Irreversibility and Entropy of an Isolated System

5. Irreversibility and Entropy of an Isolated System

Since δQ = 0 for isolated systems, and δσ > 0 for irreversible processes, Eq. (28) yields that dS > 0. When warm water is exposed to ambient air, as illustrated in Example 9, the

system “drifts” in the absence of internal constraints towards an equilibrium state. We saw from Example 9 that for a composite system consisting of (1) warm water directly losing heat to air (cf. Figure 16a ) and (2) that water being supplied with heat equal to the lost value through a Carnot heat pump, a net work loss occurred. This loss is called the irreversibility I of the composite system. In the case discussed in Example 9,

I=T 0 σ.

A rigorous proof of this equality is contained in Chapter 4. j. Example 10

An uninsulated coffee pot is maintained at a temperature of 350 K in a 300 K ambient –2 by supplying 1050 W of electrical work. The heat transfer coefficient is 0.2 kW m –1 2

K , and heat transfer occurs over a pot surface area of 0.5 m . Determine the entropy generated:

In the system contained within the boundary cs 1 , as illustrated in Figure 20a (i.e., for only the coffee within the pot), assuming the pot boundary temperature to be 350 K.

The matter contained within cs 2 , as illustrated in Figure 20b (i.e., for the system in- cluding both the coffee and pot) For the system containing the coffee, pot, and the ambient (i.e., bounded by the sur-

face cs 3 illustrated in Figure 20c ) for which T b =T 0 which is the ambient temperature. Solution

Selecting the control surface internally, and applying the First law, ˙ Q – W ˙ elec = dE/dt. At steady state, dE/dt = 0 so that ˙ Q = W ˙ elec = –1050 W. Applying the entropy balance equation in rate form dS/dt –

Q /T ˙ b = σ ˙ , we obtain

0 – (–1050/T b )= σ ˙ . Since T b = 350 K,

= 1050÷350 = 3 W K –1 . Selecting the control surface

cs 2 to be flush with the pot walls, the boundary tempera- ture T b must be determined. Applying the convection heat transfer relation

h A (T b –T 0 )= ˙ Q = 1050

W, the boundary temperature is determined as,

T b = 1.05 ÷ (0.2 × 0.5) + Figure 20: Entropy generation within a coffee pot.

300 = 310.5 K, and σ ˙ =

–(–1050 ÷ 310.5) = 3.382 W K –1 . Upon comparison with the previous solution, we find that irreversible heat transfer

between the coffee and pot walls causes an entropy generation of 3.382 – 3 = 0.382 W K –1 .

Selecting the control surface cs 3 such that the boundary exists outside the pot, T b =T 0 ,

0 – (–1050 ÷ 300) = -1 σ, i.e., σ ˙ = + 3.5 W K . No irreversibilities exist outside the boundary of the control surface cs 3 . The entropy

change in this composite system (using T b =T 0 ) equals the entropy change in an iso- lated system, since there is no entropy production within the ambient. Remarks For the matter contained within the surfaces cs 2 and cs 3 which include the pot wall, σ ˙ = 0 – (1050 ÷ 310.5) – (–1050 ÷ 300) = 0.118 W K –1 due to the heat transfer between the ambient and pot walls. By a suitable choice of the boundary, we are able to determine contributions to over- all σ. The major contribution is due to destruction of electrical work into heat called

electrical frictional work. The change in entropy due to:

destruction of electrical work within the coffee pot = 3 W K –1 . irreversible heat transfer between coffee and pot walls = 0.38 W K –1 . irreversible heat transfer between pot walls and ambient = 0.12 W K –1

. The change in entropy of the isolated system = 3.5 W K .

k. Example 11 An uninsulated coffee pot is maintained at a temperature of 350 K in a 300 K ambi- ent. Instead of supplying electrical work, we can compensate for the heat loss by

placing a heat pump between the coffee pot and ambient, as shown in Figure 21 . What is the electrical work required to operate the heat pump?

Solution: COP = 350 ÷ (350–300) = 7, i.e., W ˙ elec = 1050 ÷ 7 = 150 W. The pot can be maintained at 350 K by providing 150 W of electrical power to a heat

engine, rather than directly supplying 1050 W as in the previous example.