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slamgirl [31]
3 years ago
7

A bearing is to be used as shaft support to carry both radial and thrust forces. Calculations show that a 02 series ball bearing

with a bore size of 20 mm would work well for this application. What are the static and dynamic load ratings of this bearing in kN
Engineering
1 answer:
shepuryov [24]3 years ago
4 0

Answer:

The correct answer is "20.8 kN" and "31 kN". A further explanation is given below.

Explanation:

  • The angular touch bearing seems to be a fine replacement while accommodating radial and even some displacement pressures.
  • You may receive static as well as dynamic scores from either the manufacturer's collections.  

The load ratings should be for the SKF bearing including its predetermined distance:

Static load

= 20.8 kN

Dynamic load

= 31 kN

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Propane is to be compressed from 0.4 MPa and 360 K to 4 MPa using a two-stage compressor. An interstage cooler returns the tempe
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Answer:

a. 81 kj/kg

b. 420.625K

c.  101.24kj/kg

Explanation:

\frac{t2}{t1} =[\frac{p2}{p1} ]^{\frac{y-1}{y} }

t1 = 360

p1 = 0.4mpa

p2 = 1.20

y = 1.13

substitute these values into the equation

\frac{t2}{360} =[\frac{1.20}{0.4} ]^{\frac{1.13-1}{1.13} }

\frac{t2}{360} =[\frac{1.2}{0.4} ]^{0.1150}\\\frac{t2}{360} =1.1347

when we cross multiply

t2 = 360 * 1.1347

= 408.5

a. the work required in the firs compressor

w=c(t2-t1)

c=1.67x10³

t1 = 360

t2 = 408.5

w = 1670(408.5-360)

= 1670*48.5

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b. n=\frac{t2-t1}{t'2-t1}

n = 80%

t2 = 408.5

t1 = 360

0.80 = 408.5-360 ÷ t'2-360

0.80 =\frac{48.5}{t'2-360}

cross multiply to get the value of t'2

0.80(t'2-360) = 48.5

0.80t'2 - 288 = 48.5

0.8t'2 = 48.5+288

0.8t'2 = 336.5

t'2 = 336.5/0.8

= 420.625

this is the temperature at the exit of the first compressor

c. cooling requirement

w = c(t2-t1)

= 1.67x10³(420.625-360)

= 1670*60.625

= 101243.75

= 101.24kj/kg

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