LESSON-9 FIRST LAW OF THERMODYNAMICS FOR OPEN SYSTEM AND ENTHALPY AND NUMERICAL PROBLEM

 INTRODUCTION

In previous module, we have discussed First Law of Thermodynamics to a closed system in which the working fluid does not flow in or out of the system. But in most engineering devices such as boiler, turbine, condenser, pump, steam engine, evaporator, etc., the working fluid continuously flows in and out of these devices/systems. Therefore, it is also essential that we consider the First Law of Thermodynamics for open system.

Before we define First Law of Thermodynamics for open system, it is necessary to find different form of energies associated with mass of fluid flows in and out of the open system. Because in flow process, the energy contained in the system is not only influenced by work and heat, like these do in closed system but it is also influenced by energies associated with mass of fluid flows in and out of the open system.

The energy associated with mass of fluid flows come into picture only when fluid flows in and out of the system. That is why in closed system these energies are absent and we deal with only energy interaction between system and surroundings in the form of work and heat.

9.1. ENERGY ASSOCIATED WITH MASS OF FLUID FLOWING IN AND OUT OF THE OPEN SYSTEM

The energies associated with per unit mass of the fluid flowing in and out of the system are as follows:

(a)  Internal energy, u                   (SI units:   J/kg or kJ/kg)

(b) Kinetic energy,                (SI units:   J/kg or kJ/kg)

(c)  Potential energy, g.Z              (SI units:   J/kg or kJ/kg)

(d) Flow work or flow energy:       The flow work or flow energy is the work required to cause the flow of fluid in a passage against the existing pressure of fluid. Consider 1 kg of fluid, flowing through a pipe of uniform cross-section area ‘A’, moves a distance of ΔL as shown in Fig. 9.1 and let the pressure intensity ‘p’ (N/m2) and velocity ‘C’ (m/s) over the cross-section ‘a-a’ be uniform.

Fig. 9.1. Flow work.

With these assumptions, the force ‘F’ acting at any cross-section such as ‘a-a’ is given by:

F = p × A

Flow work required to move 1 kg of fluid from section ‘a-a’ to section ‘b-b’ is given by:

Flow work  (w_f) = F. ΔL  = p . A .ΔL                                                                       

or                        = p . v                                                                      (SI units:   kJ/kg)

where, v is the specific volume of gas flowing (m³/kg).

Thus the flow work required to flow 1 kg of fluid in a passage is the product of pressure and specific volume of the fluid. This flow work is provided by the pump located somewhere else in the surrounding. 

9.2. FIRST LAW OF THERMODYNAMICS FOR OPEN SYSTEM

Consider open system/control volume (C.V.) as shown in Fig. 9.2.

9.2.1.  Unsteady flow process:

Let,

δm_i is mass of fluid entering the C.V.;

δm_e is mass of fluid leaving the C.V.;

δQ is amount of heat transferring into the C.V.;

δW is amount of work done by the C.V.  

(u_i + p_iv_i + ½C_i² + gZ_i) and (u_e + p_ev_e + ½C_e² + gZ_e) are the energies associated with fluids entering and leaving the C.V, respectively.

where, u, pv, ½C², and gZ is internal energy, flow work, kinetic energy and potential energies respectively.

 

Fig. 9.2. An open system

Since, the energies entering and leaving the system vary with time, it will result in change in total energy within the C.V. which is given by

    = d(U + ½ mC² + mgZ)_б 

Here subscript ‘б’ represents condition within the system.

By applying energy balance to the C.V.,

Difference in the energies entering and leaving the C.V. = change in total energy within the C.V.

