First law analysis of nonflow 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 = U_{2} –U_{1}
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 = mC_{v} (T_{2} –T_{1})
The path followed by the gas is shown on a PV 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 – (dW_{pdv} + dW_{s}) or dU = dW_{s} or – W_{s} = ∆U = U_{2} –U_{1}
Where dW_{pdv} is the compression /expansion work and dW_{s} 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
Application of the first law of thermodynamics gives
or Q = ∆H
dW_{s} = 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 P_{1} to P_{2} 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(v_{2}/v_{1})


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