A line source of strength $15 π m / s$ is situated within a uniform stream flowing at -12 m/s (i.e. right to left). At a distance of 0.6 m downstream from the source is an equal sink. How far will the stagnation points be from the nearest source/sink?

0.38 m

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0.46 m

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0.52 m

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0.58 m

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Solution

The correct option is A 0.38 m
(a)

$G i v e n, a = \frac{0.6}{2} = 0.3 m$

u = 12 m /sec

$N o w, r = a \sqrt{\frac{Q}{π a U_{\infty}} + 1}$

$0.3 = \sqrt{\frac{15 π}{π (0.3) (12)} + 1}$
= 0.6819 m

So, r- a = 0.6819 - 0.3

= 0.381 m

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A line source of strength 15π m/s is situated within a uniform stream flowing at -12 m/s (i.e. right to left). At a distance of 0.6 m downstream from the source is an equal sink. How far will the stagnation points be from the nearest source/sink?

The correct option is A 0.38 m(a) Given, a=0.62=0.3m u = 12 m /sec Now,r=a√QπaU∞+1 0.3=√15ππ(0.3)(12)+1 = 0.6819 m So, r- a = 0.6819 - 0.3 = 0.381 m

A line source of strength $15 π m / s$ is situated within a uniform stream flowing at -12 m/s (i.e. right to left). At a distance of 0.6 m downstream from the source is an equal sink. How far will the stagnation points be from the nearest source/sink?

The correct option is A 0.38 m
(a)

$G i v e n, a = \frac{0.6}{2} = 0.3 m$

u = 12 m /sec

$N o w, r = a \sqrt{\frac{Q}{π a U_{\infty}} + 1}$

$0.3 = \sqrt{\frac{15 π}{π (0.3) (12)} + 1}$
= 0.6819 m

So, r- a = 0.6819 - 0.3

= 0.381 m