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ivann1987 [24]
3 years ago
8

Circular fins are employed around the cylinder of a lawn mower engine to dissipate heat. The fins are made of aluminum are 0.3-m

thick, and extend 2 cm from base to tip. The outside diameter of the engine cylinder is 0.3 m. Design operating conditions are Tai 30°C and h-12 Wm2.K. The maximum allowable cylinder temperature is 300°C. Estimate the amount of heat transfer from a single fin. How many fins are required to cool a 3-kw engine operating at 30% thermal efficiency, if 50% of the total heat given off is transferred by the fins?

Engineering
1 answer:
frosja888 [35]3 years ago
8 0

Answer: heat through each fin is 291W

Total no of required fin from design calculations is 3

Explanation:

The detailed calculations and explanation is shown in the image below.

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Which statement about tensile stress is true? A. Forces that act perpendicular to the surface and pull an object apart exert a t
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Answer:

A. Forces that act perpendicular to the surface and pull an object apart exert a tensile stress on the object.

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The tensile stress is a type of normal stress, in which a perpendicular force creates the stress to an object’s surface.

Hence, the correct option is "A."

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Answer:

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8 0
3 years ago
Twenty six million gallons per day of wastewater with a DO of 1.00 mg/L is discharged into a river with a DO of 6.00 mg/L. If th
laiz [17]

Answer:

oxygen deficit = 3.851 mg/L

Explanation:

given data

flow rate of the river=  165 × 10^{6} gal/d

saturation value of dissolved oxygen = 9.17 mg/L

to find out

oxygen deficit he two flows

solution

we will apply here formula for dissolved oxygen content after dilution is

Do mix = \frac{Qw*(Do)w +Qr*(Do)r}{Qw+Qr}     ..........................1

here Qw is rate of flow of waste water  i.e 26 ×10^{6} gal/d

(Do)w is Do of waste waster i.e 1 mg/L

Qr is aret of flow of river i.e 165 ×10^{6} gal/d

(Do)r is do of river water i.e 6 mg/L

so put all value in equation 1 we get

Do mix = \frac{26*10^6*1 +165*10^6*6}{(26+165)10^6}

solve we get

Do mix = 5.319 mg/L

so

oxygen deficit =  saturation oxygen - (Do) mix     ..............2

oxygen deficit = 9.17 - 5.319

oxygen deficit = 3.851 mg/L

8 0
3 years ago
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