First law analysis of non-flow processes

First law analysis of non-flow processes

The first law of thermodynamics can be applied to a
system to evaluate the changes in its energy when it undergoes a change of
state while interacting with its surroundings. The processes that are usually
encountered in thermodynamic analysis of 
systems  can  be 
identified as any one or a combination of the following elementary processes:

  1. Constant volume (isochoric)  process

2. Constant pressure (isobaric)  process

3. Constant temperature
(isothermal)
process.

4. Adiabatic process.

1. Constant volume process

Suppose a gas
enclosed in a rigid vessel is interacting with the surroundings and absorbs
energy Q as heat. Since the vessel is rigid, the work done W due to expansion
or compression is zero. Applying the first law, we get

  dU = dQ or Q = U2 –U1

That is, heat interaction is equal to the change in
internal energy of the gas. If the system contains a mass m equal of an ideal
gas, then

Q = ∆U = mCv (T2 –T1)

The path followed by the gas is shown on a P-V diagram.
Now consider the fluid contained in   a
rigid vessel as shown. The vessel is rigid and insulated. Shaft work is done on
the system by   a paddle wheel as shown
in Fig. a. In Fig. b electric work is done on the system. Since the vessel is
rigid, the PdV work is zero. Moreover, the vessel is insulated and hence dQ = 0.

Application of the first law of thermodynamics gives

dU = dQ – dW = dQ – (dWpdv +
dW
s) or dU = -dWs or – Ws = ∆U = U2
–U
1

Where dWpdv is the compression /expansion
work and dWs is the shaft work. That is increase   in internal energy of a system at a constant
volume, which is enclosed by an adiabatic wall, is equal to the shaft work done
on the system.

2. Constant pressure process

Several industrial processes are carried out at constant
pressure. A few examples of constant pressure processes are: (a) reversible
heating/cooling of a gas (b) phase change (c) paddle
  wheel work (d) electrical work. For a constant pressure
process, the work done W is given by

 W = ∫PdV = P (V2-V1)

Application of the first law of thermodynamics gives

 dU = dQ – dW = dQ – PdV = dQ – d(PV) or dQ =
dU + d(PV) = d(U + PV) = dH

or Q = ∆H

 That is in a constant pressure process, the heat
interaction is equal to the increase in the enthalpy of the system. Now
consider the constant pressure processes in which the system is enclosed by an
adiabatic boundary. Application of the first law gives:

 dU = dQ – dW = dQ – (PdV + dWs)

 Here, the net work done (dW) consists of two parts – the
PdV work associated  with  the 
motion of the boundary and (-dWs), the shaft work (or electrical work) done by the surroundings. Since the system is enclosed by an adiabatic boundary, dQ = 0 the equation can be written as

-dWs
= dU + d(PV) = dH

That is, the increase in the enthalpy of the system is
equal to the shaft work 
done on the  system.

2. Constant temperature process

Suppose a gas enclosed in the piston
cylinder assembly is allowed to expand from P1 to P2  while the temperature is held constant. Then
application of the first law
gives:

dU = dQ – dW = dQ –PdV

It is not possible to calculate work and heat
interactions unless the relationships between the thermodynamic properties of
the gas are known. Suppose the gas under consideration is an
  ideal gas (which follows the relation Pv = RT
and u = u(T) only) then for an
 
isothermal process,

 dU = 0

dQ = PdV = RTdv/v or Q =W = RTln(v2/v1)





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