Ask question

Ask question

All categories

Citrus2011 [14]

3 years ago

8

Water at 20°C flows by gravity through a smooth pipe from one reservoir to a lower one. The elevation difference is 60 m. The pi

pe is 360 m long, with a diameter of 12 cm. For the given system a pump is used at night to drive water back to the upper reservoir. If the pump delivers 15,000 W to the water, estimate the flow rate. For water at 20°C, take rho = 998 kg/m3 and μ = 0.001 kg/m-s. Round the answer to the nearest whole number.

1 answer:

Serga [27]3 years ago

7 0

Answer:

Flow Rate = 80 m^3 /hours (Rounded to the nearest whole number)

Explanation:

Given

Hf = head loss
f = friction factor
L = Length of the pipe = 360 m
V = Flow velocity, m/s
D = Pipe diameter = 0.12 m
g = Gravitational acceleration, m/s^2
Re = Reynolds's Number
rho = Density =998 kg/m^3
μ = Viscosity = 0.001 kg/m-s
Z = Elevation Difference = 60 m

Calculations

Moody friction loss in the pipe = Hf = (f*L*V^2)/(2*D*g)

The energy equation for this system will be,

Hp = Z + Hf

The other three equations to solve the above equations are:

Re = (rho*V*D)/ μ

Flow Rate, Q = V*(pi/4)*D^2

Power = 15000 W = rho*g*Q*Hp

1/f^0.5 = 2*log ((Re*f^0.5)/2.51)

We can iterate the 5 equations to find f and solve them to find the values of:

Re = 235000

f = 0.015

V = 1.97 m/s

And use them to find the flow rate,

Q = V*(pi/4)*D^2

Q = (1.97)*(pi/4)*(0.12)^2 = 0.022 m^3/s = 80 m^3 /hours

You might be interested in

How much work would be needed to raise the payload from the surface of the moon (i.e., x = r) to an altitude of 5r miles above t

Cloud [144]

Let the data is as following

mass of payload = "m"

mass of Moon = "M"

now we know that we place the payload from the position on the surface of moon to the position of 5r from the surface

So in this case we can say that change in the gravitational potential energy is equal to the work done to move the mass from one position to other

so it is given by

$W = U_f - U_i$

we know that

$U_f = -\frac{GMm}{6r}$

$U_i = -\frac{GMm}{r}$

now from above formula

$W = -\frac{GMm}{6r} + \frac{GMm}{r}$

$W = \frac{5GMm}{6r}$

so above is the work done to move the mass from surface to given altitude

7 0

3 years ago

Is 40 mph possible to run?

Bogdan [553]

NO

The likelihood of reaching 40 miles per hour is very slim. Several factors are related to how quickly somebody can run, which is why only one person has ever run 28 miles per hour.Catch a sprinter smashing a speed record and they look like they could keep up with a car chase. 40 MPH: The fastest speed humans can run. The current fastest human in the world is Usain Bolt, who can run at nearly 28 miles per hour.Steve in Davis, Calif. So far, the fastest anyone has run is about 27½ miles per hour, a speed reached (briefly) by sprinter Usain Bolt just after the midpoint of his world-record 100-meter dash in 2009.

Find more about run :-brainly.com/question/17889385?referrer=searchResults

#SPJ4

8 0

1 year ago

What is the current in mA through a 500 Ω resistor that is connected to a 1.5 V battery? Show all calculations using Ohm’s law (

Ludmilka [50]

Answer:

3 mA.

Explanation:

The following data were obtained from the question:

Resistor (R) = 500 Ω

Potential difference (V) = 1.5 V

Current (I) =.?

Using the ohm's law equation, we can obtain the current as follow:

V = IR

1.5 = I x 500

Divide both side by 500

I = 1.5 / 500

I = 3×10¯³ A.

Therefore, the current in the circuit is 3×10¯³ A.

Finally, we shall convert 3×10¯³ A to milliampere (mA).

This can be obtained as follow:

Recall:

1 A = 1000 mA

Therefore,

3×10¯³ A = 3×10¯³ × 1000 = 3 mA

Therefore, 3×10¯³ A is equivalent to 3 mA.

Thus, the current in mA flowing through the circuit is 3 mA.

3 0

3 years ago

A blue-green photon (λ = 488 nm ) is absorbed by a free hydrogen atom, initially at rest. What is the recoil speed of the hydrog

Natalka [10]

Answer:

The recoil speed is $2.207\times 10^{4} m/s$

Solution:

Wavelength of a blue-green photon, $\lambda_{BG} = 488 nm = 488\times 10^{- 9} m$

Now, the energy associated with the blue-green photon:

$E_{BG} = \frac{hc}{\lambda_{BG}}$

where

h = Planck's constant

C = speed of light ion vacuum

$E_{BG} = \frac{6.626\times 10^{- 34}\times 3\times 10^{8}}{488\times 10^{- 9}}$

$E_{BG} = 4.07\times 10^{- 19} J$

Also, we know that the recoil speed can be calculated by the KInetic energy which is equal to the Energy of the blue-green photon:

$KE_{H} =\frac{1}{2}m_{p}v_{H}$

where

$v_{H}$ = velocity of Hydrogen atom

$m_{p} = 1.67\times 10^{- 27} kg$ = mass of H-atom

Now,

$KE_{H} =\frac{1}{2}m_{p}(v_{H})^{2}$

$4.07\times 10^{- 19} =\frac{1}{2}\times 1.67\times 10^{- 27}\times (v_{H})^{2}$

$v_{H} = \sqrt(4.87\times 10^{8}) = 2.207\times 10^{4} m/s$

7 0

3 years ago

A bullet has a speed of 500 m/s as it leaves a rifle. If it is fired horizontally from a cliff 12m above a lake, how far does th

dem82 [27]

Because you don’t have it

3 0

3 years ago