Problem Number 1
University Physics Problem: Use Bernoulli's equation to determine the velocity with which water will exit from a hole in a uniform cylinder when the cylinder is filled to a point 3.1 meters above the hole, assuming that the water in the cylinder moves with negligible velocity. If the cylinder has a circular cross-section with radius 7 meters while the hole has radius .075 meters, then what is the total kinetic energy of the water in the cylinder above the hole? Describe what would happen if the hole was suddenly plugged. Explain why, if the hole is instantly unplugged, the exiting water requires a short time interval to reach its maximum velocity.
KE=1/2(1000kg/m^3)v^2 since v is negligible at top wouldn’t that mean the KE would be approx 0?
However, I don’t think you would ask this if this was the case so do I do the following:
V_exit=sqrt(2(9.8m/s^2)(3.1m))=7.8m/s
KE=1/2(1000kg/m^3)(7.8m/s)^2
KE=30,420 wouldn’t this be in Pa which doesn’t make much sense
The water obviously could not flow out of the hole if it is plugged so would an immediate but brief pressure build??? Also, I think the water would reach maximum velocity quickly because the volume of water is substantially larger than the area of the cross section of the hole.
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Bernoulli's Equation states that
rho g y + 1/2 rho v^2 + P
is constant, so that for two points in a continuous fluid
rho g y_1 + 1/2 rho v_1^2 + P_1 = rho g y_2 + 1/2 rho v_2^2 + P_2
The term rho g y is referred to as the 'potential energy term', due to its similarity with gravitational PE equal to m g y. However rho g y has units of Pascals, not Joules, as does the 'kinetic energy term' 1/2 rho v^2 (analogous but not identical to 1/2 m v^2).
Note that Joules / m^3 have units of Pascals.
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Problem Number 2
Find an expression for the average pressure exerted by a single particle of mass m traveling always at speed v on one end wall of a cylindrical container of length L meters and cross-sectional area A m^2, provided the particle travels always along the axis of the cylinder and collides elastically with the ends of the container.
• Use your expression to show whether the pressure is proportional to the velocity v of the particle, inversely proportional to v, proportional to v^2 or inversely proportional to v^2.
• Show whether the pressure is proportional to the KE of the particle, inversely proportional to this KE, proportional to KE^2 or inversely proportional to KE^2.
Im getting confused on this problem and would like some assistance????
2L=round trip
v1--v2=2v
2mv= change in momentum
(mv^2)/l=change in F
(Mv^2)/(l A)=(mv^2)/vol=aver press
Based on this would pressure by proportional to v^2 and the pressure would be proportional to KE????
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You are correct. Pressure is proportional to v^2, by your steps, and KE = 1/2 m v^2 is also proportional to v^2, so pressure is proportional to v^2.
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Problem Number 3
A wall is made of a substance whose thermal conductivity is .27 J / (m sec Celsius). What is the temperature gradient through the wall if its cross-sectional area is 38 m^2 and if thermal energy flows through the wall at a rate of 55 watts?
Q=-kA(temp gradient)
55watts= -(.27J/(m*s*Cels))(38m^2)(Temp Grad)
We know J/s =watts
Temp grad=(55watts)/((-.27Watts/(mCels))*m^2)
Temp grad=-5.36Cels/m
Is this correct and are the units okay????
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Everything looks good.
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Problem Number 4
A tube 3.6 mm in diameter is run through the stopper of a sealed 1.5-liter container. The tube outside the container forms a U, then runs in a straight line with slope .027 with respect to horizontal. Alcohol is introduced into the tube, and fills the U, extending into the linear section of the tube. Both ends of the tube are open. The container is slightly heated, and the alcohol column is observed to move .76 meters along the linear section of the tube. The material of which the container is constructed has coefficient of linear expansion `alpha = 86 * 10^6 / C. If the temperature of the air in the container was originally 15 Celsius, what is the new temperature?
Can you please give me somewhat of a solution for this or number 9???? These two questions are giving me a lot of trouble and seem to be very important since they come up so frequently.
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You should also thoroughly review Class Notes 8-12.
