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sweet-ann [11.9K]
3 years ago
15

A 0.39 percent Carbon hypoeutectoid plain-carbon steel is slowly cooled from 950 oC to a temperature just slightly below 723 oC.

Calculate the weight percent proeutectoid ferrite in the steel. Please do not include any numbers after the decimal and give the percentage as "xx" and not "0.xx"
Engineering
1 answer:
belka [17]3 years ago
8 0

Answer:

53%

Explanation:

To find the weight % in proeutectoid ferrite we use:

Wt % in proeutectoid ferrite=

[(0.80 - 0.39) / (0.80 - 0.02)] * 100=

= (0.41 /0.78) * 100 =

52.56%

which is approximately 53%

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Air is used as the working fluid in a simple ideal Brayton cycle that has a pressure ratio of 12, a compressor inlet temperature
7nadin3 [17]

Answer:

A) m' = 351.49 kg/s

B) m'= 1036.91 kg/s

Explanation:

We are given;

Pressure Ratio;r_p = 12

Inlet temperature of compressor;T1 = 300 K

Inlet temperature of turbine;T3 = 1000 K

cp = 1.005 kJ/kg·K

k = 1.4

Net power output; W' = 70 MW = 70000 KW

A) Now, the formula for the mass flow rate using the total power output of the compressor and turbine is given as;

m' = W'/[cp(T3(1 - r_p^(-(k - 1)/k)) - T1(r_p^((k - 1)/k))

At, 100% efficiency, plugging in the relevant values, we have;

m' = 70000/(1.005(1000(1 - 12^(-(1.4 - 1)/1.4)) - 300(12^((1.4 - 1)/1.4)))

m' = 70000/199.1508

m' = 351.49 kg/s

B) At 85% efficiency, the formula will now be;

m' = W'/[cp(ηT3(1 - r_p^(-(k - 1)/k)) - (T1/η) (r_p^((k - 1)/k))

Where η is efficiency = 0.85

Thus;

m' = 70000/(1.005(0.85*1000(1 - 12^(-(1.4 - 1)/1.4)) - (300/0.85)(12^((1.4 - 1)/1.4)))

m' = 70000/(1.005*(432.09129 - 364.9189)

m'= 1036.91 kg/s

4 0
3 years ago
Wqqwfqwfqwfqfqfqffqwffqwqfqqfqfqffqqfqfwccc
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Answer:

?

Explanation:

4 0
3 years ago
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Musya8 [376]
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5 0
3 years ago
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Why is the lubrication system of an internal combustion engine equipped with an oil filter?
Ierofanga [76]

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6 0
3 years ago
Determine the combined moment about O due to the weight of the mailbox and the cross member AB. The mailbox weighs 3.2 lb and th
koban [17]

Answer:

Attached is the complete question but the weight of the mailbox and cross bar differs from the given values which are : weight of mail box = 3.2 Ib, weight of the uniform cross member = 10.3 Ib

Answer : moment of inertia = 186.7 Ib - in

Explanation:

Given data

weight of the mailbox = 3.2 Ib

weight of the uniform cross member = 10.3 Ib

The origin is of mailbox and cross member is 0

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= 4 + (25/2) = 16.5"

The centroid of the bar =( ( 1 + 25 + 4 + 4 ) / 2 ) - 4  = 13"

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8 0
4 years ago
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