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erik [133]
3 years ago
14

1. A thin-walled cylindrical pressure vessel is capped at the end and is subjected to an internal pressure (p). The inside diame

ter of the vessel is 6 ft and the wall thickness is 1.5 inch. The vessel is made of steel with tensile yield strength and compressive yield strength of 36 ksi. Determine the internal pressure required to initiate yielding according to (a) The maximum-shear-stress theory of failure, and (b) The maximum-distortion-energy theory of failure, if a factor of safety (FS) of 1.5 is desired.
Engineering
1 answer:
Vesna [10]3 years ago
6 0
I DONT KNOW OKAY UGHHH
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Determine the percent increase in the nominal moment capacity of the section in Problem 2 when including compression steel at to
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Explanation:

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Discuss the difference between a dual split and a diagonal split master cylinder.
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2 years ago
A lake with an area of 525 [acre] was monitored during a one-month period. The average inflow was 30 [cfs] for the month and the
Alex73 [517]

Answer:

\Delta S = 1581663.5ft^3

Explanation:

We need to calculate the change in storage through the changes given,

That is,

\Delta S = P+I-O-O_{seepage}-E

Where the loss are representing by,

P= Precipitation\\I= Inflow\\O= Outflow\\O_{Seepage}= Outflow by seepage\\E=Evaporation

So calculating the values we have

\Delta S = P+(I-0)-O_{seepage}-E

\Delta S = 4.25+ (30ft^3/s-27ft^3/s)-1.5in-6in

The values inside the are parenthesis need to be konverted as I note here.

(30days(24hr/1day)(3600s/1hr)(1acre.ft/43560ft^3)(1/525acres)(12in/1ft)

That is,

\Delta S = 0.83in\Delta S= (0.83in*1ft/12in)(525acres)\\\Delta S=36.31 acres.ft\\\Delta S=36.31acres.ft*(43560ft^3/acrees.ft)\\\Delta S = 1581663.5ft^3

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A steam power plant receives heat from a furnace at a rate of 280 GJ/h. (that is
krek1111 [17]

Answer:

a) \dot W_{out} = 35.278\,MW, b) \eta_{th} = 45.357\,\%

Explanation:

a) The net power output is:

\dot W_{out} = \dot Q_{in} - \dot Q_{out} - \dot Q_{loss}

\dot W_{out} = \left(280\,\frac{GJ}{h} - 145\,\frac{GJ}{h} - 8\,\frac{GJ}{h}\right)\cdot \left(\frac{1000\,MJ}{1\,GJ}\right)\cdot \left(\frac{1\,h}{3600\,s} \right)

\dot W_{out} = 35.278\,MW

b) The thermal efficiency of the power plant is:

\eta_{th} = \frac{\dot Q_{in}-\dot Q_{out} - \dot Q_{loss}}{\dot Q_{in}}\times 100\,\%

\eta_{th} = \frac{280\,\frac{GJ}{h}-145\,\frac{GJ}{h}-8\,\frac{GJ}{h} }{280\,\frac{GJ}{h} } \times 100\,\%

\eta_{th} = 45.357\,\%

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3 years ago
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