Confusion about the interpretation of specific volume

In summary: Yes, agreed with the first part. And this would hold even for the compressed liquid state, correct?The second part, what I'm thinking is: imagine a piston pushing down on a liquid contained in a cylinder with some pressure. This compresses the liquid. So wouldn't the the specific volume for a given pressure be equal to the volume of this liquid divided by the mass. And this would be the value reported in...The specific volume for a given pressure would be the same as the volume of liquid divided by the mass of the liquid at that pressure.
  • #1
EddiePhys
131
6
So I'm reading through Cengel's thermodynamics textbook, and came across this solved example:

Screenshot_20230109-145757.jpg


Firstly, pressure in this context I'm assuming is vapour pressure? Since we're dealing with pure substances in this chapter.

But what's confusing me is, here's the diagram I have:

Screenshot_20230109-150159.jpg


They've not mentioned that the water fills the tank. So I don't understand why the specific volume times the mass of the water equals the volume of the tank?

They've used similar reasoning in following solved examples as well where it's unclear that the liquid will occupy the tank volume
 
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  • #2
EddiePhys said:
They've not mentioned that the water fills the tank.
1673261437500.png

I think you can safely take it that there is nothing else present, i.e. no water vapour...

##\ ##
 
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  • #3
BvU said:
View attachment 320034
I think you can safely take it that there is nothing else present, i.e. no water vapour...

##\ ##

They have used similar lines of reasoning in future solved examples where it's not so clear that the tank will be filled with liquid

For example, here, when they compare the ##v## value with ##v_g## and ##v_f##, to determine the state, it would only make sense if, when it is in liquid state, it is occupying the entire vessel:

Screenshot 2023-01-09 at 5.51.12 PM.png
 
  • #4
EddiePhys said:
They have used similar lines of reasoning in future solved examples where it's not so clear that the tank will be filled with liquid

For example, here, when they compare the ##v## value with ##v_g## and ##v_f##, to determine the state, it would only make sense if, when it is in liquid state, it is occupying the entire vessel:

View attachment 320044
I guess you weren't satisfied with my answers in that other forum.
 
  • #5
Yes, there is a subtle difference between "4 kg of refrigerant" and "50 kg of saturated liquid water" .
You'd better learn to distinguish the difference in these simple exercises -- practice for later when it's not so easy any more. :wink:

##\ ##
 
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  • #6
Chestermiller said:
I guess you weren't satisfied with my answers in that other forum.

I really appreciated your answers. Something just wasn't clicking with me. So I thought getting different perspectives would perhaps help. Please don't take it personally, I really do appreciate your help.
 
  • #7
The largest volume that 4 kg of refrigerant 134a can occupy as a liquid at 160 kPa is (0.0007435)(4)=0.0030 m^3 = 3.0 liters
 
  • #8
Chestermiller said:
The largest volume that 4 kg of refrigerant 134a can occupy as a liquid at 160 kPa is (0.0007435)(4)=0.0030 m^3 = 3.0 liters

Yes. So what I'm not getting is why they're dividing the volume of the vessel by 4kg. They're then comparing this value with 0.0007435. The two would only be equal if they were the volume of the liquid by 4kg (or equivalently, that the vessel was fully filled, in which case dividing the volume of the vessel by 4kg is the same as dividing the volume of the liquid).

So in the case that the liquid only occupies a small portion of the volume of the vessel, volume of the vessel divided by 4kg would not be suitable means to assess if it's in compressed liquid state
 
  • #9
EddiePhys said:
Yes. So what I'm not getting is why they're dividing the volume of the vessel by 4kg. They're then comparing this value with 0.0007435. The two would only be equal if they were the volume of the liquid by 4kg (or equivalently, that the vessel was fully filled, in which case dividing the volume of the vessel by 4kg is the same as dividing the volume of the liquid).
The mass of refrigerant is way less than sufficient to fill the tank with liquid alone.. So there must be mixture of liquid and vapor in the tank. At equilibrium, we must have that : $$m_Lv_L+m_Vv_V=V$$Doo you agree with this?
EddiePhys said:
So in the case that the liquid only occupies a small portion of the volume of the vessel, volume of the vessel divided by 4kg would not be suitable means to assess if it's in compressed liquid state
This would mean that it can't be in a state of compressed liquid (i.e., liquid at a pressure greater than the vapor pressure at the tank temperature).
 