 [δm_i (u_i+ P_iv_i+ ½ C_i²+ gZ_i) + δQ] – [δm_e(u_e+ P_ev_e+  ½ C_e²+gZ_e) + δW ] = d(U + ½ mC² + mgZ)_б

  or    δQ + δm_i (u_i+ P_iv_i+ ½ C_i²+ gZ_i) =  δm_e(u_e+ P_ev_e+ ½ C_e²+gZ_e) + δW + d(U + ½ mC² + mgZ)_б

By considering analysis for time period of ‘δt’, this law may also be written as a rate equation

        ……………….(9.1)

or                   ……………….(9.2)

Here, we introduced new thermodynamic property, enthalpy. Enthalpy is an extensive property (point function) and is given by symbol ‘H’ (SI unit: kJ) or ‘h’(SI unit: kJ/kg). Enthalpy is defined as the sum of internal energy and pressure-volume product.

i.e.  H = U+ p.V                 and           h = u+ pv         

By introducing enthalpy, the equations (9.1) and (9.2) can be written as

                             ……………….(9.3)

or                       ……………….(9.4)

The equations (9.2) and (9.4) are the First Law of Thermodynamics for the open system for unsteady flow process.

9.2.2.  Steady flow process

In many engineering applications, some of the processes can be treated as steady flow processes. A steady flow process is one in which the following conditions exist.

(i)       The mass rate of flow entering and leaving the system does not vary with time. Also mass rate of flow at entrance and exit of the C.V. is the same, then

           i.e.  = constant =  

(ii)      The state, velocity, and elevation of the fluid entering and leaving the C.V. do not vary with time.

    i.e. h_i, C_i, Z_i, h_e, C_e, and Z_e are  not functions of time.

(iii)     The state, velocity, and elevation of the fluid at each point within the C.V. do not vary with time. This results in no change of total energy within  the C.V.

          i.e.   d(U + ½ mC² + mgZ)_б =0

(iv)     The rate of heat and rate of work crossing the boundary of C.V. do not vary with time.

          i.e.   are not functions of time.

Steam turbine is a typical example of a steady flow process that operates at constant load, constant steam flow rate and constant rate of heat transfer.

For steady flow process, the First Law of Thermodynamics in equation (9.4) is reduced to

                         ( SI units: J/s or kJ/s)                  ………….(9.5)

all the above terms indicate energy per unit time                                        

Divide the above equation by  in kg/s, the equation (9.5) is reduced to

                                  ( SI units: J/kg or kJ/kg)                ………….(9.6)

all the terms indicate energy per unit mass.

Note:- All forms of energies should be expressed in one unit. For example all should be either in kJ/kg   or   kJ/s (kW).

The equation (9.6) can also be written in the form

                                                              ……………………(9.7)

The equations (9.5), (9.6)  and (9.7) are the First Law of Thermodynamics for the open system for steady flow process or the steady flow energy equations.

9.2.3.  When more than one fluid enters and leaves the system in a steady flow, we can write

or  

      where   are masses of different fluids entering the system in a given time interval

    are masses of different fluids leaving the system in a   given time interval

and from equation (9.6), the steady flow process becomes

[h_i + C_i²/2 + gZ_i]_a +  [h_i + C_i²/2 + gZ_i]_b +  [h_i + C_i²/2 + gZ_i]_c +…. =  [ h_e + C_e²/2 + gZ_e]_d +  [ h_e + C_e²/2 + gZ_e]_e +  [ h_e + C_e²/2 + gZ_e]_f  + ….+ 

or     

9.3.  MECHANICAL WORK IN A STEADY FLOW SYSTEM

We have seen that in a non-flow process (i.e. in a closed system), the work done = δW =  p .dV  or δw =  p .dv  (on per unit mass basis). For a steady flow reversible process, a similar expression for work done (w_sf)  is also feasible. From equation (9.2), the steady flow energy equation per unit mass in differential form can be written as, 

δq = dh + d(C²/2) + d(gZ) + δ w_sf  ;

δq = dh + d(C²/2) + d(gZ)  + δ w_sf   ;

δq = du + d(Pv) + d(C²/2) + d(gZ) + δ w_sf  ;           [  h = u + pv]

