need some help on this

[al zami]

a math problem goes like this

a gas initially at p1=512 kpa and v1=142 litre undergoes a process to p2=172 kpa and v2=274 litre

durig which enthalpy decreases 65.4 kj.specific heats are constant,Cv=3.175 kj/kg-k.determine dU,Cp and R.

here my question is

1) I cant understand which thermodynamic process is this,isothermal,polytropic or adiabetic.and why??

Edited June 7, 2013 by al zami

[studiot]

Ask yourself what changes to the gas state variables from the initial state to the final one.

 

The enthalpy changes so what happens to the energy ?

 

Is any work done?

 

Is there any change to the temperature ? How does this affect the change to internal energy

[al zami]

here pressure decreases and volume increases

 

and enthalpy is negetive so that means work is done by the gas

no changes in temperature is mentioned

its not making any sence.

[daniton]

here pressure decreases and volume increases

 

and enthalpy is negetive so that means work is done by the gas

no changes in temperature is mentioned

its not making any sence.

Check whether there is temperature change ,you have ideal gas equation.

Now.......

[studiot]

and enthalpy is negetive so that means work is done by the gas

 

 

Is this a condition of work done?

 

Work is only done when the volume changes.

This is true in this case.

Since the volume expands work is done by the gas on the surroundings.

 

As a result of doing work the gas suffers a loss of internal energy.

This loss of internal energy shows up as the enthalpy change given.

 

Many problems with several parts are set so that you use the first thing you calculate to calculate the second answer required and so on.

 

Hint 1 You are first asked for dU

 

Do you know ( can you state) the relationship between enthalpy and internal energy?

 

Hint 2 You are given dH and are asked to calculate C_p

Do you know (can you state) a relationship between dH and C_{p .}You will also need to look at Danitons post.

 

Hint 3 You are asked to calculate C_p and given C_v and then asked to calculate R

Do you know (can you state) a relationship between C_v , C_p and R

 

Hint 4 beware of the units C_v is given in

Edited June 8, 2013 by studiot

[al zami]

i've solved this like this

 

we know y=cp/cv

1.4=cp/3.175
cp=4.445
R=cp-cv
=1.27
as for ideal gas
p1v1=mRT1
and
p2v2=mRT2

from here
m(T2-T1)=-20130.24
from here
dU=m*Cv*(T2-T1)
=-89478.92

 

but i have some question

1) ive never seen a R (characterstic gas constant)value 1.27 before.is it ok?

 

Do you know ( can you state) the relationship between enthalpy and internal energy?

actually i dont know.can you explain ?

[studiot]

actually i dont know.can you explain ?

 

 

 

I have been editing my post# 5 whilst you came online so have another look.

 

Meanwhile

 

The defintion of enthalpy is

 

H = U + PV

 

or at any state 1

H₁ = U₁ + P₁V₁

 

Can you see how to get dU = (U₂ - U₁) from this?

[al zami]

Check whether there is temperature change ,you have ideal gas equation.

 

Now.......

in question there is no mention of its being ideal gas or not. do enthalpy change means temperature change??

i ve done the math in comments section.please check it whether it is ok or not.

 

are you saying

H₁ = U₁ + P₁V₁

_H2=U2+P2V2

_{H2-H1=(U2-U1)+(P2V2-P1V1)}

_{dH=dU+(P2V2-P1V1)}

_{so from here we can find dU.right?}

 

and relation between cp and H is dH=m*Cp*(T2-T1)

 

and we can find m(T2-T1) from p1v1=mRT1 and p2v2=mRT2

 

and then cp-cv=R

 

how i did it is a little bit different from this ,so please check it,especially the R value ,i am confused about its value

Edited June 8, 2013 by al zami

[studiot]

are you saying

 

 

H₁ = U₁ + P₁V₁

 

 

_H2=U2+P2V2

 

 

_{H2-H1=(U2-U1)+(P2V2-P1V1)}

 

 

_{dH=dU+(P2V2-P1V1)}

 

 

_{so from here we can find dU.right?}

 

Yes that's right.

 

and we can find m(T2-T1) from p1v1=mRT1 and p2v2=mRT2

 

 

 

 

 

Once you have dH and dU you can use

 

dH = mC_pdT

 

dU = mC_vdT

 

dH/dU = C_p/C_v

 

to find C_p from C_v

 

You were not justified using gamma for this ratio that is for reversible adiabatic expansion, which this is not. Do you know why not?

 

 

Remember that you are working per kg so the R you are calculating will be the characteristic gas constant for this particular gas, not the universal gas constant.

 

The value of C_v given is certainly not for air, which has C_v around 0.7. Perhaps it is an air/hydrogen mix or a heavy hydrocarbon at around 3.

 

https://www.google.co.uk/#sclient=psy-ab&site=&source=hp&q=characteristic+gas+constant&oq=characteristic+gas+constant&gs_l=hp.12..0l5j0i22i30j0i22i10i30j0i22i30l3.1593.9187.1.27500.27.18.0.9.9.3.1031.6140.0j9j1j1j2j4j0j1.18.0...0.0...1c.1.16.hp.HP1GTVG4_2g&psj=1&bav=on.2,or.&bvm=bv.47534661,d.d2k&fp=82c5307a998d7faf&biw=1024&bih=559

Edited June 8, 2013 by studiot

[al zami]

You were not justified using gamma for this ratio that is for reversible

adiabatic expansion, which this is not. Do you know why not?

