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tatyana61 [14]
3 years ago
14

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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Type the correct answer in the box. Spell all words correctly. Mike is constructing a project in which he uses a motor. How can
tankabanditka [31]
If it is. DC, direct current reverse the polarity of power leads on the motor.

If it is a 3 phase ac alternating current, reverse any of the two of three leads.

Disconnect power before attempting.
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WHICH ONE DO YOU LIKE *CREDIT 2 sbrianna3606*
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Considering the analogy between electrical circuit and thermal circuit, show your approach to derive an expression for the therm
Simora [160]

Answer:

Thermal resistance for a wall depends on the material, the thickness of the wall and the cross-section area.

Explanation:

Current flow and heat flow are very similar when we are talking about 1-dimensional energy transfer. Attached you can see a picture we can use to describe the heat flow between the ends of the wall. First of all, a temperature difference is required to flow heat from one side to the other, just like voltage is required for current flow.  You can also see that R_{th} represents the thermal resistance. The next image explains more about the parameters which define the value of the thermal resistances which are the following:

  1. Wall Thickness.  More thickness, more thermal resistance.
  2. Material thermal conductivity (unique value for each material). More conductivity, less thermal resistance.
  3. Cross-section Area. More cross-section area, less thermal resistance.

A expression to define  the thermal resistance for the wall is as follows:  R_{th} =\frac{l}{Ak}, where  l is the distance between the tow sides of the wall, that is to say the wall thickness; A is the cross-section area and k is the material conducitivity.

5 0
2 years ago
In a production facility, 3 cm thick large brass plates (k = 110 W/mC, α = 33.9 × 10-6 m2 /s) that are initially at a uniform
zysi [14]

Answer:

Explanation:

Given.

Thickness of brass plate t = 3 cm

Thermal conductivity of brass k = 110 W/m.°C

Density of brass \rho = 8530 kg/m^3

Specific heat of brass C_p =380J/kg.°C

Thermal diffusivity of brass \alpha = 33.9\times 10^{-6} m^2/s

Temperature of oven T_{\infty} = 700°C

The initial temperature T_i= 25°C

Plate remain in the oven t =10 min  

Heat conduction in the plate is one-dimensional since the plate is large relative to its thickness  and there is thermal symmetry about the center plane.

The thermal properties of the plate are constant.

The heat transfer coefficient is constant and uniform over the entire surface.

The Fourier number is > 0.2 so that the one-term approximate solutions (or the transient  temperature charts) are applicable (this assumption will be verified).

The Biot number for this process Bi = \frac{hL}{k}\\\\Bi=\frac{(80 W/m^2.°C)(0.015 m)}{(110 W/m.°C)}\\=Bi =0.0109

The constants \lambda_1 and A_1 corresponding to this Biot are, from 11-2 tables.

The interpolation method used to find the

\lambda_1=0.1039  and A_1=1.0018
  

The Fourier number \tau=\frac{\alpha t}{L^2}\\\\\tau=\frac{(33.9\times 10^{-6} m^2/s)(10 min \times 60 s/min)}{(0.015m)^2}
\\\\\tau=90.4>0.2

Therefore, the one-term approximate solution (or the transient temperature charts) is applicable.

Then the temperature at the surface of the plates becomes

\theta(L,t)_{wall}=\frac{T(x,t)-T_{\infty}}{(T_i-T_{\infty})}\\\\\theta(L,t)_{wall}=A_1e^{-\lambda_1^2\tau}\cos(\lambda_1L/L)\\\\\theta(L,t)_{wall}=(1.0018)e^{-(0.1039^2(90.4))}\cos(0.1039)\\\\\theta(L,t)_{wall}=0.378\\\\\frac{T(L,t)-700}{25-700}=0.378\\\\T(L,t)=445°C

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2 years ago
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