LESSON-7 APPLICATION OF FIRST LAW IN HEATING AND EXPANSION OF PERFECT GASES IN NON-FLOW PROCESSES (CLOSED SYSTEM) AND THEIR NUMERICAL PROBLEMS

7.1. CONSTANT VOLUME PROCESS (ISOCHORIC PROCESS):

Refer Fig. 7.1.

 

Fig. 7.1. Arrangement for constant volume heat addition and its p-V diagram.

p-V-T relationship:

     V1 = V2 = V = constant ( c )     

Applying the first law of thermodynamics for a change of state, we get

     1Q2 = (U 2 – U 1) +  1W2                                                                                                    ………..…….….(7.1)    

     V1 = V2 = V = constant      or   dV = 0   

Work done:

         1W2 =   =  0

Heat transfer:

 From equation (7.1) we have,         1Q2 = U2 – U1 + 0     

1Q2 = U2 – U1                

On per unit mass basis,        

1q2 = u2 – u1        

       where u is specific internal energy of the gas, kJ/kg

Hence during constant volume process the heat interaction is equal to the change in the internal energy of the system.

For a perfect gas,    dU = m Cv dT      

The total quantity of energy transferred during a constant volume process when the system temperature changes from T1 to T2 can be given as

           1Q2 = U2 – U1 = m Cv (T2 – T1)                               (kJ)

On per unit mass basis,       

          1q2 = CV (T2 – T1)                                                  (kJ/kg)

7.2. CONSTANT PRESSURE PROCESS (ISOBARIC PROCESS):

Refer Fig. 7.2.

Fig. 7.2. Arrangement for constant pressure heat addition and its p-V diagram.

p-V-T relationship:

p1 = p2 = p = constant

Applying the first law of thermodynamics for a change of state, we get

1Q2 = (U 2 – U 1) +  1W2                      …….(7.2)    

Work done:

1W2 =    = p(V2 – V1) ;                              ( p1 = p2 = p = constant)

Heat transfer:

From equation (7.2), we have

  1Q2 = U2 – U1 + P(V2 – V1)

           = U2 – U1 + P2V2 – P1V1 

        = (U2 + P2V2) – (U1– P1V1) 

          1Q2 = H2 – H1                                      (  H = U + PV)

where, H is the thermodynamic property

On per unit mass basis,       1q2 = h2 – h1         

where h is specific enthalpy of the gas, kJ/kg

 Hence during constant pressure process the heat interaction is equal to the change in enthalpy of the system.

For a perfect gas,    dH = m Cp dT      

The total quantity of energy transferred during a constant pressure process when the system temperature changes from T1 to T2 can be expressed as

1Q2 = H2 – H1 = m Cp (T2 – T1)      (kJ)

On per unit mass basis,         1q2 = Cp (T2 – T1)      (kJ/kg)

 7.3. CONSTANT TEMPERATURE PROCESS (ISOTHERMAL PROCESS):

 Refer Fig. 7.3.

                              

                                                               Fig. 7.3. Arrangement for isothermal process and its p-V diagram.

p-V-T relationship:

T1 = T2 = T = constant 

For a perfect gas     pV = mRT    

PV = c    or  p1V1 = p2V2 = Constant

Applying the first law of thermodynamics for a change of state, we get

1Q2 = (U 2 – U 1) +  1W2                    …….(7.3)    

Work done:

1W2 =  =        (  pV = c  or p = )

or    1W2  =  

On per unit mass basis,        

         where v is specific volume, m3/kg

Heat transfer:

For perfect gas,    dU = m Cv dT      

 or      (U2-U1) = m Cv (T2 – T1) ;

 or      (U2 - U1) = 0                                  ( T1 = T2  )

Substituting values of (U2 – U1) and 1W2   in equation (7.3), we get

Therefore, 1Q2 = 0 + p1V1 ln  

1Q2   =  p1V1 ln     

PV = c =  or     P1V1 = P2V2 = mRT1

Therefore,      1Q2  = p2V2 ln    = mRT1 ln   =     , kJ   

       On per unit mass basis,         1q2 = p1v1 ln    = p1v1 ln               , kJ/kg     

7.4. REVERSIBLE ADIABATIC PROCESS:              

Refer Fig. 7.4.

                               

 

                                                                                       Fig. 7.4. Arrangement for adiabatic process and its p-V diagram.