  • #10
Chestermiller said:
The mass of refrigerant is way less than sufficient to fill the tank with liquid alone.. So there must be mixture of liquid and vapor in the tank. At equilibrium, we must have that : $$m_Lv_L+m_Vv_V=V$$Doo you agree with this?

This would mean that it can't be in a state of compressed liquid (i.e., liquid at a pressure greater than the vapor pressure at the tank temperature).

Yes, agreed with the first part. And this would hold even for the compressed liquid state, correct?

The second part, what I'm thinking is: imagine a piston pushing down on a liquid contained in a cylinder with some pressure. This compresses the liquid. So wouldn't the the specific volume for a given pressure be equal to the volume of this liquid divided by the mass. And this would be the value reported in tables.

In our case we're dividing the volume of an arbitrary vessel by the mass of the liquid and comparing it to the above value.
 
  • #11
EddiePhys said:
Yes, agreed with the first part. And this would hold even for the compressed liquid state, correct?

The second part, what I'm thinking is: imagine a piston pushing down on a liquid contained in a cylinder with some pressure. This compresses the liquid. So wouldn't the the specific volume for a given pressure be equal to the volume of this liquid divided by the mass. And this would be the value reported in tables.
Yes, reported in tables for compressed liquid.
EddiePhys said:
In our case we're dividing the volume of an arbitrary vessel by the mass of the liquid and comparing it to the above value.
The value we calculate is between that of a saturated liquid at 160 kPa and a saturated vapor at 160 kPa. So we must have a combination of the two.
 
  • #12
Chestermiller said:
Yes, reported in tables for compressed liquid.

The value we calculate is between that of a saturated liquid at 160 kPa and a saturated vapor at 160 kPa. So we must have a combination of the two.

But even if it was a compressed liquid, we wouldn't know because we're not dividing the volume of the liquid by its mass, and this is the value that is given in tables and that we're using as a standard for comparison.
 
  • #13
EddiePhys said:
But even if it was a compressed liquid, we wouldn't know because we're not dividing the volume of the liquid by its mass, and this is the value that is given in tables and that we're using as a standard for comparison.
If it was a compressed liquid, the tank volume divided by the msss of refrigerant in tank would be less than the specific volume of the saturated liquid.
 
  • #14
Chestermiller said:
If it was a compressed liquid, the tank volume divided by the msss of refrigerant in tank would be less than the specific volume of the saturated liquid.

Why though. Let's say V1 is the volume of the compressed liquid and V2 the tank. And V2>>V1. In this case, the specific volume of the compressed liquid would be calculated V1/M in the tables while a saturated liquid at that pressure might have a volume in the vicinity of V1, let's say V1 plus a since liquids aren't that compressible. But we're calculating it as V2/M and V2>>V1, so it why shouldn't the tank volume give the larger answer
 
  • #15
EddiePhys said:
Why though. Let's say V1 is the volume of the compressed liquid and V2 the tank. And V2>>V1. In this case, the specific volume of the compressed liquid would be calculated V1/M in the tables while a saturated liquid at that pressure might have a volume in the vicinity of V1, let's say V1 plus a since liquids aren't that compressible. But we're calculating it as V2/M and V2>>V1, so it why shouldn't the tank volume give the larger answer
If it is "compressed liquid," then there is no vapor in the tank, and ##V_1=V_2##. In this case, $$\frac{V_1}{M}=\frac{V_2}{M}<v_L(T)$$where ##v_L(T)## is the specific volume of saturated liquid at the tank contents temperature T. So the specific volume is a little less than the specific volume of the saturated liquid, and the tank is filled with liquid if the pressure is above the saturation vapor pressure at T.
 
