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deff fn [24]
2 years ago
8

Compute the mean and maximum velocities for a liquid with a flow rate of 20 L/min in a 1.5-in nominal diameter sanitary pipeline

. The liquid has a density of 1030 kg/m3 and viscosity of 50 cP. Is the flow laminar or turbulent?
Physics
1 answer:
vlada-n [284]2 years ago
7 0

Answer:

Mean velocity = 0.292 m/s

Maximum velocity = 0.584 m/s

The flow is laminar as Re = 229.2

Explanation:

D = 1.5 inches = 0.0381 m

Q = volumetric flow rate = 20 L/min = 0.000333 m³/s

Q = A × v

A = Cross sectional Area = πD²/4 = π(0.0381)²/4 = 0.00114 m²

v = average velocity

v = Q/A = 0.000333/0.00114 = 0.292 m/s

For flow in circular pipes, maximum velocity = 2 × average velocity = 2 × 0.292 = 0.584 m/s

To check if the flow is laminar or turbulent, we need its Reynolds number

Re = (ρvD)/μ

ρ = 1030 kg/m

v = 0.292 m/s

D = 1.5 inches = 0.0381 m

μ = 50 cP = 0.5 poise = 0.05 Pa.s

Re = (1030 × 0.292 × 0.0381)/0.05 = 229.2

For laminar flow, Re < 2100

For turbulent flow, Re > 4000

And 229.2 < 2100, hence, this flow is laminar.

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Suppose you are driving a car and your friend, who is with you in the car, tosses a softball up and down from her point of view.
Sholpan [36]

Answer:

No, i disagree.

Explanation:

If the car is moving, it only has a velocity with a component in the horizontal direction. If we use galilean relativity, the velocity of the ball observed by my friend standing in the ground should only be affected in the horizonal direction, while the vertical stays the same for both observers.

3 0
2 years ago
What is the total kinetic energy of a 0.15 kg hockey puck sliding at 0.5 m/s and rotating about its center at 8.4 rad/s? The dia
ycow [4]
The mass of the puck is
m = 0.15 kg.
The diameter of the puck is 0.076 m, therefore its radius  is
r = 0.076/2 = 0.038 m
The sliding speed is
v = 0.5 m/s
The angular velocity is
ω = 8.4 rad/s

The rotational moment of inertia of the puck is
I = (mr²)/2
  = 0.5*(0.15 kg)*(0.038 m)²
  = 1.083 x 10⁻⁴ kg-m²

The kinetic energy of the puck is the sum of the translational and rotational kinetic energy.
The translational KE is
KE₁ = (1/2)*m*v²
       = 0.5*(0.15 kg)*(0.5 m/s)²
       = 0.0187 j

The rotational KE is
KE₂ = (1/2)*I*ω²
       = 0.5*(1.083 x 10⁻⁴ kg-m²)*(8.4 rad/s)²
       = 0.0038 J

The total KE is
KE = 0.0187 + 0.0038 = 0.0226 J

Answer: 0.0226 J


4 0
1 year ago
An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t
gtnhenbr [62]

1. tan \theta = \frac{v^2}{Rg}

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

N cos \theta = mg (1)

N sin \theta = m \frac{v^2}{R} (2)

where in the second equation, the term m\frac{v^2}{R} represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the  following expression:

tan \theta = \frac{v^2}{Rg}

2. F=\frac{m}{R}(w^2-v^2)

In this second situation, the cars moves around the track at a speed

w>v

This means that the centripetal force term

m\frac{v^2}{R}

is now larger than before, and therefore, the horizontal component of the normal reaction, N sin \theta, is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

N sin \theta + F = m\frac{w^2}{R}

And re-arranging for F,

F=m\frac{w^2}{R}-N sin \theta (3)

But from eq.(2) in the previous part we know that

N sin \theta = m \frac{v^2}{R}

So, susbtituting into eq.(3),

F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)

4 0
1 year ago
Use the drop-down menus to complete the statements. The atoms in a solid . The arrangement of atoms in a solid causes it to have
Natalija [7]

Answer:The atoms in a solid  .

remain in fixed position

The arrangement of atoms in a solid causes it to have a definite  .

 shape and value

Solids in which the atoms have no particular order or pattern are called   solid

noncrystalline

Explanation:

8 0
1 year ago
Read 2 more answers
Two thin lenses with a focal length of magnitude 12.0cm, the first diverging and the second converging, are located 9.00cm apart
attashe74 [19]

Answer:

Explanation:

b ) First is concave lens with focal length f₁ = - 12 cm .

object distance u = - 20 cm .

Lens formula

1 / v - 1 / u = 1 / f

1 / v + 1 / 20 = -1 / 12

1 / v =  - 1 / 20  -1 / 12

= - .05 - .08333

= - .13333

v = - 1 / .13333

= - 7.5 cm

first image is formed before the first lens on the side of object.

This will become object for second lens

distance from second lens = 7.5 + 9 = 16.5 cm

c )

For second lens

object distance u = - 16.5 cm

focal length f₂ = + 12 cm ( lens is convex )

image distance = v

lens formula ,

1 / v - 1 / u = 1 / f₂

1 / v + 1 / 16.5 = 1 / 12

1 / v =   1 / 12 -  1 / 16.5

= .08333- .0606

= .02273

v = 1 /  .02273

= 44 cm ( approx )

It will be formed on the other side of convex lens

distance from first lens

= 44 + 9 = 53 cm .

magnification by first lens = v / u

= -7.5 / -20 = .375 .

magnification by second lens = v / u

= 44 / - 16.5

= - 2.67

d )

total magnification

= .375 x - 2.67

= - 1.00125

height of final image

= 2.50 mm x 1.00125

= 2.503mm

e )

The final image will be inverted with respect to object  because total magnification is negative .

6 0
1 year ago
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