 

actually i don't know.it would be very helpful if you explain it to me?

 

how i can can know the process is reversible or not from any problem given??

 

Remember that you are working per kg

 

there was no mention that i am working with per kg or not?? how can i come by this information???

[studiot]

there was no mention that i am working with per kg or not?? how can i come by this information???

 

 

 

Hint 4 noted the units of C_v in post#5.

 

This question is about the most general case where all three of the fundamental directly observable (variables, temperature, pressure and volume) vary. Further the system does both work on the surroundings and gains heat from it.

 

Yes this could be the polytropic process if the same process took place throughout the change, but we are not told this. The general polytropic process is not reversible and has an index usually denoted by n. Gamma is reserved for the reversible adiabatic case.

You should re read the derivations of the relationships to consolidate this.

 

We are told that values of system variables at two defined states so we can use this to obtain changes in state variables.

 

But we are not told the quantity of gas nor the starting or finishing temperatures.

 

So we cannot assume an adiabatic or isothermal process.

 

You are entitled to assume that the perfect gas laws hold because you are not told anything else and no information is supplied that would allow any other law to be employed.

 

The property of the perfect gas we employ here is that both internal energy and enthalpy depend only on the temperature of the gas and lead to the heat capacity equations I gave.

 

Does this help?

Edited June 9, 2013 by studiot

[al zami]

You were not justified using gamma for this ratio that is for reversible adiabatic expansion, which this is not

how can i assume this is reversible of irreversible ???

 

this leads to another question that in the book we always find reversible process is that which returns to its initial state on contrary to irreversible.not more than that

 

but what are the actual physical significance of those two process ??i mean basic difference other than ''returns to its original state or not"

 

as you told we can not use gamma here as it is not reversible? how can i make sure that it is reversible or irreversible.

Edited June 9, 2013 by al zami

[studiot]

this leads to another question that in the book we always find reversible process is that which returns to its initial state on contrary to irreversible.not more than that

 

 

 

 

 

but what are the actual physical significance of those two process ??i mean basic difference other than ''returns to its original state or not"

 

 

To have reached this problem you posted you must have been studying thermodynamics for some time, it is not at elementary level.

 

But you need to revise some elementary concepts.

 

A reversible process is one that can be reversed, not one that reverses of its own accord.

Reversible means that the after reversal the system is put back to the same state it started in, and all the work done or heat transferred is also returned. None of the work done is lost in additional processes such as friction.

 

An irreversible process is one which is not reversible.

 

Which textbook are you using?

[al zami]

actually i am following thermodynamics by hollman

in books usually there are big mathemetical proofs but no explanation in plain english which leads to better understanding the concepts

 

 

But we are not told the quantity of gas nor the starting or finishing temperatures.

 

So we cannot assume an adiabatic or isothermal process

and here though there is no mention of temperature how that means it will not be adiabetic.as we found m(T2-T1) there is temperature involved in two state for which it can be a adiabetic curve ,isn't it?( here confusion arrises)

so for that we can use cp/cv=y (if we assume adiabetic)

from cv value we can find

1)it is not for air;

2)from its unit ,we find we are dealing with per kg of substance

 

what other we can assume from the value of C_p or C_v????i mean its significance

Edited June 9, 2013 by al zami

[studiot]

What do you understand adiabatic to mean?

 

It is a good exercise to describe what happens during the process.

By this I mean the energy transfers between the system and surroundings and the changes to the system internal energy, pressure, volume and temperature. You don;t need numbers.

 

Try it and then I will say how I would describe it.

[al zami]

i solved the problem following your hints .adiabatic means there will be no heat transfer between the system and surrounding.neither take nor give away.

It is a good exercise to describe what happens during the process.

temperature isn't given here.but if in this problem the mass m and molecular mass M of gas was given,we can find temperatures T₁ and T₂ from PV=mRT₁

if that so can we assume the process adiabatic and so we can use c_p/c_v=y; which is 1.4

but its incorrect for this problem

how can i tell from a given problem the process is adiabatic or not.its not clear to me.

 

are we assuming it is not adiabatic for our own convenience ???or there is exact reason that the process is not adiabatic?

as i made a mistake solving the problem assuming it adiabatic and taking y=1.4

 

here c_v is given 3.175. so it is not air

Edited June 9, 2013 by al zami

[studiot]

how can i tell from a given problem the process is adiabatic or not.its not clear to me.

 

A problem will make it clear if a process is adiabatic. Usually by stating this explicitly.

 

In this case at the outset we do not have enough information to say if the process is adiabatic or if it is not so we have to allow that it may not be and not use properties specific to adiabatic processes.

If our calculations later show it to be adiabatic, so much the better, but we can't 'guess'.

 

The important thing with this problem is to realise that whatever the nature of the process differences between state variable values depend only on the state and not on the process by which that state was reached. So we can calculate dH, dU, d(PV), dT and so on.

 

You complained earlier that your book has more maths than explanation in words.

Well this thread is your opportunity for words.

I really do recommend trying to work through what happens in the process as I suggested.

It will definitely help.

Edited June 9, 2013 by studiot

[al zami]

got it.this entire discussion was very helpful for me.thanx a lot to you .