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Citrus2011 [14]
1 year 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]1 year 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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Answer:

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Two friends, barbara and neil, are out rollerblading. with respect to the ground, barbara is skating due south at a speed of 5.9
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2 years ago
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Answer:

C

Explanation:

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d = d₀ + d₀αT

for the sphere, we were given

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α = 1.1 x 10⁻⁵/degrees celsius

we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T             EQN 1

Similarly for the Aluminium ring we have

we were given

d₀ = 3.994 cm

α = 2.4 x 10⁻⁵/degrees celsius

we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T       EQN 2

Since @ the temperature T at which the sphere fall through the ring, d=D

Eqn 1 = Eqn 2

4 + (4.4x10⁻⁵)T =3.994 + (9.58x10⁻⁵)T, collect like terms

0.006=5.18x10⁻⁵T

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8 0
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Answer:

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Explanation:

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c) in the center of mass all external forces are applied, they relate it is

                      a = ½ g

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1 year ago
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