Mr. Hugo, your explanation was fantastic! I understood more with this than in hours of lecture! Best regards from a Mechanical Engineering student from Italy
Compression work is the area at the left of curve. Since isentropic process has largest area, isentropic compression requires the most input work. How is that possible when steady flow device suppose to consume the LEAST work when the process is reversible(isentropic)?
MrZealot1 if you consider Bernoulli generalized equation, and suppose the process is reversible (so that there are no pressure losses due to friction or other irreversibilities), you’ll notice that the power required to power the compressor is the product between the air mass flow rate and the integral from 1 to 2 of v*dP
Pdv is the work done on a moving boundary of a closed system. vdP is the isentropic work for an open system. Remeber enthalpy: h = u + Pv Change of enthalpy (infinitesimal) dh = du + Pdv + vdP So you see when enthalpy changes it can basically be attributed to the fluid changing it's internal energy, making boundary work (Pdv) or flow work (vdP). To prove that you need Gibbs Equation δq + δw = Δu (First law closed system) δq = Tds (Second law) δw = -Pdv Substituting: Tds = du + Pdv This is Gibbs equation. However in open systems we relate everything to enthalpy not internal energy, it eases calculations. We use the change in enthalpy above and find: dh - vdP = du + Pdv Substituting into Gibbs equation: Tds = dh - vdP And if we consider an isentropic process: dh = vdP To find out an expression for work we use the first law for open systems, considering that because we assumed an isentropic process it is adiabatic (heat q must be 0) and the work has to be reversible, Kinetic and Potential energy are neglected as well. Then it simply is δw_(reversible) = dh δw_(reversible) = vdP
Mr. Hugo, your explanation was fantastic! I understood more with this than in hours of lecture!
Best regards from a Mechanical Engineering student from Italy
awesome explanation
Compression work is the area at the left of curve. Since isentropic process has largest area, isentropic compression requires the most input work.
How is that possible when steady flow device suppose to consume the LEAST work when the process is reversible(isentropic)?
Why are compressors not considering Pdv? The specific volume changes during the process.
MrZealot1 if you consider Bernoulli generalized equation, and suppose the process is reversible (so that there are no pressure losses due to friction or other irreversibilities), you’ll notice that the power required to power the compressor is the product between the air mass flow rate and the integral from 1 to 2 of v*dP
Pdv is the work done on a moving boundary of a closed system. vdP is the isentropic work for an open system.
Remeber enthalpy:
h = u + Pv
Change of enthalpy (infinitesimal)
dh = du + Pdv + vdP
So you see when enthalpy changes it can basically be attributed to the fluid changing it's internal energy, making boundary work (Pdv) or flow work (vdP).
To prove that you need Gibbs Equation
δq + δw = Δu (First law closed system)
δq = Tds (Second law)
δw = -Pdv
Substituting:
Tds = du + Pdv
This is Gibbs equation. However in open systems we relate everything to enthalpy not internal energy, it eases calculations. We use the change in enthalpy above and find:
dh - vdP = du + Pdv
Substituting into Gibbs equation:
Tds = dh - vdP
And if we consider an isentropic process:
dh = vdP
To find out an expression for work we use the first law for open systems, considering that because we assumed an isentropic process it is adiabatic (heat q must be 0) and the work has to be reversible, Kinetic and Potential energy are neglected as well. Then it simply is
δw_(reversible) = dh
δw_(reversible) = vdP
Can someone please explain his sign convention for work.