Tuesday, September 6, 2011

MrsDrPoe: Conservation of Momentum, Part VI

Good morning to you all and welcome to Thesis Tuesday on the blog!  Today we'll be discussing the last segment of the conservation of momentum equation: an example using the inviscid reduction of its differential form (i.e. the Bernoulli Equation).

Example: Water flows from the faucet on the first floor of the building shown in the figure below with a maximum velocity of 20 ft/s.  For steady, inviscid flow, determine the maximum water velocity from the basement faucet and from the faucet on the second floor (assume each floor is 12 feet tall).

Given: V1max = 20 ft/s, hf = 12 ft, hgf = 4 ft, gammawater = 62.4 lbf/(ft*ft*ft), rhowater = 1.94 slugs/(ft*ft*ft)

Find: V2max, Vbmax

Assumptions: inviscid, steady, incompressible flow along a streamline, no outside forces, water exits faucet as a free jet


Because of our assumptions, we can immediately employ the Bernoulli equation between each set of two points without consideration for the continuity equation, etc.

a) Between floors 1 and 2

p1 + 0.5*rhowater*V1*V1 + gammawater*z1 = p2 + 0.5*rhowater*V2*V2 + gammawater*z2

We can rearrange the equation algebraically to solve for V2 (the velocity from the second floor faucet):

V2 = sqrt((p1 - p2 + 0.5*rhowater*V1*V1 + gammawater*z1 - gammawater*z2)/0.5*rhowater)

Since water from the faucets exits like a free jet, we are essentially saying that the pressure at each exit is atmospheric (0 gage pressure); thus:

p1 = p2 = 0 psi

Since we are given information about the height between the floors:

z1 = 0 ft, z2 = 12 ft

Plugging these and our Given values in to our V2 expression, we find that the maximum velocity from the second floor faucet is: 19.28i ft/s.

The i indicates that this number is an imaginary number; since this number is impossible to achieve physically, we know that water will not flow from the upstairs faucet. 

b) Between floor 1 and the basement 

To determine the velocity from the basement faucet, we can use the manipulated equation above for V2, replacing the subscript 2 with the subscript b (i.e. using the basement values instead of the second floor values):

Vb = sqrt((p1 - pb + 0.5*rhowater*V1*V1 + gammawater*z1 - gammawater*zb)/0.5*rhowater)

Again the free jet assumption tells us:

p1 = pb = 0 psi

And the height between floors information yields:

z1 = 0 ft, zb = -12 ft

Plugging in all our values, we find that the maximum velocity from the basement faucet is: 34.23 ft/s.

That concludes our discussion on momentum.  You should now have the basic tools necessary to work a problem by employing the integral form of the momentum equation, the viscous differential form (Navier Stokes) of the momentum equation, and the inviscid differential form (Bernoulli Equation) of the momentum equation.  Feel free to revisit any of the previous equation derivations or example problems here, and until next time...happy studying!


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