1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ratelena [41]
3 years ago
14

Two sections, A and B, are 0.5 km apart along a 0.05 m diameter rough concrete pipe. A is 115 m higher than B, the water tempera

ture is 4 °C, and the pressure heads measured at A and B are 21.7 m and 76.1 m, respectively. Assume minor losses are negligible. Determine the flow rate through the pipe
Physics
1 answer:
KonstantinChe [14]3 years ago
3 0

Answer:

Q = 178.41 m^3 / s

Explanation:

Given:

  • Length of the pipe L = 0.5 km
  • Diameter D = 0.05 m
  • Pressure head @ A (P_a / γ )= 21.7 m
  • Pressure head @ B (P_b / γ )= 76.1 m
  • Elevation head Z_a = 115 m
  • Elevation head Z_b = 0 m
  • Minor Losses = 0 m
  • Major Losses = f*L*V^2 / 2*D*g = (500/2*0.05*9.81) *f*V^2 = 509.684*f*V^2
  • Velocity at cross section A and B: V_a = V_b m/s
  • Roughness e = 2.5 mm
  • Dynamic viscosity of water u = 8.9*10^-4 Pa-s
  • Density of water p = 997 kg/m^3

Find:

Flow Rate Q = pi*V*D^2/4  m^3/s  ??

Solution:

We will use the Head Balance as derived from Energy Balance:

(P_a / γ ) - (P_b / γ ) + (V_a^2 - V_b^2) / 2*g + (Z_a - Z_b) = Major Losses

21.7 - 76.1 + 0 + 115-0 = 509.684*f*V^2

f*V^2 = 0.18897199

To find correction factor f which is a function of e / D = 0.05, and Reynold's number which is unknown. In such cases we will guess a value of f and perform iterations as follows:

Guess: f_o = 0.072 (Moody's Chart @ e / D = 0.05 and most turbulent function).

V_o = sqrt(0.18897199 / 0.072) = 1.28504 m/s

Re_o = p*V_o*D / u =  997*1.28504*0.05 / 8.9*10^-4 = 71976.6

1st iteration

f_1 = g (Re_o , e/d) = 0.0718702 (Moody's Chart)

V_1 = sqrt(0.18897199 / 0.0718702) = 1.621528 m/s

Re_1 = p*V_1*D / u =  997*1.621528*0.05 / 8.9*10^-4 = 90823.81698

2nd iteration

f_2 = g (Re_1 , e/d) = 0.0718041 (Moody's Chart)

V_2 = sqrt(0.18897199 / 0.0718041) = 1.662273585 m/s

Re_2 = p*V_2*D / u =  997*1.662273585*0.05 / 8.9*10^-4 = 90865.54854

3rd iteration

f_3 = g (Re_2 , e/d) = 0.0718040  (Moody's Chart)

V_3 = sqrt(0.18897199 / 0.0718040) = 1.622274714 m/s

Re_3 = p*V_3*D / u =  997*1.622274714*0.05 / 8.9*10^-4 = 90865.61182

We can observe the convergence of V to 1.6222 m /s. Hence, the required velocity will be used to calculate the Flow rate Q:

Q =  pi*V*D^2/4 = pi*1.6222*0.05^2 / 4

Q = 178.41 m^3 / s

You might be interested in
Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at
PSYCHO15rus [73]

Answer:

|\Delta v |=\sqrt{\frac{4Gm}{d} }

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

\mu= \frac{m_1m_2}{m_1+m_2}

now, since both the masses have mass m

therefore,

\mu= \frac{m^2}{2m}

= m/2

The final K.E. of the particles is

KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2

Distance between two particles is d and the gravitational potential energy between them is given by

PE_{Gravitational}= \frac{Gmm}{d}

By law of conservation of energy we have

KE_{initial}+KE_{final}= PE_{gravitaional}

Now plugging the values we get

0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}

|\Delta v |=\sqrt{\frac{4Gm}{d} }

=\sqrt{\frac{Gm}{d} }

This the required relation between G,m and d

5 0
3 years ago
How would clearing a forest to plant corn affect an enviroment
Oksanka [162]
It would destroy animals homes shelter etc. it also would make global warming go faster. Hope this helped :D
4 0
3 years ago
Suppose the student in (Figure 1) is 68kg, and the board being stood on has a 12kg mass. What is the reading on the left scale?
lesantik [10]

The equilibrium conditions allow to find the results for the balance forces are:

  • F₁ = 225.4 N
  • F₂ = 558.6 N

When the acceleration is zero we have the equilibrium conditions for both linear and rotational motion.

