LESSON-10 VARIOUS STEADY FLOW PROCESSES FOR AN IDEAL GAS AND THEIR NUMERICAL PROBLEMS

 VARIOUS STEADY FLOW PROCESSES FOR AN IDEAL GAS

Assume that fluid a undergoes a steady flow process from inlet to exit, the energy equation for steady flow process on per unit mass basis is given as

q = (h_e - h_i) + (C_e²/2 - C_i²/2) + (gZ_e -  gZ_i )+ w_s ;                                        

If the kinetic energy and potential energy changes are negligible, the above equation reduces to

q = (h_e - h_i) + w_s ;                                                                           ……………………(10.1)

Work done or shaft work, w_s = -vdp - (C_e²/2 - C_i²/2) - (gZ_e -  gZ_i )

If the kinetic energy and potential energy changes are negligible, then the above equation is reduced to

        w_s = -vdp                                                                            ……………………(10.2)

10.1. Constant volume process (Isochoric process):   v_i = v_e

v = constant = c    

From eqn. (10.2),  w_s = - [v.(p_e - p_i)]  =   v.(p_i - p_e)      kJ/kg                          ……….(10.3)

For an ideal gas    (h_e - h_i) = C_p (T_e - T_i)                                       ………...…………(10.4)

Substituting equation (10.3) and (10.4) in eqn.  (10.1), we get

            q = C_p (T_e-T_i)  + v (p_i - p_e)    

10.2. Constant pressure process (Isobaric process):  p_i = p_e

p = constant (c)

From eqn. (10.2),   w_s = 0                                                                   ……..………. (10.5)               

For an ideal gas    (h_e - h_i) = C_p (T_e-T_i)                                               …..………… (10.6)

 Put (10.5) and (10.6) in eqn.  (10.1)

            q = C_p (T_e-T_i) 

10.3. Constant temperature process (Isothermal process): T_e = T_i

pv = constant (c)

and  pv = RT_i     or          v =

         From eqn. (10.2),    w_s ==  =                   ……….(10.7)               

For ideal gas, we have     (h_e - h_i) = C_p (T_e -T_i)              

       Therefore,  (h_e - h_i) = C_p (T_e -T_i) = 0                                                                        …………(10.8)

 Using equations (10.7) and (10.8) in eqn.  (10.1), we get

                   

        or        

 10.4. Adiabatic process 

  q = 0

        or      

From eqn. (10.2),  

 w_s                                 

=                                                                                              …….…. (10.9)

For ideal gas,    (h_e  h_i) = C_p (T_e-T_i)                                                                       …………(10.10)

Using equations (10.9) and (10.10) in eqn.  (10.1), we get

                        q = C_p (T_e-T_i)  +   = 0       

10.5. Polytropic process  

pvⁿ = c        or      

From eqn. (10.2),

w_s    

                                                                          ………....(10.11)

For ideal gas,    (h_e - h_i) = C_p (T_e-T_i)                                                   …………(10.12)

 put (10.11) and (10.12) in eqn.  (10.1)

        

10.6. Throttling Process

It is a process in which fluid flows across some restriction, as shown in Fig. 10.1, in such a manner that there is always a drop of pressure without any change in kinetic energy, potential energy of the fluid.

During throttling process, there is

• no heat transfer as the process is so fast that there is no time for the heat to transfer i.e. = 0

• no work done i.e.  = 0

Refer. Gas is made to pass through a restriction from high to low pressure side.

 

Fig. 10.1. The throttling process.

Energy Equation for steady flow process:

( h_i + C_i²/2 + gZ_i) =   ( h_e + C_e²/2 + gZ_e) +                                  .............................……....(10.13)

By using throating conditions, the equation (10.13) is reduced to      

h _i  =  h_e 

The enthalpy before and after throttling remain equal. Therefore, the throttling process is also called an equal-enthalpy process.

