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Flauer [41]
3 years ago
10

Two pipes move the same amount of ideal fluid in the same amount of time. One pipe has a 2 in. diameter; the other has a 3 in. d

iameter. In which pipe is the fluid pressure higher, assuming that all other conditions are the same?
a)3-in. pipe
b)2-in. pipe
c)same pressure in both
d)cannot be determined for an ideal fluid
Physics
1 answer:
xxTIMURxx [149]3 years ago
6 0
By Bernoulli Theorem, We know, that for any two points in the pipe, total energy would be same. 

Pressure Energy + K.E. of 1st pipe = Pressure Energy + K.E. of 2nd pipe
[ No need to consider Potential energy as height isn't mentioned] 
P₁ + mv₁²/2 = P₂ + mv₂²/2

So, we know, P + mv²/2 = constant

As value of kinetic energy is larger in larger in 2 in. pipe [ 'cause area is indirectly proportional to velocity ], Value of pressure energy would be smaller in it. So, pressure energy will be higher in the pipe of 3 in.

In short, option A would be your correct answer.

Hope this helps!



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Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure (P_{atm}), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:

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Where:

P_{1}, P_{2} - Water total pressures inside the tank and at ground level, measured in pascals.

\rho - Water density, measured in kilograms per cubic meter.

g - Gravitational acceleration, measured in meters per square second.

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z_{1}, z_{2} - Heights of the tank and ground level, measured in meters.

Given that P_{1} = P_{2} = P_{atm}, \rho = 1000\,\frac{kg}{m^{3}}, g = 9.807\,\frac{m}{s^{2}}, v_{1} = 0\,\frac{m}{s}, z_{1} = 6.9\,m and z_{2} = 4.9\,m, the expression is reduced to this:

\left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m) = \frac{v_{2}^{2}}{2} + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.9\,m)

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v_{2} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m-4.9\,m)}

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