# Having trouble with thermo.

1. Jun 3, 2006

### Ibn_Nizar

Hi,

I'm stuck. Can someone help me out?

Problem:
The properties of a certain fluid are related as follows:

u=196+0.718t

pv=0.287(t+273)

(a)where 'u' is the specific internal energy(Kj kg), t is in degress Celsius, p is pressure(Kn/m2) and v is specific volume (m3 kg).For this fluid. Find Cp and Cv

(b)A system composed of 2 kg of the above fluid expands in a frictionless piston and cylinder from an initial state of 1 Mpa, 100degrees Celsius to a final temp of 30degress Celsius.If there is no heat transfer, find the net work for the process.
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This is what i have tried:

in (a), i tried to link the two equations together, so that i could get 'u' and 'pv' in the same equation:

u=196+2.5(pv-78.35)

I'm stuck now.How do i proceed? i know the formula for cp and cv, which is Cp=(do)u/(do)t

cv=(do)Q/(do)T .How do i use these formulas with respect to the above equation?

(b)I'm totally confused on this one. I don't know where to start.All i could understand is to find the work done, and that Q=0.

Thanks.

2. Jun 9, 2006

### Tom Mattson

Staff Emeritus
Hi, sorry I haven't responded before now.

You should have the following relations handy:

$$c_v=\left(\frac{\partial u}{\partial T}\right)_P$$

$$c_p=\left(\frac{\partial h}{\partial T}\right)_v$$

You should also know the definition of enthalpy.

Start by writing down the first law for the system.

$$Q_{NET}+W_{NET}+E_{mass,NET}=\Delta U + \Delta KE + \Delta PE$$

Now you've already noted that you can set $Q_{NET}=0$. What else can you set equal to zero? And what else do you know?

3. Jun 9, 2011

### sacmohelay

Hi,
just do these things:
for Ques a)
Cv=du/dt=d(196+0.718t)/dt

after differentiating wrt to t u will get Cv=0.718 Ans

And Cp= dh/dt=d(u+pv)/dt=du/dt+d(pv)/dt
=d(196+0.718t)/dt + d(0.287(t+273))/dt
Cp= 0.718 + 0.287
Cp=1.005 Ans

For Ques b):

dQ=dU+dW as per first law of thermodynaimcs
but as there is no heat transfer therefore dQ=0. So above equation becomes
dW= - dU= - (u2-u1)= u1-u2

at t=100 C, u1= 196+(0.718*100)
& at t=30 C, u2= 196+ (0.718*30)

so dW= u1-u2= 50.26 (By solving above two equations of u1 & u2)

therefore for 2 Kg of fluid dW= 2 x 50.26 KJ = 100.52 KJ Ans