E-Storethermodynamic process
| Identifier | Description | Units (typical) |
| c p | Specific Heat Capacity at Constant pressure | kJ/(kg K) |
| c v | Specific Heat Capacity at Constant Volume | kJ/(kg K) |
| P | Absolute Pressure | N / m 2 |
| T | Absolute Temperature | K |
| V | volume | m 3 |
| m | mass | kg |
| W | Work Output per unit mass | kJ/kg |
| M | Molecular Weight | - |
| R o | Universal Gas Constant = 8,31 | kJ /(kg mole.K) |
| Q | Heat Quantity | kJ |
| R | Gas Constant = R o / M | kJ /kg.K |
| U | Internal energy (thermal) | kJ |
| γ | Ratio cp / cv | - |
Thermodynamic Process Relationships
The majority of frictionless processes for ideal gases are called polytropic processes and are in accordance with the following relationship
PV n = constant
That is PV n = c
therefore P = cV -n
Equation of state for and Ideal Gas
PV = mRT
Thermodynamic Relationships between P,V & T
Consider a piston in a frictionless cylinder
The work done on/by the gas in moving the piston δx = (PA)δx = P δ V = δ W
The gas is assumed to be expanding in balanced resisted reversible process.
The equation of state for an ideal gas is assumed to apply i.e PV = mRT
The total work done in moving the piston from state 1 to state 2 =
For a perfect gas - The relationship between Temperature , Pressure and Volume over a cycle
Adiabatic Process.
For an adiabatic process with no transfer of heat across the system boundary.(Q = 0 )
Consider a fixed mass of gas in a cylinder which is expanding in a reversible manner...
For an adiabatic process there is no heat transfer.
Therefore applying the first law of thermodynamics ..heat transfer (= 0) = increase in internal energy +external work done by gas...
Therefore the increase in internal energy = - External work done by gas
It is shown below that cp - cv = R = cv ( cp / cv -1) and therefore
γ = 1.4 for Air, H 2, O 2, CO, NO, Hcl
γ = 1.3 for CO 2, SO 2, H 2O, H 2S, N 2O, NH 3, CL 2, CH 4, C 2H 2, C 2H 4
Isothermal Process
In a isothermal process the temperature = constant and therefore
PV = c and P = c / V
Internal Energy, Cp and Cv
Although it is not possible to determine the absolute value of the internal energy of a substance. The internal energy change between the initial and final equilibrium states of any process is definite and determinable.
It can be easily proved that the internal energy of a fluid depends on the temperature alone and not upon changes in the pressure or volume.
Heating at constant volume....
If a definite mass of gas (m) at constant volume is a closed system is heated from initial conditions P1, V, T1, U1 to P2, V , T2,U2. As the volume is fixed then no work has been done. Then in accordance with the First Law of Thermodynamics (δQ = δU + δW ).
mCv (T2 - T1) = (U2 - U1) + 0
or U2 - U2 = mCv (T2 - T1)
Heating at constant pressure....
If a definite mass of gas (m) at constant volume is a closed system is heated from initial conditions P, V1, T1, U1 to P, V1 , T2,U2. As the volume is fixed then no work has been done. Then in accordance with the First Law of Thermodynamics (δQ = δU + δW ).
mcp (T2 - T1)
= (U2 - U1) + P (V2 - V1)
= (U2 - U1) + mR (T2 - T1)
mcv (T2 - T1) = U2 - U1 therefore mcp (T2 - T1) = mcv (T2 - T1) + mR (T2 - T1) therefore
cp = cv + R... and ..
cp - cv = R = PV/mT
Comments (8)
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