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3 years ago

Find roughness ε

Engineering

1 answer:

iris [78.8K]3 years ago

3 0

Hope that helps had to do it myself http://160592857366.free.fr/joe/ebooks/Mechanical%20Engineering%20Books%20Collection/FLUID%20MECHANICS/Fundamentals%20of%20Fluid%20Mechanics%204th%20Edition%20-%20Munson%20-%20John%20Wiley%20and%20Sons/MUNSON%20-%20livro.pdf

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An automobile having a mass of 1100 kg initially moves along a level highway at 110 km/h relative to the highway. It then climbs

Greeley [361]

Answer:

Change in kinetic energy=-513.652 KJ

Change in potential energy=431.64KJ

Explanation:

We are given that

Mass of an automobile , m=1100 kg

Initial speed, u=110 km/h= $110\times \frac{5}{18}=30.56 m/s$

Where $1 km/h=5/18 m/s$

Height , $h_2=40 m$

$g=9.81 m/s^2$

Final speed, v=0

Change in kinetic energy, $\Delta K.E=\frac{1}{2}m(v^2-u^2)$

$\Delta K.E=\frac{1}{2}(1100)(0-(30.56)^2)=-513652.48 J$

$\Delta K.E=-\frac{513652.48}{1000}=-513.652 KJ$

Where 1 KJ=1000 J

Change in potential energy, $\Delta P.E=mgh(h_2-h_1)$

Initially height, h1=0

Using the formula

$\Delta P.E=1100\times 9.81(40-0)$

$\Delta P.E=431640J$

$\Delta P.E=431.64KJ$

6 0

3 years ago

The amplitudes of the displacement and acceleration of an unbalanced motor were measured to be 0.15 mm and 0.6*g, respectively.

ehidna [41]

Answer:

The speed of shaft is 1891.62 RPM.

Explanation:

given that

Amplitude A= 0.15 mm

Acceleration = 0.6 g

we can say that acceleration= 0.6 x 9.81

$acceleration,a=5.88\ \frac{m}{s^2}$

We know that

$a=\omega ^2A$

So now by putting the values

$a=\omega ^2A$

$5.88=\omega ^2 \0.15\times 10^{-3}$

$\omega =198.09\ \frac{rad}{s}$

We know that

ω= 2πN/60

198.0=2πN/60

N=1891.62 RPM

So the speed of shaft is 1891.62 RPM.

4 0

3 years ago

a cantilever beam 1.5m long has a square box cross section with the outer width and height being 100mm and a wall thickness of 8

djverab [1.8K]

Answer:

a) 159.07 MPa

b) 10.45 MPa

c) 79.535 MPa

Explanation:

Given data :

length of cantilever beam = 1.5m

outer width and height = 100 mm

wall thickness = 8mm

uniform load carried by beam along entire length= 6.5 kN/m

concentrated force at free end = 4kN

first we determine these values :

Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m

Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N

A) determine max bending stress

б = $\frac{MC}{I}$ = $\frac{13312.5 ( 0.112)}{1/12(0.1^4-0.084^4)}$ = 159.07 MPa

B) Determine max transverse shear stress

attached below

ζ = 10.45 MPa

C) Determine max shear stress in the beam

This occurs at the top of the beam or at the centroidal axis

hence max stress in the beam = 159.07 / 2 = 79.535 MPa

attached below is the remaining solution

6 0

3 years ago

2. Determine the surface area of a primary settling tank sized to handle a maximum hourly flow of 0.570 m3/s at an overflow rate

Hitman42 [59]

Answer:

The surface area of the primary settling tank is 0.0095 m^2.

The effective theoretical detention time is 0.05 s.

Explanation:

The surface area of the tank is calculated by dividing the volumetric flow rate by the overflow rate.

Volumetric flow rate = 0.570 m^3/s

Overflow rate = 60 m/s

Surface area = 0.570 m^3/s ÷ 60 m/s = 0.0095 m^2

Detention time is calculated by dividing the volume of the tank by the its volumetric flow rate

Volume of the tank = surface area × depth = 0.0095 m^2 × 3 m = 0.0285 m^3

Detention time = 0.0285 m^3 ÷ 0.570 m^3/s = 0.05 s

7 0

4 years ago

Read 2 more answers

Any change in the system from one equilibrium state to another is called: A) Path B) Process C) Cycle D) None of the above

dexar [7]

Answer:

B) Process

Explanation:

In thermodynamics a process is a passage of a thermodynamic system from an initial to a final state of thermodynamic equilibrium.

A thermodynamic process path is the series of states through which a system passes from an initial to a final state.

Cycle is a process in which initial and final state are identical.

7 0

4 years ago