What is the equation for Cp in terms of R for an adiabatic charging process?

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SUMMARY

The discussion focuses on the adiabatic charging process of a 250mm diameter insulated cylinder with a frictionless piston, utilizing the first law of thermodynamics. Key calculations include the final pressure of the air in the cylinder at 636kPa, the final temperature at 340.8K, and the mass of air injected at 0.0479kg. Participants emphasize the importance of using the correct thermodynamic equations, particularly in non-steady flow situations, and clarify that the process is adiabatic, necessitating specific equations for work and energy changes.

PREREQUISITES
  • Understanding of the first law of thermodynamics for open systems
  • Knowledge of adiabatic processes in thermodynamics
  • Familiarity with ideal gas laws and equations of state
  • Basic principles of work and energy in mechanical systems
NEXT STEPS
  • Study the derivation of the first law of thermodynamics for non-steady flow systems
  • Learn about adiabatic processes and their implications in thermodynamic cycles
  • Explore the calculation of work done in compressible fluid systems
  • Investigate the relationship between enthalpy, temperature, and pressure in ideal gases
USEFUL FOR

Mechanical engineers, thermodynamics students, and professionals involved in fluid mechanics and energy systems will benefit from this discussion, particularly those working with adiabatic processes and gas compression systems.

  • #31
Chestermiller said:
As you responded in a private conversation: W=[kx2/2+PaAx]. So now what do we get if we substitute this into our equation:
$$m(h-h_{in})=-W+kx^2+P_aAx$$
Chet
m(h−hin)=−[kx2/2+PaAx]+kx2+PaAx ?
 
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  • #32
Tomdickjerry said:
m(h−hin)=−[kx2/2+PaAx]+kx2+PaAx ?
Collect terms?
 
  • #33
Chestermiller said:
Collect terms?
m(h−hin)=1/2kx2?
 
  • #34
Correct. Pretty simple, huh?

This is all that needs to be done on the right hand side of the equation. I'll be back in a little while to get you started working on the left hand side of the equation. Right now, I'm on my iPhone.
 
  • #35
Chestermiller said:
Correct. Pretty simple, huh?

This is all that needs to be done on the right hand side of the equation. I'll be back in a little while to get you started working on the left hand side of the equation. Right now, I'm on my iPhone.
alright thanks! for the left side do i just use cv(t2-t1) i figured since the it isn't constant pressure process
 
  • #36
Tomdickjerry said:
alright thanks! for the left side do i just use cv(t2-t1) i figured since the it isn't constant pressure process
Actually, no, but you're on the right track. The molar enthalpy of a pure single phase substance can be expressed uniquely as a function of temperature T and pressure P: h = h(T,P). The molar heat capacity Cp is, by definition, the partial derivative of h with respect to T at constant P:
$$C_p\equiv \left(\frac{\partial h}{\partial T}\right)_P$$For an ideal gas, molar enthalpy is a function only of temperature T. So the partial derivative becomes a total derivative: $$\frac{dh}{dT }=C_p$$For an ideal gas, this equation applies even if the pressure is not constant. The quantity ##C_p## is called the heat capacity at constant pressure only because, in measuring ##C_p## experimentally, not only is the change in enthalpy dh at constant pressure equal to ##C_pdT##, but so also is the amount of heat added dQ. So, under constant pressure conditions, ##C_p## can be measured experimentally by directly determining the amount of heat added. If the pressure is not constant, then dQ is not equal to ##C_pdT## but dh (for an ideal gas) is still equal to CpdT.

So, since for an ideal gas, dh is always equal to ##C_pdT##, irrespective of whether the pressure is constant, we have:
$$(h-h_{in})=C_p(T-T_{in})$$where h is the final molar enthalpy of the air in the cylinder, ##h_{in}## is the molar enthalpy of the air in the tank, T is the final temperature of the air in the cylinder, and ##T_{in}## is the temperature of the air in the tank (27 C = 300 K).

Now m is the final number of moles of air in the cylinder. From the ideal gas law, please (algebraically) express m as a function of PV, R, and T, where R is the universal gas constant.

Air is a diatomic gas. In the ideal gas region, how is the molar heat capacity ##C_p## of a diatomic ideal gas related to the ideal gas constant R?

Chet
 
  • #37
Chestermiller said:
Actually, no, but you're on the right track. The molar enthalpy of a pure single phase substance can be expressed uniquely as a function of temperature T and pressure P: h = h(T,P). The molar heat capacity Cp is, by definition, the partial derivative of h with respect to T at constant P:
$$C_p\equiv \left(\frac{\partial h}{\partial T}\right)_P$$For an ideal gas, molar enthalpy is a function only of temperature T. So the partial derivative becomes a total derivative: $$\frac{dh}{dT }=C_p$$For an ideal gas, this equation applies even if the pressure is not constant. The quantity ##C_p## is called the heat capacity at constant pressure only because, in measuring ##C_p## experimentally, not only is the change in enthalpy dh at constant pressure equal to ##C_pdT##, but so also is the amount of heat added dQ. So, under constant pressure conditions, ##C_p## can be measured experimentally by directly determining the amount of heat added. If the pressure is not constant, then dQ is not equal to ##C_pdT## but dh (for an ideal gas) is still equal to CpdT.

So, since for an ideal gas, dh is always equal to ##C_pdT##, irrespective of whether the pressure is constant, we have:
$$(h-h_{in})=C_p(T-T_{in})$$where h is the final molar enthalpy of the air in the cylinder, ##h_{in}## is the molar enthalpy of the air in the tank, T is the final temperature of the air in the cylinder, and ##T_{in}## is the temperature of the air in the tank (27 C = 300 K).

Now m is the final number of moles of air in the cylinder. From the ideal gas law, please (algebraically) express m as a function of PV, R, and T, where R is the universal gas constant.

Air is a diatomic gas. In the ideal gas region, how is the molar heat capacity ##C_p## of a diatomic ideal gas related to the ideal gas constant R?

Chet
m=pv/rt.
i don't know what a diatomic gas is it wasn't covered in my syllabus but I am assuming you're referring to the k=1.4?
 
  • #38
Tomdickjerry said:
m=pv/rt.
i don't know what a diatomic gas is it wasn't covered in my syllabus but I am assuming you're referring to the k=1.4?
Yes. So, if ##\frac{C_p}{C_v}=1.4## and ##C_p-C_v=R## (you are familiar with this equation, correct?), what is the equation for Cp expressed exclusively in terms of R?
 

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