First law analysis of non-flow processes ~ MECHTECH GURU

# 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.

## 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 + dWs) or dU = -dWs or – Ws = ∆U = U2 –U1

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) Previous
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