Query 07

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course PHY 202

3/17 about 2:30 pm

007. `query 6

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Question: query introset How do we find the change in pressure due to diameter change given the original velocity of the flow and pipe diameter and final diameter?

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Your Solution:

Final velocity can be found by

(v2/v1) = (d1/d2)^2

V2 = v1*(d1/d2)^2

Assume h is constant, ‘rho*g*h can be dropped.

.5*’rho*v1^2 + P1 = .5*’rho*v2^2 + P2

‘dP = .5*”rho*v1^2 - .5*’rho*(v1*(d1/d2)^2)

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Given Solution:

** The ratio of velocities is the inverse ratio of cross-sectional areas.

Cross-sectional area is proportional to square of diameter. So velocity is inversely proportional to cross-sectional area:

v2 / v1 = (A1 / A2) = (d1 / d2)^2 so

v2 = (d1/d2)^2 * v1.

Since h presumably remains constant we have

P1 + .5 rho v1^2 = P2 + .5 rho v2^2 so

(P2 - P1) = 0.5 *rho (v1^2 - v2^2) . **

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Question: query video experiment terminal velocity of sphere in fluid. What is the evidence from this experiment that the drag force increases with velocity?

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Your Solution:

As more weight is added, the rate of change in the velocity decreases for the additional increments of mass added.

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Given Solution:

** When weights were repetitively added the velocity of the sphere repetitively increased. As the velocities started to aproach 0.1254 m/sec the added weights had less and less effect on increasing the velocity. We conclude that as the velocity increased so did the drag force of the water. **

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Question: `q001. If you know the pressure drop of a moving liquid between two points in a narrowing round pipe, with both points at the same altitude, and you know the speed and pipe diameter in the section of pipe with the greater diameter, how could you determine the pipe diameter at the other point?

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Your Solution:

The velocity of the liquid in the smaller diameter find can first be found using Bernoulli’s equation involving .5*’rho*v^2 and P. Solve the equation P2-P1=’dP=.5*’rho*v1^2 - .5*’rho*v2^2 for v2. Using this value for v2, we can now find the diameter of the smaller pipe by using v2/v1 = (d1/d2)^2.

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