Ask question

Ask question

All categories

Svetllana [295]

3 years ago

9

A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length 2l. The flow is

laminar and fully developed. The pressure drop for the first pipe is 1.789 times greater than it is for the second pipe. If the diameter of the first pipe is D, determine the diameter of the second pipe.

1 answer:

Alexxx [7]3 years ago

6 0

Answer:

Explanation:

For rate of flow of liquid in a pipe , the formula is

Q = π p r⁴ / 8 η l

where Q is volume of liquid flowing per unit time , p is pressure drop across pipe , r is radius of pipe , η is coefficient of viscosity and l is length of pipe

If η and l are constant for two pipes having equal rate of flow as in the given case

p₁ r₁⁴ = p₂ r₂⁴

p₁ = 1.789 p₂ ( given )

r₁ = D/2

1.789 p₂ x (D/2)⁴ = p₂ x r₂⁴

r₂ = $\frac{D}{2}\times\sqrt[4 ]{1.789}$

2 x r₂ = D x 1.1565

D₂ = D x 1.1565

You might be interested in

A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with

Julli [10]

Explanation:

Given that,

Capacitor $C=1.66\ \mu F$

Resistor $R=80.0\ \Omega$

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

$f_{c}=\dfrac{1}{2\pi R C}$

Where, R = resistor

C = capacitor

Put the value into the formula

$f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}$

$f_{c}=1198.45\ Hz$

(B). We need to calculate the $V_{R}$ when $f = \dfrac{1}{2f_{c}}$

Using formula of $V_{R}$

$V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})$

Put the value into the formula

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=2.280\ Volt$

(C). We need to calculate the $V_{R}$ when $f = f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=3.606\ Volt$

(D). We need to calculate the $V_{R}$ when $f = 2f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=4.561\ Volt$

Hence, This is the required solution.

8 0

4 years ago

When the space shuttle coasts in a circular orbit at constant speed about the earth, is it accelerating? if so, in what directio

pochemuha

Yes. Acceleration means any change in speed or direction of motion. When an object coasts in a circular orbit at constant speed around the Earth, its direction is constantly changing. The acceleration is "CENTRIPETAL", which points toward the center of the circle.

5 0

4 years ago

Read 2 more answers

Why are some substances dissolve both polar and nonpolar substances?

dlinn [17]

Answer:

Because water is polar and oil is nonpolar, their molecules are not attracted to each other.<em> polar solvents dissolve polar solutes, and nonpolar solvents dissolve nonpolar solutes.</em>

<em></em>

<em></em>

<em>please mark me as the brainliest :) </em>

<em></em>

<em>use your brain :)</em>

8 0

3 years ago

Calculating ph how is ph related to the concentration of hydronium ions. True or False

Dmitry_Shevchenko [17]

I think this is TRUE. Ph is calculating the acidic acids in water. And you need to know the concentration of hydronium ions. Hope this helped !

7 0

3 years ago

An object is made of glass and has the shape of a cube 0.13 m on a side, according to an observer at rest relative to it. Howeve

Vlad1618 [11]

Answer:

$v=0.9833\ c$

Explanation:

The density changes means that the length in the direction of the motion is changed.

Therefore,

$\text{Density} = \frac{m}{lwh}$

Given :

Side, b = h = 0.13 m

Mass, m = 3.3 kg

Density = 8100 $kg/m^3$

So,

$8100=\frac{3.3}{l \times 0.13 \times 0.13}$

$l=\frac{3.3}{8100 \times 0.13 \times 0.13}$

l = 0.024 m

Then for relativistic length contraction,

$l= l' \sqrt{1-\frac{v^2}{c^2}}$

$0.024= 0.13 \sqrt{1-\frac{v^2}{c^2}}$

$0.184= \sqrt{1-\frac{v^2}{c^2}}$

$0.033= 1-\frac{v^2}{c^2}}$

$\frac{v^2}{c^2}= 0.967$

$\frac{v}{c}=0.9833$

$v=0.9833\ c$

Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).

3 0

3 years ago