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Kruka [31]
4 years ago
6

WANT POINTS? JUST ANSWER ME:)

Engineering
2 answers:
bulgar [2K]4 years ago
7 0

Answer:

mcfndvkernfnkvfvjnfdkvfv

Explanation:

Lapatulllka [165]4 years ago
6 0

Answer:

yes why thank you

snarfs :)

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How does heat conduction differ from convection?
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3 years ago
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Atmospheric air at 25 °C and 8 m/s flows over both surfaces of an isothermal (179C) flat plate that is 2.75m long. Determine the
vekshin1

Answer:

Re=100,000⇒Q=275.25 \frac{W}{m^2}

Re=500,000⇒Q=1,757.77\frac{W}{m^2}

Re=1,000,000⇒Q=3060.36 \frac{W}{m^2}

Explanation:

Given:

For air      T_∞=25°C  ,V=8 m/s

  For surface T_s=179°C

     L=2.75 m    ,b=3 m

We know that for flat plate

Re⇒Laminar flow

Re>30\times10^5⇒Turbulent flow

<u> Take Re=100,000:</u>

 So this is case of laminar flow

  Nu=0.664Re^{\frac{1}{2}}Pr^{\frac{1}{3}}

From standard air property table at 25°C

  Pr= is 0.71  ,K=26.24\times 10^{-3}

So    Nu=0.664\times 100,000^{\frac{1}{2}}\times 0.71^{\frac{1}{3}}

Nu=187.32   (\dfrac{hL}{K_{air}})

187.32=\dfrac{h\times2.75}{26.24\times 10^{-3}}

     ⇒h=1.78\frac{W}{m^2-K}

heat transfer rate =h(T_∞-T_s)

                           =275.25 \frac{W}{m^2}

<u> Take Re=500,000:</u>

So this is case of turbulent flow

  Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}

Nu=0.037\times 500,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}

Nu=1196.18  ⇒h=11.14 \frac{W}{m^2-K}

heat transfer rate =h(T_∞-T_s)

                             =11.14(179-25)

                           = 1,757.77\frac{W}{m^2}

<u> Take Re=1,000,000:</u>

So this is case of turbulent flow

  Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}

Nu=0.037\times 1,000,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}

Nu=2082.6  ⇒h=19.87 \frac{W}{m^2-K}

heat transfer rate =h(T_∞-T_s)

                             =19.87(179-25)

                           = 3060.36 \frac{W}{m^2}

7 0
3 years ago
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