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4 years ago

Give two methods on how powder is produced in powder metallurgy.

Engineering

2 answers:

Katen [24]4 years ago

7 0

Answer:

Explanation:

<u>Atomization using gas stream</u>

Molten metal is forced through a small orifice and is shatter by a jet of compressed air,inert gas .

In Atomization, the particles shape is analysed by the rate of solidification and varies from spherical to highly irregular shape.

Reduction

oxide of metals are transformed to pure metal powder when undefended to under melting point gases results in a product of spongy material.

It is used for Iron,copper,tungsten,Nickel etc.

daser333 [38]4 years ago

7 0

Answer:

(1)Atomizing process

(2)Gaseous reduction

Explanation:

The first step in powder metallurgy is the production of powder,because the property of the final product depends on the powder.

The methods for the production of powder are as follow

(1)Atomizing process

(2)Gaseous reduction

(1)Atomizing process:

In the Atomizing process the molten metal is passing through an orifice into a stream of inert gas.Due to this rapid cooling of metal occurs and then it will in very fine particle .

(2)Gaseous reduction:

In this process powder is producing by grinding of metallic oxide to a fine state,after that reducing it by carbon mono oxide.

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kari74 [83]

Answer:

The minimum diameter for each cable should be 0.65 inches.

Explanation:

Since, the load is supported by two ropes and the allowable stress in each rope is 1500 psi. Therefore,

(1/2)(Weight/Cross Sectional Area) = Allowable Stress

Here,

Weight = 1000 lb

Cross-sectional area = πr²

where, r = minimum radius for each cable

(1/2)(1000 lb/πr²) = 1500 psi

500 lb/1500π psi = r²

r = √1.061 in²

r = 0.325 in

Now, for diameter:

Diameter = 2(radius) = 2r

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<u>Diameter = 0.65 in</u>

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Refrigerant-134a at 700 kPa, 70°C, and 7.2 kg/min is cooled by water in a condenser until it exists as a saturated liquid at the

alex41 [277]

Answer:

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Explanation:

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$\dot Q_{ref} = \dot Q_{w}$

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The mass flow rate of the cooling water is now cleared:

$\dot m_{w} = \dot m_{ref }\cdot \frac{h_{ref,in}-h_{ref,out}}{h_{w,out}-h_{w,in}}$

Given that $h_{ref,in} = 808.34\,\frac{kJ}{kg}$ , $h_{ref, out} = 88.82\,\frac{kJ}{kg}$ , $h_{w,out} = 104.83\,\frac{kJ}{kg}$ and $h_{w,in} = 62.98\,\frac{kJ}{kg}$ , the mass flow of the cooling water is:

$\dot m_{w} = \left(7.2\,\frac{kg}{min} \right)\cdot \left(\frac{808.34\,\frac{kJ}{kg}-88.82\,\frac{kJ}{kg} }{104.83\,\frac{kJ}{kg}-62.98\,\frac{kJ}{kg} } \right)$