δq = du + dw_f + d(C²/2) + d(gZ)  + δ w_sf

where,  dwf  = change in flow work/unit mass,	du  = change in internal energy/unit mass,
d(V2/2)   = change in kinetic energy /unit mass,	d(gZ) = change in potential energy /unit mass

According to first law of thermodynamics, for the moving system, to an observer moving with the fluid, the attention gets focused on a seemingly stationary system, and therefore,

δq = du + Pdv

i.e. the same equation holds good for flow as well as non-flow processes. Therefore, the expression for enthalpy can be reduced to,

du + Pdv = du + d(Pv)  + d(C²/2) + d(gZ)  + δ w_sf

or    du + Pdv = du + Pdv + vdP + d(C²/2) + d(gZ)  + δ w_sf

or    0 = vdP + d(C²/2) + d(gZ)  + δ w_sf

δw_sf  = - vdP - d(C²/2) - d(gZ) 

Therefore    w_sf =  − vdP - (C_e²/2 - C_i²/2) - (gZ_e-  gZ_i )

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

              

Work done per unit mass for steady flow process (w_sf ) is also called shaft work and is represented by w_s.

 9.4.  ENTHALPY FOR AN IDEAL GAS

 The enthalpy for an ideal gas is a function of temperature alone i.e. H = f(T) only.

 Therefore for an ideal gas, the change in enthalpy is given by

                    dH = m C_p dT                                             (SI unit: J or kJ)

 i.e. For a process from inlet to exit, it is   H_e-H_i = m C_p (T_e-T_i)

 Based on unit mass the change in enthalpy for an ideal gas is given by

                    dh = C_p dT                                                 (SI unit: J/kg  or kJ/kg)

For a process from inlet to exit, it is     h_e-h_i = C_p (T_e-T_i)

 where,    m = mass of a gas, kg

dH  =  change in enthalpy of a gas during a process,

dh   = change in specific enthalpy of a gas during a process,

             dT = change in absolute temperature of a gas between two states of a process, K

             C_p = Specific heat of a gas at constant pressure, J/kg K   or  kJ/kg K

Problem 9.1. Steam is supplied to a fully loaded 100HP turbine at 1470KN/m² and internal energy of 2944.2kJ/kg and volume of 0.16 m³/kg and velocity of 110m/s. Exhaust takes place at 4.9kN/m² with internal energy of 1890 kJ/kg and specific volume equal to 26 m³/kg and velocity of 300m/s. Heat loss from steam turbine is 21 kJ/kg. Potential energy change is negligible. Determine:

i) Shaft work out put/kg

ii) Steam flow rate in kg/h

Solution:

Given:   Turbine power output at full load = 100 H.P.; 

             p_i = 1470 KN/m²;  u_i = 2944.2 kJ/kg;   v_i= 0.16 m³/kg;   C_i = 110 m/s;

             p_e = 4.9 KN/m²;  u_e = 1890 kJ/kg; v_e = 26 m³/kg; C_e = 300 m/s;

              q = - 21 kJ/kg;    Z_i = Z_e

(a) Determine the shaft workout put/kg, w_s:

          Formula: From the first law for flow process,  

       h_i + ½ C_i²+ gZ_i+q = h_e + ½ C_e²+ gZ_e + w_s

or   u_i+ P_iv_i + ½ C_i²+ gZ_i + q = u_e+ P_ev_e + ½ C_e²+ gZ_e + w_s

or   w_s = q + (u_i – u_e) + (P_iv_i – P_ev_e) + ½ (C_i² – C_e²)

Answer:    w_s = (– 21) + (2944.2 – 1890) + (1470 x 0.16 – 4.9x 26) + ½ (110 – 300)

                          = 934.05 kJ/Kg

(b) Determine the steam flow rate in kg/h:

              Formula:  Steam flow rate, 

Answer:   Steam flow rate, 

    1 H.P. = 736     (W or J/s)   or    =    = 2649.6 KJ/h