Question

A body is shaped as a portion of a half-body as shown in the figure 2 below. Air flows over this half –body and the value of the stream function at the stagnation point is (ψ_st = 15.7 )

it ρ_air = 1.23 kg/m³, determine:

a) the value of ( b) , see the figure

b) the pressure difference p₂ – p₃

_{A body is shaped as a portion of a half-body as shown in the the stream funetion at the stagnation point is Ca 15.7) it . Dair 1.23 kg/m3, determine: a) the value of (b), see the figure b) the pressure difference p2 p3 10.45 m U-90 km/hr 2 0.6 m}

Answer 1

SOLUTION: -

Given- U = 90 km/hr = 90 x 10³/3600 = 25 m/s

= 15.7

$\large \rho_{air}$ = 1.23 kg/m³

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a) Calculation:- for the value of b, we can find this value by-

let us consider,

$\large \theta = \pi$

at this position, $\large \psi_s_t$ = m/2

so that, m = 15.7 x 2 = 31.4 Kg

Now we can find the value of b, by the formula-

$\large \Rightarrow b = \frac {m}{2\pi U}$

now substitute all the values and we get -

  $\large \Rightarrow b = \frac {31.4}{2\pi(90 *10^3)}$

b = 0.06 m

So form the calculation we have the value of b is "0.06m".

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b) Calculation:- now we have to calculate the pressure difference P₂ - P₃ by -

now we can find the value of r by-

$\large \Rightarrow r = \sqrt {0.45^2 + (0.6 - 0.06)^2}$

$\large \Rightarrow$ r = 0.702 m

and we can find the angle as-

$\large \Rightarrow tan \theta =\frac{0.45}{0.6 - 0.06}$

so that, tan $\large \theta$ = 0.833 $\large \Rightarrow \theta$ = 39.8

now we can find the value of V as -

$\large V^2 = U^2( 1+2\frac{b}{r}cos \theta + \frac {b^2}{r^2})$

Substituting all the values and we get -

V = 26.6 m/s = 96 km/hr

so we can find the pressute difference by-

$\large P_2 - P_3 = \frac{1}{2} \rho (V^2 - U^2)$

By putting all the values we get-

P₂-P₃ = 50,774. 4 N/m² = 50.7 Kpa

therefore the finally preesure difference P₂ - P₃ = 50.7 KPa .