# Find the pressure coefficient at a point on wing placed in a wind tunnel.

The velocity at a point on the wing placed in a wind tunnel is $$251 m/s$$. The pressure and temperature of the test section is $$1\,atm$$ and $$285\,K$$. If the velocity of the flow in the test section is $$235\;m/s$$, find the pressure coefficient at the point.

Find the pressure coefficient at the same point if the velocity of flow in the test section is reduced to $$34\,m/s$$.

• 1 Answer(s)

We need to calculate the Mach number of the flow. Speed of sound,

${a_\infty } = \sqrt {\gamma R{T_\infty }} = \sqrt {1.4 \times 287 \times 285} = 338.398\,m/s$

Therefore, Mach number of the flow is

${M_\infty } = \frac{{{V_\infty }}}{{{a_\infty }}} = \frac{{235}}{{338.398}} = 0.694$

Mach number of $$0.694$$ shows that the flow is compressible. Density of flow is obtained on applying the equation of state.

${p_\infty } = {\rho _\infty }R{T_\infty } \Rightarrow {\rho _\infty } = \frac{{{p_\infty }}}{{R{T_\infty }}} = \frac{{1.01 \times {{10}^5}}}{{287 \times 285}} = 1.235\,kg/{m^3}$

Temperature at the point on the wing can be calculated from the energy equation,

${C_p}T + \frac{{{V^2}}}{2} = {C_p}{T_\infty } + \frac{{V_\infty ^2}}{2}$

$$C_p$$ for air is

${C_p} = \frac{{\gamma R}}{{\gamma – 1}} = \frac{{1.4 \times 287}}{{1.4 – 1}} = 1004.5\,kJ/kg.K$

Therefore,

$\left( {1004.5 \times T} \right) + \frac{{{{\left( {251} \right)}^2}}}{2} = \left( {1004.5 \times 285} \right) + \frac{{{{\left( {235} \right)}^2}}}{2}$

$\Rightarrow 1004.5T + \frac{{{{\left( {251} \right)}^2}}}{2} = 286282.5 + \frac{{{{\left( {235} \right)}^2}}}{2}$

$\Rightarrow 1004.5T = 286282.5 + \frac{{\left( {{{235}^2} – {{251}^2}} \right)}}{2}$

$\Rightarrow T = 281.129\,K$

Pressure at the point will be calculated as

$\frac{p}{{{p_\infty }}} = {\left( {\frac{T}{{{T_\infty }}}} \right)^{\frac{\gamma }{{\gamma – 1}}}}$

$\Rightarrow p = 1.01 \times {10^5}{\left( {\frac{{281.129}}{{285}}} \right)^{\frac{{1.4}}{{1.4 – 1}}}} = 96279.568\,N/{m^2}$

Therefore, coefficient of pressure at the point will be,

${C_p} = \frac{{p – {p_\infty }}}{{{q_\infty }}} = \frac{{96279.568 – 1.01 \times {{10}^5}}}{{\frac{1}{2} \times 1.235 \times {{\left( {235} \right)}^2}}} = \, – 0.1384$

When the velocity of flow is reduced to $$34\,m/s$$, It is a low speed flow. Therefore, pressure coefficient at the same point can be calculated from Prandtl-Glauert rule,

${C_p} = \frac{{{C_{p0}}}}{{\sqrt {1 – M_\infty ^2} }} \Rightarrow\, – 0.1384 = \frac{{{C_{p0}}}}{{\sqrt {1 – {{\left( {0.694} \right)}^2}} }} \Rightarrow {C_{p0}} = \,-0.099644$

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