Two Reversible Adiabatic Paths Cannot Intersect Each Other

# Let us consider two reversible adiabatic paths 1-3 and 2-3 intersect each other at point 3. 

# Also a reversible constant temperature processes 1-2 be drawn in such a way that it intersects the reversible adiabatic paths at 1 and 2.

# These three reversible processes 1-2, 2-3, 3-1 constitutes a reversible cycle.

# We know that the area under the p-v plot represents the net work output in a cycle. Therefore the area under the three reversible paths represents the net work output in a cycle.

# But such a cycle is not possible, since net work is being produced ia a cycle by a heat engine by exchanging heat with a single reservoir in the processes 1-2, violates the Law of Thermodynamics (Kelvin-Plank). Therefore the consideration of the intersection of the two reversible adiabatic process is wrong.

# Through one point, only one reversible adiabatic can pass.

# As two constant property lines can never intersect each other, it is observed that a reversible adiabatic path must denote some property, which is found later that it is entropy.

COMBINED FIRST AND SECOND LAW OF THERMODYNAMICS

Combined 1^st and 2^nd Law of Thermodynamics.

By first law of thermodynamics, ΔQ = dU + ΔW

Since ΔW = pdV

ΔQ = dU + pdV ----------------- 1.

From second law of thermodynamics, (Entropy concept)

ΔQ = T. dS ----------------------- 2.

Put equation 2. in 1.

We get, TdS = dU + pdv ---------------- 3.

We know that, enthalpy h = u + pv

On differentiating, we get

dh = du + pdv + vdp,

From equation 3. , dh = Tds + vdp

Tds = dh - vdp -------------- 4.

The equations 3. and 4. are the thermodynamic equations relating the properties of system.

The following are the relations obtained from the first and second laws.

1. dQ = dE + dW: Holds good for all process, reversible or irreversible and for all systems.

2. dQ = dU + dW: Holds good for any process undergone by a closed system.

3. dQ = dU + pdV: It is good for a closed system , where pdV work is present. This relation true only for quasi-static* process.

4. dQ = TdS: This equation is true only for a reversible processes.

5. TdS = dH –Vdp: This relation hold good for any process, since there is no path function term in the equation. 

6. TdS = dU + pdV: It is good for any reversible or irreversible process, undergone by a closed system. Since the properties in the relation which are independent of the path. 

Note:

* A quasi-static process is a thermodynamic process that happens very slowly  for the system to be in equilibrium. It is reversible.

Slope of constant volume and constant pressure processes in T-s plot

# For constant pressure processes(p=C),

ds = m . Cp . dT/T ---------------------- 1.

T-s Plot

# The slope of the curve in T-s plot is dT/dS

Therefore from equation 1.,  dT/dS = T/(m . Cp) ---------------- 2.

For 1 kg of perfect gas,  

Equation 2. becomes

dT/dS = T/Cp    {Since m = 1 kg} -------------------- 3.


# Similarly for constant volume processes(v=C),  

dT/dS = T/Cv --------------------- 4.

# For perfect gas, we know that

Cp > Cv 

or

1/Cp < 1/Cv  --------------------- 5.

# Compare equation 5. with equation 2. & 4.

T/Cp  <  T/Cv

or

{dT/dS} of v=C > {dT/dS} of p=C ----------------6.

Here {dT/dS} is slope of the curve in T-s plot.

# Therefore the slope of the curve for constant volume process (a-b) is higher than that of constant pressure processes (a-b’). 




Note ;

Constant volume process , v=C.

Constant pressure process , p=C.

Cp, Cv - Specific heats at constant pressure and constant volume.

ds - Change in entropy.

Unit of Refrigeration

Tonne of Refrigeration (TR) 

The amount of heat removed (i.e., refrigeration effect) from 1 tonne* of water from 0 °C to make 1 tonne of ice at 0 °C in 24 hours.

It can also be explained as the amount of heat extracted to make 1 tonne of ice from and at 0 °C in 24 hrs.

To make ice from water at 0 °C, it is important to remove latent heat**.

For water, latent heat = 335 kJ/kg.

    Therefore, one tonne of refrigeration = 1000 X 335 kJ in 24 hrs.

   =  (1000 X 335 )/24 kJ/hr

     

   = 13,958.3 kJ/hr

   =  (1000 X 335 )/(24 X 60) kJ/min

   = 232.6 kJ/min

In practice, 1 tonne of refrigeration is equivalent to 210 kJ/min or 3.5 kJ/Sec of heat.

Note:

*1 tonne is equivalent to weight of 1000 kg.

**Amount of heat transfer required to cause a change of phase in mass of a substance at constant temperature and pressure.