If we have a non-flow process occurring, say, within an engine cylinder, then, inevitably, some heat energy is transferred to or from the gas through the cylinder walls. Usually, this heat energy is transferred out of the cylinder walls to the engine cooling water.

The quantity of heat energy transferred affects the values of pressure and temperature achieved by the gas within the cylinder. Only in an adiabatic process is there no transfer of heat energy to or from the gas, and this situation is impossible to achieve in practice.

In order to find how much heat energy has been transferred during a non-flow process, we use the non-flow energy equation (NFEE) (see page 17, ‘The non-flow energy equation’.

Q = W + (U2 – U1)

In words, this means that the heat energy transferred through the cylinder walls is the work done during the process added to the change in internal energy during the process. Remember that the internal energy of the gas is the energy it has by virtue of its temperature, and that if the temperature of the gas increases, its internal energy increases, and vice versa.

Using the NFEE is straightforward, in that we have already seen the equations for work done, and the change in internal energy for any process (Internal energy) is given by

U2 – U1 = m.cv (T2 – T1)

We can calculate each and add them together. Applying the NFEE to each process gives,

Constant volume
Q = W + (U2 – U1). W = 0 in a constant volume process.
Q = (U2 – U1)
Q = m.cv (T2 – T1)
Constant pressure
Q = W + (U2 – U1). W = p. V in a constant pressure process.
Q = p(V2 – V1) + m.cv (T2 – T1)
It is useful to digress here to establish an important expression
concerning the gas constant, R, and values of cp and cv.
We know that the heat energy supplied in the constant pressure
process is
Q = m.cp (T2 – T1)
m.cp (T2 – T1) = p(V2 – V1) + m.cv (T2 – T1)

Substituting from p.V = m.R.T), p(V2 – V1) = m.R(T2 – T1), and dividing
by (T2 – T1),
cp = R + cv for unit mass
R = cp – cv
Q = W + (U2 – U1)
Q = 0.
This is the definition of an adiabatic process.
Note: if W and U were calculated and put in the formula, the answer
would be 0.
Q = W + (U2 – U1)
Q =
(p1V1 – p2 V2)
n – 1
+ m.cv (T2 – T1)
Q = W + (U2 – U1). No change in temperature, therefore no change in U.
Q = p1V1 ln V2
Note that the ‘m.R.T’ versions of the work done expressions could be
used instead.

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