Governing equation:

Applying the first law of thermodynamics for a change of state, we get

1Q2 = (U 2 – U 1) +  1W2                                                    ………………. (7.4)  

1Q2 = 0 ;

Therefore,   0 =  U2 – U1 + 1W2

 dU + δW = 0        

 dU + pdV = 0                     ( δW = pdV)

      For perfect gas,  dU = m Cv dT

m Cv dT + pdV = 0                                          ……………………… (7.5)        

For perfect gas,    pV = mRT  

pdV + Vdp = mRdT  

dT =              

Substitute dT from the above eqn. into eqn. (7.5)

Cv + pdV = 0 ;         

Cv (pdV + Vdp) + R pdV = 0

Cv (pdV + Vdp)+ (Cp – Cv)pdV = 0                                       ( Cp – Cv = R)

Cv pdV + Cv Vdp + Cp pdV – Cv pdV = 0             

Cv Vdp+ Cp p dV = 0        , Divide by  (p V Cv)

     ;  

 

or       

ln(p) + ln(V) = constant 

= c   

Therefore an adiabatic process is mathematically expressed as = c

or     = Constant =  c

p-V-T relationship:

Now   = c   and

  pV = m RT   

or      p1V1 = mRT1     &       p2V2 = mRT2

Therefore,   

or               

Work done:

Now 1W2 =

                                            

            

or    1W2 =                ,   kJ                                          

   pV= mRT   

 or   p1V1 = mRT1     &      p2V2 = mRT2

1W2 =          kJ                                                               …………………..(7.6)

On per unit mass basis,         1w2 =               , (kJ/kg)     

             where v is specific volume, m3/kg

Heat transfer:

         1Q2 = 0

 As   1Q2 = 0 ;  From equation (7.4) we have

0 =  U2 – U1 + 1W2              or           U2 – U1 = - 1W2

From eqt (iv), put value of 1W2

U2 – U1 = −

 7.5. POLYTROPIC PROCESS:

Polytropic process is given by the following equation

p V n = constant = c                                                              ………..………..   (7.7)          

or   p1 V1n = p2 V2n

Differentiate equation (7.7)

Vndp + p nVn-1dV = 0 

or  

Therefore the slope of the p-V curve increases in negative direction with increase in 'n' (polytropic index). This is shown in Fig. 7.5.

From equation (7.7) we can conclude that

If   n = 1, then the process is isothermal (temperature constant);

If   n = , then the process is adiabatic (No heat transfer);

If   n = 0, then the process is isobaric (pressure is constant);

If   n =  ± ∞, then the process is isochoric (volume is constant).

All these processes are shown in Fig. 7.5.


Fig. 7.5. Various thermodynamic processes

p-V-T relationship:

Polytropic process is given by

 p V n =  c      and     pV = mRT   

or    p1V1 = mRT1   &       p2V2 = mRT2

Therefore,     =    

or         

Work done:

1W2 = pdV = 

or   

     pv = mRT 

or   p1v1 = mRT1  &  p2v2 = mRT2

Therefore,    1W2 =                     ( kJ)                                              ………………(7.8)

On per unit mass basis,         1w2 =               (kJ/kg)     

             where, v is specific volume, m3/kg

Heat transfer:

Applying the first law of thermodynamics for a change of state, we get

        1Q2 = (U 2 – U 1) +  1W2                                                                                                         ..............…….(7.9)  

 From equation (7.8), using the value of 1w2

                      ( For perfect gas,  U2 – U1 = m Cv (T2 – T1))

                          ( R = Cp – Cv)

      &         R = Cp – Cv  ,       from these two relations, we get 

      ,  Using this in the above equation we get

  

                            (kJ)

 From equation (7.8)        

                       

On per unit mass basis, 1q2 = =           , (kJ/kg)     

7.6. EQUAL-INTERNAL ENERGY (FREE EXPANSION) PROCESS:

To analyze this process, consider a system having insulated vessel in which there are two compartments separated by a membrane as shown in Fig. 7.6 (a). There is a gas in one compartment while the other compartment is in a state of complete vacuum. Let the membrane be ruptured and gas fills the entire volume of the compartment (Fig. 7.6 (b)). Neglect any work associated with the rupturing of the membrane.  The free expansion process on U-V diagram is shown in Fig. 7.7.   

Since the tank is insulated there is no heat transfer for change of state from state 1 to state 2 in the vessel i.e.  1Q2 = 0.

Since the expansion is free from any resistance, no work is done by the system for change of state in the vessel. i.e.  1W2 = 0.

                      

             Fig. 7.6. Example of a process involving a change of volume for which the work is zero.

                                

                          Fig. 7.7. U-V diagram of free expansion process.  

Applying the first law of thermodynamics for a change in state, we get

       1Q2 = (U2 – U1) +  1W2             

      0 = (U2 – U1) +  0              

Therefore, (U2 – U1) = 0 ;               

U1 = U2

       It is  constant internal energy process.

       For perfect gas   (U2 – U1)  = m Cv (T2 – T1)

              Therefore, T1 = T2                    

Last modified: Saturday, 7 September 2013, 5:28 AM