be the radius of the smaller pipe (8.50 cm in diameter = 4.25 cm in radius).

Now, we can rewrite the equation as:

𝜌1𝜋𝑟1^2𝑣 = 𝜌2𝜋𝑟2^2𝑣

Since the density of water is constant, we can cancel it out from both sides of the equation:

𝜋𝑟1^2𝑣 = 𝜋𝑟2^2𝑣

Now, we can solve for the velocity 𝑣:

𝑟1^2𝑣 = 𝑟2^2𝑣

𝑟1^2 = 𝑟2^2

(8.50 cm)^2 = (4.25 cm)^2

Let's calculate the value of 𝑣:

𝑣 = 𝜋𝑟1^2 / 𝑟2^2

𝑣 = 𝜋(8.50 cm)^2 / (4.25 cm)^2

Convert the units from cm to m:

𝑣 = 𝜋(8.50 cm / 100)^2 / (4.25 cm / 100)^2

𝑣 = 𝜋(0.085 m)^2 / (0.0425 m)^2

Now, we have the velocity 𝑣, but we need to convert the answer to kg/s (kilograms per second).

To convert the mass flow rate from 𝑚̇ in kg/s, we need to multiply it by the density of water (𝜌). The density of water at room temperature is approximately 1000 kg/m^3.

Finally, the rate at which water flows through the pipes in kg/s is:

𝑚̇ = 𝜌𝐴𝑣

𝑚̇ = 1000 kg/m^3 * 𝜋(0.085 m)^2 * 𝑣

Calculate 𝑚̇ using the value of 𝑣 obtained earlier.