line source of strength 180 L/s is placed in a uniform flow of velocity 0.1 m/s.
(a) Plot the streamlines above the x-axis.
(b) Obtain the pressure distribution on the streamline denoted by ψ = 0.

Question

Accepted Answer

(a) It is convenient to consider the uniform flow to be in the x-direction. This results in a streamline having the value ψ = 0, which may be interpreted as the boundary of a solid body.

[latex]\Psi =u y + \frac{q heta }{2\pi }[/latex] (Equation 10.12)

or

[latex]\Psi = u r \sin heta + \frac{q heta }{2\pi }[/latex]

Setting ψ successively to 0,−0.01,−0.02,…and noting that u=−0.1, the coordinates (r, 𝜃) along the streamlines can be obtained from Equation 10.12.

The pairs of coordinates (𝜃, r) are plotted to give the streamline pattern. The graphical method gives identical results and is clearly quicker; however, the mathematical approach is appropriate in computer applications where computer graph-plotting facilities are available (see Figure 10.10).
The graphical construction proceeds as follows: the uniform flow is represented by a series of equidistant parallel straight lines (defined by [latex]\psi_{u} [/latex] in Figure 10.10). Discharge between the streamlines was chosen to be 0.010 m³/s m in the analytical solution; thus, the spacing of the uniform flow streamlines in the diagram represents to scale a distance of 0.10 m. It is convenient to choose the streamlines of the source such that the discharge between them is also 0.010 m³/(s m); the angle Δ𝜃 is simply obtained since

[latex]\Delta \Psi _{s} = \frac{q \Delta heta }{2 \pi}[/latex]

that is, 0.010 = [latex]\frac{0.180 imes \Delta heta }{2 \pi}[/latex]

whence Δ𝜃 = 20°

At the points of intersection of the uniform flow and source streamlines, the individual stream functions are added algebraically and the resultant flow pattern is obtained by sketching the streamlines through points of equal resultant stream function. The construction and final flow pattern are illustrated in Figure 10.10.
(b) The pressure distribution in the disturbed flow field at (r, 𝜃) is obtained from

[latex]\frac{p_{r, heta } - p_{0}}{ ho g} = \frac{V^{2}_{0} - V^{2}_{r} heta }{2g}[/latex] (i)

where [latex]V_{0} [/latex] is the undisturbed velocity and [latex]V_{r, heta} [/latex] the velocity vector at (r, 𝜃).

[latex]V_{r, heta } = \sqrt{ u ^{2}_{r} + u ^{2}_{ heta } }[/latex]

[latex] u _{r} = \frac{1}{r} \frac{ \partial \psi }{ \partial heta }[/latex] (Equation 10.3)

[latex]= \frac{ur\cos heta }{r} + \frac{q}{2\pi r}[/latex]

[latex] u _{ heta } = − \frac{\partial \psi }{\partial r }[/latex] (Equation 10.3)

[latex]=−u\sin heta[/latex]

[latex]\Rightarrow V ^{2}_{r heta } = u^{2} + 2u\cos heta \frac{q}{2\pi r} + \left(\frac{q}{2\pi r} ight)^{2}[/latex] (ii)

Around the ‘body’ (Ψ = 0),

[latex]ur\sin heta + \frac{q heta }{2\pi } =0[/latex]

i.e. [latex]\frac{q }{2\pi } = − \frac{ur\sin heta}{ heta }[/latex]

Thus

[latex]V ^{2}_{r heta } = u^{2} \left(1- \frac{\sin 2 heta }{ heta } +\frac{\sin^{2} heta }{ heta^{2} } ight) [/latex]

whence, from Equation (i), since [latex]V_{0} [/latex] = u in this case

[latex]\frac{p_{r, heta } - p_{0}}{ ho g} = \frac{u^{2}}{2g heta } \left(\sin 2 heta - \frac{\sin^{2} heta } { heta } ight) [/latex] (iii)

Substitution of u = −0.1 m/s into Equation (iii) yields the pressure distribution for a range of 𝜃.

Question 10.1: line source of strength 180 L/s is placed in a uniform flow ...

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