Thermodynamics, work interaction, pressure, power

In summary: As for (a), P=F/A, and we know it's a circular bearing pad with diameter 1.22, thus radius is 0.61m, which means each pad has an area of 1.169 m^2 which is A. There are 46 pads, and it's asking for the water pressure under *each* pad, so I have to divide 3.92x10^7 N by 46 which yields 8.5x10^5 N, thus my answer is 7442 N/m^2.Correct again!I think this is correct? And only for part (a) do I have to consider 46 pads, whereas in parts (b)
  • #1
xzibition8612
142
0

Homework Statement


Denver's Mile High football stadium incorporates a system by which a section of stands can be moved to accommodate different sports on the field. The grandstand, which has a mass of 4x10^6 kg, can be moved a distance of 45 m on a thin film of water. The movable grandstand rests on 46 bearing pads that are each 1.22m in diameter. Water at high pressure is pumped into each of the pads until the stands are lifted a vertical distance of 3.8 cm. Excess water forms a lubricating film over which the grandstand is moved. The force required to move the stands is approximately 4.45 N per 450 kg of staduym mass. Calculate the following:

a. The pressure of the water under each bearing pad

b. The power of the motor required to move the grandstands over the distance of 45 m if the job takes 1 h.

c. the work required to raise the grandstand 3.8 cm.




Homework Equations



See attachment formulas. Here is a brief description of each:
1. Force-displacement work
2. Force-displacement power
3. PdV work (piston)
4. PdV power

The Attempt at a Solution


Am I picturing this problem correctly (see "picture" attachment)?
I don't understand the bearing pads part. Do the bearing pads move or are the bearing pads wet and the grandstand slides on the bearing pads? And for part (a) asking the pressure of the water under each bearing pad, does this mean the water pressure inside each pad? I'm very confused about this problem and don't even know where to start. Any help would be appreciated. Thanks!
 

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  • #2
xzibition8612 said:

Homework Equations



See attachment formulas. Here is a brief description of each:
1. Force-displacement work
2. Force-displacement power
3. PdV work (piston)
4. PdV power
Equations 1 & 3 are correct, but 2 & 4 are not.
For #2, recall that W = F·d, where d is the displacement.

The Attempt at a Solution


Am I picturing this problem correctly (see "picture" attachment)?
I don't understand the bearing pads part. Do the bearing pads move or are the bearing pads wet and the grandstand slides on the bearing pads? And for part (a) asking the pressure of the water under each bearing pad, does this mean the water pressure inside each pad? I'm very confused about this problem and don't even know where to start. Any help would be appreciated. Thanks!
I'm also not clear on whether the bearing pads move or are stationary, but I think that's not necessary for answering the questions. Yes, they would mean the water pressure inside each pad. Each pad is acting as a hydraulic lift, similar to in this picture:

yg.13.41.jpg

To help get you started: how much force must each bearing pad exert on the grandstand in order to lift it?
 
  • #3
For (a), P= density x gravity x height

So P = 1000 kg/m^3 x 9.8 m/s^2 x 0.032 m
P = 313.6 N/m^2

(b) Power (my equation 2):

Power = F x V (V=velocity)
Since 4.45N is required to move 450 kg, then (4x10^6 kg / 450 kg) x 4.45 N = 39555 N = F
Velocity = 45m/60s
Power = F x V = 39555 N x 0.75 m/s
Power = 29666.25 N-m/s

(c) Use equation 1.
Integrate from 0 to 3.8 cm. Only need to find F, so would that be my answer in (a) 313.6 N/m^2?
So work would then be 11.9 N/m...but the units are not of energy, so it's wrong. Don't know how to go on.

I think the 46 bearing pads can be treated as one single pad, as the effect is the same. Also the 1.22m diameter is extra unnecessary information right?
 
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  • #4
xzibition8612 said:
For (a), P= density x gravity x height

So P = 1000 kg/m^3 x 9.8 m/s^2 x 0.032 m
P = 313.6 N/m^2
Not quite. That is the change in water pressure over a depth of 0.032 m.

