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d1i1m1o1n [39]
1 year ago
10

____ grinders are used to grind diameters, shoulders, and faces much like the lathe is used for turning, facing, and boring oper

ations.
Engineering
1 answer:
skelet666 [1.2K]1 year ago
8 0

Answer:

Cylindrical

Explanation:

<em>A cylindrical grinder </em><em>is a tool for shaping the exterior of an item. Although cylindrical grinders may produce a wide range of forms, the item must have a central axis of rotation. Shapes such as cylinders, ellipses, cams, and crankshafts are examples of this.</em><em> Cylindrical grinding</em><em> machines are specialized grinding machines that are used to process cylinders, rods, and similar workpieces. The cylinders revolve in one direction between two centers, while the grinding wheel or wheels are close together and rotate in the other direction.</em>

You might be interested in
A hollow aluminum sphere, with an electrical heater in the center, is used in tests to determine the thermal conductivity of ins
Stella [2.4K]

Answer:

K_{ins}=\frac {0.157892}{2.854263}=0.055318 W/m.K

Explanation:

Generally, thermal resistance for conduction heat transfer in a sphere.

R_{cond} = \frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi K}}  

Where R_{cond} is the thermal resistance for conduction, K is the thermal conductivity of the material, r_{i} is the inner radius of the sphere, and r_{o} is the outer radius of the sphere.

The surface area of sphere, A_{s} is given by

A_{s}=4\pi {r^2}

For aluminum sphere, the thermal resistance for conductive heat transfer is given by

Calculate the thermal resistance for conductive heat transfer through the aluminum sphere.

R_{cond,s{\rm{ - 1}}} = \frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi {K_{Al}}}}

Where K_{Al} is aluminum’s thermal conductivity at T_{s}

Thermal resistance for conductive heat transfer through the insulation.

R_{cond,1{\rm{ - 2}}} = \frac{{\left( {1/{r_o}} \right) - \left( {1/r} \right)}}{{4\pi {K_{ins}}}}

Thermal resistance for convection is given by

R_{conv} = \frac{1}{{hA}}

Where h is convective heat transfer coefficient, R_{conv} is thermal resistance for convection and A is the cross-sectional area normal to the direction of flow of heat energy

Thermal resistance for convective heat transfer in-between the outer surface of the insulation and the ambient air.

R_{conv,2{\rm{ - }}\infty } = \frac{1}{{h{A_s}}}

Where h represents convective heat transfer coefficient at the outer surface of the insulation. Since A_{s} is already defined, substituting it into the above formula yields

R_{conv,2{\rm{ - }}\infty } = \frac{1}{{h\left( {4\pi {r^2}} \right)}}

To obtain radial distance of the outer surface of the insulation from the center of the sphere.

r = r_{o} + t where t is thickness of insulation

r=0.21+0.15=0.36m

Total thermal resistance

R_{eq} = {R_{cond,s{\rm{ - 1}}}} + {R_{cond,1{\rm{ - 2}}}} +{R_{conv,2{\rm{ - }}\infty }}

Where R_{eq} is total thermal resistance

R_{eq} = \frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi {K_{Al}}}} + \frac{{\left( {1/{r_o}} \right) - \left( {1/r} \right)}}{{4\pi {K_{ins}}}} + \frac{1}{{h\left( {4\pi {r^2}} \right)}}

Consider the thermal conductivity of aluminum at temperature T_{s} as 234W/m.K

Rate of heat transfer for the given process

\dot Q_{s - \infty } = \frac{{{T_s} - {T_\infty }}}{{{R_{eq}}}}

Where \dot Q_{s - \infty }} is the steady state heat transfer rate in-between the inner surface of the sphere and the ambient air.

Substituting \left( {\frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi {K_{Al}}}} + \frac{{\left( {1/{r_o}} \right) - \left( {1/r} \right)}}{{4\pi {K_{ins}}}} + \frac{1}{{h\left( {4\pi {r^2}} \right)}}} \right) for R_{eq} we obtain

\dot Q_{s - \infty } = \frac{{{T_s} - {T_\infty }}}{{\left( {\frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi {K_{Al}}}} + \frac{{\left( {1/{r_o}} \right) - \left( {1/r} \right)}}{{4\pi {K_{ins}}}} + \frac{1}{{h\left( {4\pi {r^2}} \right)}}} \right)}}

\begin{array}{l}\\80{\rm{ W}} = \frac{{250{\rm{ }}^\circ {\rm{C}} - 20{\rm{ }}^\circ {\rm{C}}}}{{\left( {\frac{{\left( {\frac{1}{{0.18{\rm{ m}}}}} \right) - \left( {\frac{1}{{0.21{\rm{ m}}}}} \right)}}{{4\pi \left( {234{\rm{ W/m}} \cdot {\rm{K}}} \right)}} + \frac{1}{{30{\rm{ W/}}{{\rm{m}}^2} \cdot {\rm{K}}\left( {4\pi {{\left( {0.36{\rm{ m}}} \right)}^2}} \right)}}\frac{{\left( {\frac{1}{{0.21{\rm{ m}}}}} \right) - \left( {\frac{1}{{0.36{\rm{ m}}}}} \right)}}{{4\pi {K_{ins}}}} + } \right)}}\\\\80{\rm{ W}}\left( {{\rm{0}}{\rm{0.020737 K/W}} + \frac{{{\rm{0}}{\rm{0.157892/m}}}}{{{K_{ins}}}}} \right) = 230{\rm{ K}}\\\\\frac{{{\rm{0}}{\rm{0.157892/m}}}}{{{K_{ins}}}} = \frac{{230{\rm{ K}}}}{{80{\rm{ W}}}} - {\rm{0}}{\rm{0.020737 K/W}}\\\\\frac{{{\rm{0}}{\rm{0.157892/m}}}}{{{K_{ins}}}} = {\rm{2}}{\rm{.854263 K/W}}\\\end{array}

K_{ins}=\frac {0.157892}{2.854263}=0.055318 W/m.K

7 0
3 years ago
Given two alphabet strings str1 and str2. You can change the characters in str1 to any alphabet characters in order to transform
Keith_Richards [23]

Answer:

Explanation:

5

6 0
3 years ago
A thermoelectric refrigerator is powered by a 16-V power supply that draws 2.9 A of current. If the refrigerator cools down 3.1
Viktor [21]

Answer:

COP = 0.090

Explanation:

The general formula for COP is:

COP = Desired Output/Required Input

Here,

Desired Output = Heat removed from water while cooling

Desired Output = (Specific Heat of Water)(Mass of Water)(Change in Temperature)/Time

Desired Output = [(4180 J/kg.k)(3.1 kg)(25 - 11)k]/[(12 hr)(3600 sec/hr)]

Desired Output = 4.199 W

And the required input can be given as electrical power:

Required Input = Electrical Power = (Current)(Voltage)

Required Input = (2.9 A)(16 V) = 46.4 W

Therefore:

COP = 4.199 W/46.4 W

<u>COP = 0.090</u>

8 0
2 years ago
What the phat is this
Alex17521 [72]

Answer:

It's Brainly ;)

8 0
3 years ago
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Drum brakes are usually designed so that the condition of the lining can be checked even if the drum has not been
artcher [175]

Answer:

no it has to be removed

Explanation:

8 0
2 years ago
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