            ∑ F = 0

            ∑ τ = 0

           

Where F are the forces and τ the torques.

The torque  is the product of the force and the perpendicular distance to the point of support,

The free-body diagrams are diagrams of the forces without the details of the bodies, see attached for the free-body diagram of the system.

We write the translational equilibrium condition.

           F₁ - W₁ - W₂ + F₂ = 0

We write the equation for the rotational motion, set our point of origin at scale 1, and the counterclockwise turns are positive.

         F₂ 2 - W₁ 1 - W₂ 1.5 = 0\frac{W_1  \ 1 + W_2 \ 1.5}{2}

Let's calculate F₂

         F₂ = \frac{W_1 \ 1 + W_2 \ 1.5 }{2}  

         F₂ = (m g + M g 1.5)/ 2

         F₂ = \frac{(12 + 68 \ 1.5 ) \  9.8}{2}  

         F₂ = 558.6 N

We substitute in the translational equilibrium equation.

         F₁ = W₁ + W₂ - F₂

         F₁ = (m + M) g - F₂

         F₁ = (12 +68) 9.8 - 558.6

         F₁ = 225.4 N

In conclusion using the equilibrium conditions we can find the forces of the balance are:

  • F₁ = 225.4 N
  • F2 = 558.6 N

Learn more here:  brainly.com/question/12830892

5 0
2 years ago
In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1
asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, f=5\times 10^{14}\ Hz

Separation between slits, d=2.2\times 10^{-5}\ m

(a) The condition for maxima is given by :

d\ sin\theta=n\lambda

For third maxima,

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{nc}{fd})  

\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})  

\theta=4.69^{\circ}

(b) For second dark fringe, n = 2

d\ sin\theta=(n+1/2)\lambda

\theta=sin^{-1}(\dfrac{5\lambda}{2d})

\theta=sin^{-1}(\dfrac{5c}{2df})

\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})

\theta=3.90^{\circ}

Hence, this is the required solution.

8 0
3 years ago
Rosa pours a cup of boiling water into a pot of room-temperature water. According to the second law of thermodynamics, what will
11111nata11111 [884]

<u>Thermal energy</u><u> from the room-temperature water will continuously flow to the boiling water.</u>

  • The second law states, in a straightforward manner, that heat cannot naturally go "uphill."
  • When a pan of boiling water and a pan of ice are in touch, the hot water cools and the ice melts and warms up.

<h3>THE FIRST LAW OF THERMODYNAMICS</h3>
  • Adiabatic Process - is a procedure that is carried out without the system's heat content changing.
  • Water is heated to a temperature of 1000C during the boiling process, making it an isothermal process. As steam, the excess heat leaves the system.

Learn more about first law of thermodynamics brainly.com/question/3808473

#SPJ4

8 0
2 years ago
Other questions:
  • Once all the hydrogen is gone, the star starts to collapse. if it collapses enough, what fuel will now be used? what color will
    8·1 answer
  • Suppose you ride your bike to the library traveling at 0.5 km/min. It takes you 25 minutes to get to the library. How far did yo
    5·2 answers
  • An inquisitive physics student and mountain climber climbs a 48.0-m-high cliff that overhangs a calm pool of water. He throws tw
    9·1 answer
  • (1 point) A horizontal clothesline is tied between 2 poles, 20 meters apart. When a mass of 5 kilograms is tied to the middle of
    15·1 answer
  • Need help with 6 , 7 , and 8.
    10·1 answer
  • What is Pascal principle
    15·1 answer
  • During a flood, a stream overflows its banks, and water covers the adjacent ____.
    14·1 answer
  • Which of the following is equal to 1W?<br><br>A.1 J<br>B.1 J/s<br>C.1 J·s<br>D.1 N·m
    10·1 answer
  • Explain what the international research team discovered and where this discovery is located in the body.
    8·2 answers
  • A 13.3 kg box sliding across the ground
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!