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ivann1987 [24]
2 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]2 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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A vector AP is rotated about the Z-axis by 60 degrees and is subsequently rotated about X-axis by 30 degrees. Give the rotation
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Answer:

R = \left[\begin{array}{ccc}1&0&0\\0&cos30&-sin30\\0&sin30&cos30\end{array}\right]\left[\begin{array}{ccc}cos 60&-sin60&0\\sin60&cos60&60\0&0&1\end{array}\right]

Explanation:

The mappings always involve a translation and a rotation of the matrix. Therefore, the rotation matrix will be given by:

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The matrix is given by the following expression:

\left[\begin{array}{ccc}1&0&0\\0&cos30&-sin30\\0&sin30&cos30\end{array}\right]\left[\begin{array}{ccc}cos 60&-sin60&0\\sin60&cos60&60\0&0&1\end{array}\right]

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Air is saturated with water vapor at 35.0 oC and a total pressure of 1.50 atmospheres. If the molar flow rate of the dry air in
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Answer:

11.541 mol/min

Explanation:

temperature = 35°C

Total pressure = 1.5 * 1.013 * 10^5 = 151.95 kPa

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from steam table it is = 5.6291 Kpa

calculate the mole fraction of H_{2}o ( YH_{2}o )

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calculate the mole fraction of air ( Yair )

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= 1 - 0.03704 = 0.9629

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lets assume N = Total molar flow rate

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Nair = molar  flow rate of air = 300 moles /min

note : Yair * n = Nair

therefore n = 300 / 0.9629 = 311.541  moles /min

Molar flowrate of water

=  n -  Nair

= 311.541 - 300 = 11.541 mol/min

4 0
3 years ago
21.Why are throttling devices commonly used in refrigeration and air-conditioning<br> applications?
Sloan [31]

Answer is given below

Explanation:

we know that some common types of throttling devices are

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so here throttling devices commonly used in refrigeration and air-conditioning because

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