  • #16
Chestermiller said:
If it is "compressed liquid," then there is no vapor in the tank, and ##V_1=V_2##. In this case, $$\frac{V_1}{M}=\frac{V_2}{M}<v_L(T)$$where ##v_L(T)## is the specific volume of saturated liquid at the tank contents temperature T. So the specific volume is a little less than the specific volume of the saturated liquid, and the tank is filled with liquid if the pressure is above the saturation vapor pressure at T.

Okay I think now I am close to getting it. What would the case then be where V1 << V2 AND there's no vapor in the tank? Or is this a case we simply don't consider. So we only have 1) tank full of liquid,2) liquid + gas, 3) only gas as possibilities. And if I understand correctly, saturated liquid is almost like (1) i.e fully filled tank but slightly larger volume
 
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  • #17
EddiePhys said:
Okay I think now I am close to getting it. What would the case then be where V1 << V2 AND there's no vapor in the tank? Or is this a case we simply don't consider.
If there is no vapor in the tank, then ##V_1=V_2##
EddiePhys said:
So we only have 1) tank full of liquid,2) liquid + gas, 3) only gas as possibilities.
Yes.
EddiePhys said:
And if I understand correctly, saturated liquid is almost like (1) but a tiny portion is gas
Purely saturated liquid is like (1) but with essentially no vapor.
 
  • #18
Chestermiller said:
If there is no vapor in the tank, then ##V_1=V_2##

Yes.

Purely saturated liquid is like (1) but with essentially no vapor.

"If there's no vapor in the tank, then ##V_1=V_2##" So you can't simply have a liquid with a volume less than the tank volume. The second you do, you are in the region to the right of the saturated liquid phase (i.e saturated liquid-gas mixture)
 
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  • #19
I'm taking you liking the post to mean yes. In which case, I think I got it! Head's a bit hazy right now so I'll know for sure by tomorrow morning, but thank you so much!! I really appreciate you taking the time you did to answer my questions
 

FAQ: Confusion about the interpretation of specific volume

What is specific volume and how is it defined?

Specific volume is defined as the volume occupied by a unit mass of a substance. It is the reciprocal of density and is typically expressed in units of cubic meters per kilogram (m³/kg). Mathematically, specific volume (v) can be represented as v = 1/ρ, where ρ is the density of the substance.

How is specific volume different from volume?

Volume is the total space occupied by a substance, usually measured in cubic meters (m³) or liters (L). Specific volume, on the other hand, is the volume occupied by a unit mass of the substance. While volume is an extensive property dependent on the amount of substance, specific volume is an intensive property that remains constant regardless of the quantity of the substance.

Why is specific volume important in thermodynamics?

Specific volume is crucial in thermodynamics because it helps describe the state of a substance in a system. It is used in various thermodynamic equations and calculations, such as those involving the ideal gas law, phase changes, and the properties of mixtures. Understanding specific volume allows scientists and engineers to predict how substances will behave under different conditions of temperature and pressure.

How can specific volume be measured or calculated?

Specific volume can be measured directly by determining the volume and mass of a sample of the substance. Alternatively, it can be calculated using the relationship with density: v = 1/ρ. For gases, specific volume can also be calculated using the ideal gas law: v = RT/P, where R is the specific gas constant, T is the temperature, and P is the pressure.

What are some common misconceptions about specific volume?

One common misconception is that specific volume is the same as density, when in fact, they are inversely related. Another misconception is that specific volume changes with the amount of substance, whereas it is actually an intensive property that remains constant for a given substance under specific conditions. Additionally, some people might confuse specific volume with molar volume, which is the volume occupied by one mole of a substance.

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