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What is the volume of the container before and after being heated?
What additional volume of tubing has become available to the air in the container?
How much higher is the alcohol column as a result of the heating, and by how much did the pressure inside the container therefore increase?
This gives you the new volume and pressure of the system. Air can't escape so n stays constant. What therefore is the new temperature?
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Problem Number 5
Use Bernoulli's equation to determine the pressure change as water flows through a full horizontal pipe from a point where the pipe diameter is .25 meters and velocity 3.7 m/s to a point where the pipe diameter is .135 meters.
Bernoulli's equation is
.5(p1)v1^2+p1gh1+P1=.5(p2)v2^2+p2gh2+P2
We know p1 and p2 are 1000kg/m^3
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Even though you're using different cases, it's a very bad idea to use the same letter for density as for pressure.
It doesn't take much to type out rho.
And when writing this out on paper, be sure to use the right symbol for rho.
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g=9.8m/s^2
v1=3.7m/s
h1 and h2=0
v1/v2=(d2/d1)^2
v2=(3.7m/s)/(.2916)
v2=12.7m/s
dp=.5(1000kg/m^3)( 12.7m/s)^2-.5(1000kg/m^3)(3.7 m/s )^2
dp=73800Pa
I think this is wrong since shouldn’t the Pressure decrease when velocity is increased since the tube gets smaller.
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You'll be better writing out the equation using subscripts 1 and 2 for the two points.
Assuming the lower point is point 2, you want to find P2 - P1.
If you write this out correctly and rearrange to get P2 - P1 on one side, you'll get the right order of terms on the right.
And the pressure will indeed decrease, as you correctly indicate.
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Problem Number 6
A diatomic gas in a 5-liter container is originally at 16 Celsius and atmospheric pressure. It is heated at constant volume until its temperature is 142 Celsius, then at constant pressure until the gas has increased its volume by .65 liters. How much thermal energy is required? By how much does the internal energy of the gas change? How much work is done in the process?
PV=nRT
101,300Pa(5L)=(n)(8.31J/moleK)(289K)
N=211 moles
At constant volume,
Qv=(3/2)nRdT
Qv=(3/2)(211mole)( 8.31J/moleK)(415K)
Qv=1,091,498J
At constant pressure,
Qp=(5/2)nRdT
Qp=(5/2)(211moles)( 8.31J/moleK)(415K)
Qp=1,819,163J
Work=Qp-Qv=727665J
How do you find the internal energy and is the rest of this correct????
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The work is the integral of P with respect to V, and would be equal to the area beneath a graph of P vs. V.
415 K might be the final temperature, but it's not the change in temperature at constant volume, nor is it the change in termperature at constant pressure (the two temperature changes would be different).
If you get the right temperature changes then you will get the right values for `dQv and `dQp (the thermal energies required for the two processes).
When you add energy to the gas some goes into internal energy, some into work. So if you know how much energy was added, and know how much work was done, you can reason out the change in internal energy.
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Problem Number 7
Find an expression for the average pressure exerted by N identical particles each of mass m traveling always at speed v on one end wall of a cylindrical container of length L meters and cross-sectional area A m^2, provided the particle travels always along the axis of the cylinder and collides elastically with the ends of the container.
• Use your expression to show whether the pressure is proportional to the N, inversely proportional to N, proportional to N^2 or inversely proportional to N^2.
Since this is a similar problem to the other one I was getting confused on I will use the same approach and I guess you will tell me if I am way off with my reasoning.
2L=round trip
v1--v2=2v
2mv= change in momentum
(mv^2)/l=change in F
(Mv^2)/(l A)=(mv^2)/vol=aver press
So N is particles. So Pv=nRT would work for this????
If so, Pressure would be proportional to N.
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This is the expression for a single particle.
What therefore would be the expression for N particles?
The answer is as obvious as it seems.
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Number 8
What gauge pressure is necessary in a hose to spray water to a height of 20 meters?
P=pgh
P=(1000kg/m^3)(9.8m/s^2)(20m)
P=196,000Pa
P=196,000Pa*(1atm/101300Pa)
P=1.93atm
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Good.
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