I will repeat my earlier hint: how much force must one bearing pad exert in supporting the 4*106 kg grandstand?
(b) Power (my equation 2):

Power = F x V (V=velocity)
Since 4.45N is required to move 450 kg, then (4x10^6 kg / 450 kg) x 4.45 N = 39555 N = F
Velocity = 45m/60s
Power = F x V = 39555 N x 0.75 m/s
Power = 29666.25 N-m/s
Your method is correct, but note that it takes 1 hour to move it. How many seconds is that?
(c) Use equation 1.
Integrate from 0 to 3.8 cm. Only need to find F, so would that be my answer in (a) 313.6 N/m^2?
This is where you went wrong, since a force must have units of N, so that can't be the force.

So, how much force is required to lift an object of mass 4x106 kg?
So work would then be 11.9 N/m...but the units are not of energy, so it's wrong. Don't know how to go on.

I think the 46 bearing pads can be treated as one single pad, as the effect is the same. Also the 1.22m diameter is extra unnecessary information right?
 
  • #5
How much force...so F=ma, thus Force = 4x10^6 kg x 9.8 m/s^2
Force = 3.92x10^7 N. This much force is needed to lift the grandstand. Yes so this number would be used in part (c) as F.

As for (a), P=F/A, and we know it's a circular bearing pad with diameter 1.22, thus radius is 0.61m, which means each pad has an area of 1.169 m^2 which is A. There are 46 pads, and it's asking for the water pressure under *each* pad, so I have to divide 3.92x10^7 N by 46 which yields 8.5x10^5 N, thus my answer is 7442 N/m^2.

I think this is correct? And only for part (a) do I have to consider 46 pads, whereas in parts (b) and (c) I can treat the bearing pads as one single pad right?
And 3600 seconds in 1 hour : )
Thanks a lot for your help.
 
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  • #6
xzibition8612 said:
How much force...so F=ma, thus Force = 4x10^6 kg x 9.8 m/s^2
Force = 3.92x10^7 N. This much force is needed to lift the grandstand. Yes so this number would be used in part (c) as F.

As for (a), P=F/A, and we know it's a circular bearing pad with diameter 1.22, thus radius is 0.61m, which means each pad has an area of 1.169 m^2 which is A. There are 46 pads, and it's asking for the water pressure under *each* pad, so I have to divide 3.92x10^7 N by 46 which yields 8.5x10^5 N,...
You're good up to here :smile:
...thus my answer is 7442 N/m^2.
Oops. Arithmetic error? Did you use the 1.169 m^2 area to get this?
I think this is correct? And only for part (a) do I have to consider 46 pads, whereas in parts (b) and (c) I can treat the bearing pads as one single pad right?
You're correct that for (b) and (c) you just need to consider the total force. Note that in (b) the pads are irrelevant, since it is a single separate motor that moves the grandstand horizontally -- the pads are only used to lift the grandstand vertically.

And 3600 seconds in 1 hour : )
Yup!
Thanks a lot for your help.
You're welcome.
 

1. What is thermodynamics?

Thermodynamics is the branch of physics that deals with the relationship between heat, temperature, energy, and work. It studies how these factors affect the behavior of matter and the physical laws that govern them.

2. How does work interact with thermodynamics?

Work is a form of energy transfer in which a force is applied to an object to move it a certain distance. In thermodynamics, work is often used to describe the transfer of energy between a system and its surroundings.

3. What is pressure in thermodynamics?

Pressure is the amount of force applied per unit area. In thermodynamics, pressure is often used to describe the amount of force exerted by a gas or liquid on the walls of its container.

4. What is power in thermodynamics?

Power is the rate at which work is done or energy is transferred. In thermodynamics, power is often used to describe the rate at which a system can convert energy from one form to another.

5. How do thermodynamics, work, pressure, and power relate to each other?

Thermodynamics, work, pressure, and power are all interconnected concepts that describe the behavior of energy and matter. In many thermodynamic processes, work is used to transfer energy, pressure is used to describe the force exerted by a system, and power is used to describe the rate at which these processes occur.

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