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.
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 \(34\,m/s\).
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,
\[{C_p} = \frac{{{C_{p0}}}}{{\sqrt {1 – M_\infty ^2} }} \Rightarrow\, – 0.1384 = \frac{{{C_{p0}}}}{{\sqrt {1 – {{\left( {0.694} \right)}^2}} }} \Rightarrow {C_{p0}} = \,-0.099644\]
