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Citrus2011 [14]
2 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.
Physics
1 answer:
Serga [27]2 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

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Vlad1618 [11]

Answer:

B_o = 1.013μT

Explanation:

To find B_o you take into account the formula for the emf:

\epsilon=-\frac{d\Phi_b}{dt}=-\frac{dBAcos\theta}{dt}=-Acos\theta\frac{dB}{dt}

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.

By applying the derivative you obtain:

\epsilon=-Acos\theta (2\pi f) B_ocos(2\pi f t+ \alpha)

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

\epsilon=2\pi fAB_o=2\pi (100*10^3Hz)(\pi (0.1m)^2)B_o=19739.20Hzm^2B_o\\\\B_o=\frac{20*10^{-3}V}{19739.20Hzm^2}=1.013*10^{-6}T=1.013\mu T

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6 0
2 years ago
A badminton player shuffles his shoes in resins explain the scientific reason behind this​
Kobotan [32]

Answer:

To avoid slipping while playing badminton or to increase friction between the shoes and the badminton court or to enhance the grip.

Explanation:

As the question is general so I will try answer it generally.

As badminton court is usually slippery and flat with no or very little friction and while playing badminton, players have to move very fast from place to place.

So, in order to effectively move faster to take the difficult shots with accuracy badminton players increase the friction between their shoes and the badminton court by shuffling resin to their shoes. It helps to enhance the grip.

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Walt ran 5 kilometers in 25 minutes, going eastward. What was his average velocity?
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About 65 million years ago an asteroid struck Earth in the area of the Yucatán Peninsula and wiped out the dinosaurs and many ot
9966 [12]

Answer:

2.44156\times 10^{13}\ m^3

29010.53917 m

Explanation:

\rho = Density of asteroid = 2 g/cm³

V = Volume

d = Diameter = 10 km

r = Radius = \dfrac{d}{2}=\dfrac{10}{2}=5\ km

v = Velocity = 11 km/s

H_v = Heat vaporization of water = 2.26\times 10^6\ J/kg

\Delta T = Change in temperature = 100-20

Mass is given by

m=\rho V\\\Rightarrow m=\rho\dfrac{4}{3}\pi r^3\\\Rightarrow m=2000\dfrac{4}{3}\times \pi\times 5000^3\\\Rightarrow m=1.0472\times 10^{15}\ kg

The kinetic energy is

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1.0472\times 10^{15}\times 11000^2\\\Rightarrow K=6.33556\times 10^{22}\ J

Heat is given by

Q=mc\Delta T+mH_v\\\Rightarrow 6.33556\times 10^{22}=m\times (4186\times (100-20)+2.26\times 10^6)\\\Rightarrow m=\dfrac{ 6.33556\times 10^{22}}{4186\times (100-20)+2.26\times 10^6}\\\Rightarrow m=2.44156\times 10^{16}\ kg

Mass of water is 2.44156\times 10^{16}\ kg

Volume is \dfrac{2.44156\times 10^{16}}{10^3}=2.44156\times 10^{13}\ m^3

Amount of water is 2.44156\times 10^{13}\ m^3

If it were a cube

h=V^{\dfrac{1}{3}}\\\Rightarrow h=(2.44156\times 10^{13})^{\dfrac{1}{3}}\\\Rightarrow h=29010.53917\ m

The height of the water would be 29010.53917 m

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