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3 years ago

What test should be performed on abrasive wheels

Engineering

2 answers:

soldier1979 [14.2K]3 years ago

8 0

Ring test I think maybe I’m not 100%

Svet_ta [14]3 years ago

3 0

Answer:

before wheel is put on it should be looked at for damage and a sound or ring test should be done to check for cracks, to test the wheel it should be tapped with a non metallic instrument (I looked it up)

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11–17 A long, thin-walled double-pipe heat exchanger with tube and shell diameters of 1.0 cm and 2.5 cm, respectively, is used t

lana [24]

Answer:

the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C

Explanation:

Given:

d₁ = diameter of the tube = 1 cm = 0.01 m

d₂ = diameter of the shell = 2.5 cm = 0.025 m

Refrigerant-134a

20°C is the temperature of water

h₁ = convection heat transfer coefficient = 4100 W/m² K

Water flows at a rate of 0.3 kg/s

Question: Determine the overall heat transfer coefficient of this heat exchanger, Q = ?

First at all, you need to get the properties of water at 20°C in tables:

k = 0.598 W/m°C

v = 1.004x10⁻⁶m²/s

Pr = 7.01

ρ = 998 kg/m³

Now, you need to calculate the velocity of the water that flows through the shell:

$v_{w} =\frac{m}{\rho \pi (\frac{d_{2}^{2}-d_{1}^{2} }{4} )} =\frac{0.3}{998*\pi (\frac{0.025^{2}-0.01^{2} }{4}) } =0.729m/s$

It is necessary to get the Reynold's number:

$Re=\frac{v_{w}(d_{2}-d_{1}) }{v} =\frac{0.729*(0.025-0.01)}{1.004x10^{-6} } =10891.4343$

Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:

$Nu=0.023Re^{0.8} Pr^{0.4} =0.023*(10891.4343)^{0.8} *(7.01)^{0.4} =85.0517$

The overall heat transfer coefficient:

$Q=\frac{1}{\frac{1}{h_{1} }+\frac{1}{h_{2} } }$

Here

$h_{2} =\frac{kNu}{d_{2}-d_{1}} =\frac{0.598*85.0517}{0.025-0.01} =3390.7278W/m^{2}C$

Substituting values:

$Q=\frac{1}{\frac{1}{4100}+\frac{1}{3390.7278} } =1855.8923W/m^{2} C$

5 0

3 years ago

You work for a printing company, and you realize that your colleague sent incorrect price quotes to a client. You begin to write

xxTIMURxx [149]

Answer:

The sentence excerpted from the e-mail uses passive voice.

Given the purpose of your message, this voice is appropriate.

Explanation:

Because the objective is to remedy the situation a passive voice is great because it emphasizes the action and the object instead of the subject.

We want to emphasize the document and the incorrect information, not our colleague.

4 0

3 years ago

¿Qué es la masonería?

timofeeve [1]

Answer: freemasonry is Being a Mason is about a father helping his son make better decisions; a business leader striving to bring morality to the workplace; a thoughtful man learning to work through tough issues in his life.

Explanation:

3 0

3 years ago

Calculate the radius of gyration for a bar of rectangular cross section with thickness t and width w for bending in the directio

SVEN [57.7K]

Answer:

a)R= sqrt( wt³/12wt)

b)R=sqrt(tw³/12wt)

c)R= sqrt ( wt³/12xcos45xwt)

Explanation:

Thickness = t

Width = w

Length od diagonal =sqrt (t² +w²)

Area of raectangle = A= tW

Radius of gyration= r= sqrt( I/A)

Moment of inertia in the direction of thickness I = w t³/12

R= sqrt( wt³/12wt)

Moment of inertia in the direction of width I = t w³/12

R=sqrt(tw³/12wt)

Moment of inertia in the direction of diagonal I= (w t³/12)cos 45=( wt³/12)x 1/sqrt (2)

R= sqrt ( wt³/12xcos45xwt)

4 0

3 years ago

How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an

iris [78.8K]

Answer:

116.3 electrons

Explanation:

Data provided in the question:

Time, t = 2.55 ps = 2.55 × 10⁻¹² s

Current, i = 7.3 μA = 7.3 × 10⁻⁶ A

Now,

we know,

Charge, Q = it

thus,

Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)

Q = 18.615 × 10⁻¹⁸ C

Also,

We know

Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C

Therefore,

Number of electrons past a fixed point = Q ÷ q

= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]

= 116.3 electrons

4 0

3 years ago