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3 years ago

What are the main factors for the high material removal rate in broaching operations?

Engineering

1 answer:

Umnica [9.8K]3 years ago

6 0

Answer:

Explanation:

Broaching:

Broaching is a metal removal process or we can say that it is a machining process.Generally broach have three types teeth in which first one is called roughing teeth,second one semi finishing and third one finishing teeth.By using these teeth metal remove so fast and that is why broaching is the high metal removal rate operation.

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A vacuum gage connected to a tank reads 30 kPa at a location where the barometric reading is 755 mmHg. Determine the absolute pr

navik [9.2K]

Answer:

Absolute pressure=70.72 KPa

Explanation:

Given that Vacuum gauge pressure= 30 KPa

Barometer reading =755 mm Hg

We know that barometer always reads atmospheric pressure at given situation.So atmospheric pressure is equal to 755 mm Hg.

We know that P= ρ g h

Density of $Hg=13600 \frac{kg}{m^3}$

So P=13600 x 9.81 x 0.755

P=100.72 KPa

We know that

Absolute pressure=atmospheric pressure + gauge pressure

But here given that 30 KPa is a Vacuum pressure ,so we will take it as negative.

Absolute pressure=atmospheric pressure + gauge pressure

Absolute pressure=100.72 - 30 KPa

Absolute pressure=70.72 KPa

8 0

3 years ago

A room is kept at −5°C by a vapor-compression refrigeration cycle with R-134a as the refrigerant. Heat is rejected to cooling wa

Fed [463]

Answer:

note:

<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</u>

Download docx

4 0

3 years ago

Read 2 more answers

Sawing stock to reduce its thickness is known as __________ .

rjkz [21]

Answer:

resawing

Explanation:

8 0

4 years ago

Refrigerant-134a enters a 28-cm-diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. The refrigerant gains heat

Alexandra [31]

Answer:

V = 0.30787 m³/s

m = 2.6963 kg/s

v2 = 0.3705 m³/s

v2 = 6.017 m/s

Explanation:

given data

diameter = 28 cm

steadily =200 kPa

temperature = 20°C

velocity = 5 m/s

solution

we know mass flow rate is

m = ρ A v

floe rate V = Av

m = ρ V

flow rate = V = $\frac{m}{\rho}$

V = Av = $\frac{\pi}{4} * d^2 * v1$

V = $\frac{\pi}{4} * 0.28^2 * 5$

V = 0.30787 m³/s

and

mass flow rate of the refrigerant is

m = ρ A v

m = ρ V

m = $\frac{V}{v}$ = $\frac{0.30787}{0.11418}$

m = 2.6963 kg/s

and

velocity and volume flow rate at exit

velocity = mass × v

v2 = 2.6963 × 0.13741 = 0.3705 m³/s

and

v2 = A2×v2

v2 = $\frac{v2}{A2}$

v2 = $\frac{0.3705}{\frac{\pi}{4} * 0.28^2}$

v2 = 6.017 m/s

7 0

3 years ago

64A geothermal pump is used to pump brine whose density is 1050 kg/m3at a rate of 0.3 m3/s from a depth of 200 m. For a pump eff

grin007 [14]

Answer:

835,175.68W

Explanation:

Calculation to determine the required power input to the pump

First step is to calculate the power needed

Using this formula

P=V*p*g*h

Where,

P represent power

V represent Volume flow rate =0.3 m³/s

p represent brine density=1050 kg/m³

g represent gravity=9.81m/s²

h represent height=200m

Let plug in the formula

P=0.3 m³/s *1050 kg/m³*9.81m/s² *200m

P=618,030 W

Now let calculate the required power input to the pump

Using this formula

Required power input=P/μ

Where,

P represent power=618,030 W

μ represent pump efficiency=74%

Let plug in the formula

Required power input=618,030W/0.74

Required power input=835,175.68W

Therefore the required power input to the pump will be 835,175.68W

5 0

3 years ago