 

 A coefficient called Joule-Thomson Coefficient which is a thermodynamic property is defined as

         μ_J =  

 As the value of change in pressure is always –ve because p_e < p_i ,

 so  the +ve  value of Joule-Thomson Coefficient (μ_J)  means fall in temperature after throttling

and the –ve value of Joule-Thomson Coefficient (μ_J)  means rise of temperature after throttling.

10.6.1. Throttling Process for ideal gases:  

     dh = C_p dT

     Therefore,  (h_e - h_i) = C_p (T_e - T_i)                                                          ................................................…………(10.14)

      As for throttling,  h_e = h_i

      Therefore from eqn. (10.14),   T_i = T_e      

       From Joule-Thomson Coefficient definition, the  μ_J = zero.

Problem 10.1: 5 kg of air is compressed in a reversible steady flow polytropic process from 100 kpa and 40°C to 1000 kpa and during this process the law followed by the gas is pV^1.25 = C. Determine the shaft  work, heat transferred and the change in entropy C_V = 0.717 kJ/kgK , R = 0.287 kJ/kgK.

Solution:

Given: m = 5 kg;  T_i= 40°C = 40 + 273 = 313 K;   P_i = 100 kPa  = 1 x 10⁵ N/m² ;          

     P_e = 1000 kPa  = 1 x 10⁶ N/m²;   n = 1.25

            C_P = R + C_V      or    C_P = 0.287 + 0.717 = 1.005 kJ/kgK

Determine shaft work, w_s:

       Formula: Shaft work w_{s   =} 

                                            =   

                                           

                                 Finding unknown, T_e;

                                             Equation of state

Therefore,  

   or  T_e =  = 496 K

Answer:   Shaft work,   w_s =  

                                               = 

                                               = 5 x 0.287 x (-183) =  262.6 kJ/kgK

               Total shaft work, W_s = m w_s = 5 x -262.6 = 1313 kJ

Determine heat transfer, _iQ_e :

      Formula: Apply eqn. of SSSF process

                 H_i + _iQ_e = H_e + W

                         _iQ_e = (H_e – H_i)+ W

                               = m C_P(T_e – T_i) + W

Answer:   _iQ_e = 5 x 1.005 x (496 – 313) – 1313   =  393.4 kJ

Problem 10.2: An axial flow compressor of a gas turbine plant receives air from atmosphere at a pressure 1 bar, temperature 300 K and velocity 60 m/s. At the discharge of compressor the pressure is 5 bar and the velocity is 100 m/s. The mass flow rate through the compressor is 20 kg/s. Assuming isentropic compressor, calculate the power required to drive the compressor. Also calculate the inlet and outlet pipe diameters.

Solution:

Given:   p_i = 1 bar;    T_i = 300K;  C_i = 60 m/s

P_e = 5 bar;    C_e = 100 m/s

 = 20 kg/s ;    Isentropic compression

(a) Determine power required to drive compressor, ;

                 Formula: From first law of thermodynamics for steady flow process neglecting heat loss and change in P.E. and each term is expressed in kJ/kg, we have

or        h_i + C_i²/(2x1000)  =   h_e + C_e²/(2x1000)   + 

            =  - [(h_e - h_i) + (C_e²- C_i²)/(2x1000)]  = - [c_p (T_e - T_i) + (C_e²- C_i²)/(2x1000)]

                                Finding unknown, T_e ;

                                For isentropic process, the final temperature is given by

                        

                                  =  = 475.14 K

 Answer:   = - [c_p (T_e - T_i) + (C_e²- C_i²)/(2x1000)]

                         =  - [1.005 (475.14 - 300) + (100² - 60²)/(2x1000)]  = - 179.21 KJ/kg

                    

(b) Determine the inlet and outlet pipe diameters:

       Formula:  Inlet diameter,         Outlet diameter ,    

                        Finding unknown A_i and A_e,

From continuity equation,

  and      

Finding unknown, ρ_i

    or      = 1.1614 kg/m³

Finding unknown, ρ_e

     or     = 3.67 kg/m³

   

Answer:        = 0.604 m 

   = 0.263 m