ME 200 – Thermodynamics I – Spring 2020

Homework 24: Introduction to Entropy

System Sketch:

A two-phase mixture within the piston-cylinder is the system.

Given:

The two-phase mixture is undergoing an unknown process that could involve heat or work transfer.

Find:

i) What is the minimum possible change in entropy (kJ/K) that would occur for this system if the phase change occurs at constant temperature of 27°C with a heat transfer to the fluid of 100 kJ.

ii) Would it be possible for the entropy of this closed system to decrease? If so, what would be necessary? Support your answer with a basic equation and example.

iii) What if there is no heat transfer, but work is being done on the system. Would it be possible for the entropy to decrease? Explain. What would be necessary for the entropy to remain constant for this example?

iv) What if the initial two-phase mixture has a quality of 0.1 at a temperature of 27°C and a process occurs such that final state of the fluid is a superheated vapor at 200°C and 3 bar. Which case would have the greater entropy change in the fluid: a) an internally reversible process with heat transfer, b) an adiabatic process with friction and other irreversibilities, c) the same entropy change, or d) not possible to evaluate. Justify!

Assumptions:

1) closed system

Basic Equation:

$$S_2 - S_1 = \int _1 ^2 \bigg (\dfrac{\delta Q}{T} \bigg )_{int,rev}$$

Solution:

Part (i)

Based on the Clausius Inequality or an entropy balance for a closed system, the minimum possible entropy change would occur for an internally reversible process. In this case,

$$S_2 - S_1 = \int _1 ^2 \bigg (\dfrac{\delta Q}{T} \bigg )_{int,rev}$$

If the temperature is constant, then the integral on the right-hand side reduces to

$$S_2-S_1=\dfrac{Q_{12}}{T}$$

T = 300.15     # Temperature of the mixture inside the cylinder [K]
Q_12 = 100     # Heat transfer to the cylinder [kJ]

Delta_S_12 = Q_12/T

print('Delta_S_12 = ',round(Delta_S_12,3),'kJ/K')

Delta_S_12 =  0.333 kJ/K

Part (ii)

Yes, the entropy of a closed system can decrease if the heat transfer is negative (i.e, out of the system). For the internally reversible heat transfer considered in part (i),

$S_2-S_1=\dfrac{Q_{12}}{T}$ $<$ $0$ for $Q_{12}$ < $0$

Part (iii)

In the absence of any heat transfer, the minimum possible entropy change that would occur for an internally reversible process would be zero according to

$$S_2 - S_1 = \int _1 ^2 \bigg (\dfrac{\delta Q}{T} \bigg )_{int,rev} = 0$$

Any irreversibilities associated with the process would result in an energy change larger than this minimum value or

$$S_2 - S_1 > 0$$

Part (iv)

Entropy is a property and depends only on the state. Therefore, the entropy change for any process (reversible or irreversible) between two known states is the same regardless of the path followed. It doesn't matter how the system transitioned between those two states. Therefore, the answer is "c".