“THERMODYNAMIC AND HEAT TRANSFER” University of Rome – Tor Vergata Faculty of Engineering – Department of Industrial Engineering Accademic Year dr. G. Bovesecchi (7249) THERMODYNAMIC PROCESSES

Joule experiment Two rigid, insulated tanks are interconnected by a septum. Initially 1kg of hydrogen at 300 K fills one tank (0.5m 3 ). In the other one there is vacuum. The septum is removed and the gas can expand until a final equilibrium state is reached. During this process, there are no heat or work interactions between the tank content and the surroundings (Joule experiment). Assume that air is an ideal diatomic gas. Calculate: the entropy variation of the gas; the final temperature; the final pressure. Ex. 6

Joule experiment Solution This is an isothermal transformation (being the gas ideal), so T=const. 1kg Hydrogen Vacum Hydrogen Initial state Final state

Joule experiment The final pressure is given by the isothermal transformation equation: The initial pressure comes from the ideal gas state equation.

Joule experiment Now we can calculate the entropy variation: The total entropy variation is:

Adiabatic process 1 kg of nitrogen expands (closed system) from 20atm (220 K) to 1 atm. If the process is reversible adiabatic, calculate: the final temperature assuming the gas ideal (c p =3.4R); the total work done in the hypothesis of ideal gas; the above quantity if the gas is real. If the process has an entropy production of J/kg K, calculate: The final temperature (ideal gas hyp.) The final temperature (real gas hyp.) Ex. 7

Adiabatic process If the transformation is a reversible adiabatic, and the gas is assumed ideal, the final temperature can be calculated from the adiabatic equation: Solution

Adiabatic process If the gas is real the final temperature is evaluated from the T-S diagram.

Adiabatic process The transformation is described by the vertical segment between the two isobaric curve (20 and 1atm).

Adiabatic process The final temperature so is about 92 K. The ideal reversible adiabatic transformation means that: If the gas is ideal and the system is closed, ∆U = c v ∆T, so the work is:

Adiabatic process The total work is given by the relation: For a closed system the internal energy is given by the relation: So W is: pV and H can be easy determinated from the diagram.

Adiabatic process

We must convert l · atm · mol -1 to the international standard units And convert cal · mol -1 to the international standard units

Adiabatic process So the total work can now be calculated: Entropy production If the gas can be considered ideal: but

Adiabatic process so

Adiabatic process The final temperature in the real gas hypothesis can be read from the chart. We must calculate the final entropy of the real transformation. Where is the final entropy of the reversible adiabatic expansion

Adiabatic process Using and the isobaric curve at 1atm we can find the final temperature of the real transformation.

Adiabatic process 129 K

Compressor An air mass flow rate (1.5 kg/s) undergoes in a compressor (β=10, η is =0.6). The air is at 1bar and 75°C. Using the ideal gas model for air (diatomic gas), calculate the outlet temperature if: the transformation is a reversible adiabatic; the transformation is irreversible; the technical work for both transformation. Ex. 8

Compressor Solution The ideal transformation is described by the polytropical equation for ideal gas: INLET OUTLET 1 bar 75°C 10 bar ???°C η=0.6 β=10 W p2p2 p1p1 T s ideal real s1s1 s2s2

Compressor The final temperature in the ideal transformation is:

Compressor If we consider now the real transformation, we have to take into account the compression isentropic efficiency (η=0.6)

Compressor The technical work is: And the compressor power consumption is: