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Old Sep 5th 2017, 09:59 AM   #1
oya
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Post Energy being transferred to a container with air...

Hello All. I would appreciate any help you can offer on this problem!

Air is tightly contained in a closed, well-insulated tank with a volume of 0.5 m3. The tank is fitted with a paddle wheel that transfers energy to the air at a constant rate of 3 W for 1 hour. The initial density of the air is 1.2 kg/m3. If no changes in kinetic or potential energy occur, determine:
a. What additional assumptions must you make to solve this problem? (come up with at least one)
b. The specific volume at the final state, in m3/kg.
c. The energy transfer by work, in kJ.
d. The change in specific internal energy of the air, in kJ/kg.
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Old Sep 5th 2017, 10:32 AM   #2
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oya, Hi

oya,

1) "Well-insulated" is one assumption and "constant volume" is another.
2) v = m^3/1.2kg.
3) W = 3 J/s x 60 s
4) ΔU = W
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Old Sep 5th 2017, 11:27 AM   #3
oya
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For (a), Would you be so kind as to explain how you arrived at this conclusion? Don't I need to know the pressure inside the system to figure these out?
For (b), shouldn't it be x 3600 seconds because that is how much an hour has?
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Old Sep 5th 2017, 01:08 PM   #4
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a)
Well insulated is not an assumption, it is a given.
The most obvious assumption is one that gives you some equations to work with, especially since you don't know the pressure, as you have correctly observed.

I would suggest the assumption that air acts as an ideal gas, so the ideal gas laws apply.

To permit this assumption, a second assumption is necessary.

That the air is dry - moist air most certainly does not obey ideal gas laws.

c) and d)
Then fall out as TSH said since the internal energy of an ideal gas depends only on the temperature (did you know this important fact?)

b)
What changes during the experiment?
The mass of air?
The volume of air?
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Old Sep 5th 2017, 03:21 PM   #5
oya
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What if I assume that this example is using standard atmospheric conditions of T=288.15 K and P=101.325 kPa? How would I figure out the impact the work had on the air?
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Old Sep 5th 2017, 03:34 PM   #6
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Originally Posted by oya View Post
What if I assume that this example is using standard atmospheric conditions of T=288.15 K and P=101.325 kPa? How would I figure out the impact the work had on the air?
How would that help and why would they have given you the air density?

Can you state the first law of themodynamics?

Last edited by studiot; Sep 5th 2017 at 03:37 PM.
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Old Sep 5th 2017, 04:02 PM   #7
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3) W = 3 J/s x 60 s
I should perhaps point out that there are 3600 seconds in one hour.
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Old Sep 5th 2017, 06:02 PM   #8
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My error...

Sorry, 1 hour equals 3600 seconds. My bad!

However, "well insulated," taken to mean "with no heat interaction,"
cannot be a "given." There are no perfect insulators; not ours to give.
Heat happens with temperature difference. To say "no heat" is an assumption.

To assume "no heat" is a reasonable assumption for systems highly
insulated. Also "no heat" happens in the limit that a thermodynamic event is instantaneous.
So to "occur quickly" is assumed "no heat."

Heat happens the greater the temperature difference and the longer the time.

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Old Sep 6th 2017, 07:39 AM   #9
oya
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I think I figured it out. Does this look correct?

Assumptions:
This is an adiabatic process - heat and matter are not exchanged between the system and its surroundings
Specific quantities are uniform across the volume.

The specific volume at the final state, in m3/kg.
Mass (m) = density (g) x volume (V)
Final volume (v) = V / m = 0.5 / (1.2 x 0.5) = 0.833 m3/kg

The energy transfer by work, in kJ.
1 watt = 1J/sec; 1 hour = 3600 seconds
3 watt per hour = 3600 seconds x 3J/second = 10,800J or 10.8kJ

The change in specific internal energy of the air, in kJ/kg.
Because it was assumed that this is an adiabatic process, the change in internal energy is equal to work or 10.8kJ.
Mass = density x volume = 0.6kg
Therefore change in specific internal energy (u) = change in internal energy (U) / mass (m) = 10.8kJ/0.6kg = 18kJ/kg
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Old Sep 6th 2017, 08:17 AM   #10
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It will do, the figures are correct, but I suggest you rephrase this

Final volume (v) = V / m = 0.5 / (1.2 x 0.5) = 0.833 m3/kg
And well done for uisng the guidance to figure it for yourself.

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air, container, energy, energy transfer, insulated, internal energy, problem, simple, thermodynamics, transferred, volume



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