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-BARSIC- [3]
3 years ago
9

The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp

er surface: Cp constant at – 0.8 from the leading edge to 60% chord, then increasing linearly to +0.1 at the trailing edge. Lower surface: Cp constant at – 0.4 from the leading edge to 60% chord, then increasing linearly to + 0.1 at the trailing edge. Estimate the lift coefficient and the pitching moment coefficient about the leading edge due to lift.
Engineering
1 answer:
Aliun [14]3 years ago
5 0

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1

The Lower Surface Cp is given as

Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1

The difference of the Cp over the airfoil is given as

\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32

Now the Lift Coefficient is given as

C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192

Now the coefficient of moment about the leading edge is given as

C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

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Answer:

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Refrigerant-134a

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v_{w} =\frac{m}{\rho \pi (\frac{d_{2}^{2}-d_{1}^{2}  }{4} )} =\frac{0.3}{998*\pi (\frac{0.025^{2}-0.01^{2}  }{4}) } =0.729m/s

It is necessary to get the Reynold's number:

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The overall heat transfer coefficient:

Q=\frac{1}{\frac{1}{h_{1} }+\frac{1}{h_{2} }  }

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h_{2} =\frac{kNu}{d_{2}-d_{1}} =\frac{0.598*85.0517}{0.025-0.01} =3390.7278W/m^{2}C

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Q=\frac{1}{\frac{1}{4100}+\frac{1}{3390.7278}  } =1855.8923W/m^{2} C

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Answer:

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