The entropy generated when Austin's father takes his daily shower needs to be determined.

Assumptions:

• The shower operates steadily at maximum flow condition.
• Heat loss from the pipes is negligible.
• Water is an incompressible substance with a density equals 1000 kg/m³ and a specific heat equals 4.18 kJ/(kg-K).
• The kinetic and potential energies are negligible.

Take the T-elbow and
the Electric Heater together
as a Control Volume

Take the T-elbow and the electric heater together as a control volume shown on the left. Cold water at 15^oC enters the electric heater (denoted as 1) and the T-elbow (denoted as 2), and leaves the T-elbow at 40^oC. Since the flow is steady, the mass balance for this control volume is

The mass flow rate to the shower head is

The entropy balance for a control volume at steady state is

It is assumed that there is no heat transfer between the pipes and the surroundings. That means no entropy flows in or out of the control volume associated with the heat transfer. The control volume has electric work input, but the work is entropy free. Hence, the entropy balance for this steady-flow control volume can be simplified as

Since water enters the T-elbow and the electric heater at the same temperature,

      s₁ = s₂

Also, with the mass balance, the entropy generation can be determined by

Hence, a 15-minute shower will generate entropy

      S_gen = (15)(60)(58.1) = 52,290 J/K = 52.3 kJ/K