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Basics of Fluid Mechanics, 2014a

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6.2. MOMENTUM EQUATION APPLICATION 183<br />

The momentum flux is<br />

∫<br />

S c.v.<br />

ρU x U i rn dA = AρU 1j<br />

2<br />

(6.IV.b)<br />

The substituting (6.IV.b) into equation (6.IV.a) yields<br />

F f = AρU 1j<br />

2<br />

(6.IV.c)<br />

The friction can be obtained from the momentum equation in the y direction<br />

m toy g + AρU 1j 2 = F earth<br />

According to the statement <strong>of</strong> question the friction force is<br />

F f = μ d<br />

(<br />

mtoy g + AρU 1j<br />

2 )<br />

The momentum in the x direction becomes<br />

μ d<br />

(<br />

mtoy g + AρU 1j<br />

2 ) = AρU 1j 2 = Aρ (U j − U 0 ) 2<br />

The toy velocity is then<br />

√<br />

μd m toy g<br />

U 0 = U j −<br />

Aρ (1 − μ d )<br />

Increase <strong>of</strong> the friction reduce the velocity. Additionally larger toy mass decrease the<br />

velocity.<br />

End Solution<br />

6.2.1 Momentum for Unsteady State and Uniform Flow<br />

The main problem in solving the unsteady<br />

state situation is that the control<br />

volume is accelerating. A possible<br />

way to solve the problem is by expressing<br />

the terms in an equation (6.10).<br />

This method is cumbersome in many<br />

cases. Alternative method <strong>of</strong> solution<br />

is done by attaching the frame <strong>of</strong> reference<br />

to the accelerating body. One<br />

such example <strong>of</strong> such idea is associated<br />

with the Rocket <strong>Mechanics</strong> which<br />

is present here.<br />

F R<br />

m f<br />

m R<br />

U g<br />

U R<br />

Fig. -6.6. A rocket with a moving control volume.<br />

6.2.2 Momentum Application to Unsteady State<br />

Rocket <strong>Mechanics</strong>

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