Review on Thermodynamic Processes with Sample Problem: Isometric Process

Hi folks!

In the field of Thermodynamics, all of us mechanical engineers and mechanical engineering students need to familiarize ourselves with the 5 different thermodynamic processes that are being incorporated and applied in studying the behaviours of substances like gases and steam. These five thermodynamic processes are as follows:

• Isometric Process (Constant Volume Process)
• Isobaric Process (Constant Pressure Process)
• Isothermal Process (Constant Temperature Process)
• Isentropic Process (Constant Entropy Process)
• Polytropic Process

So today, allow me to discuss about the constant volume process which is known as the Isometric process, also known as the isochoric process and isovolumetric process. This process are greatly exhibited especially on tanks or cylinders wherein substances like gases are often stored and whatever is the volume of the rigid tank or cylinder is just equal to that fluid being stored. And bear in mind that isometric process can be reversible (the term itself says that it can be brought back to its original state) or irreversible (opposite to reversible and will no longer be able to came back to its original state).

Image Source
And the favourite of engineers and engineering students especially those in the field of mechanical engineering, and that is formulas and relations. And here are the formulas and relations:

• Pressure (P) and Temperature (T) relations
  
• Nonflow Work (W_n)
  Since this process is under constant volume, W_n = 0.
  
• Change in Internal Energy (Change in U)
  
• Heat Transferred (Q)
  For isometric process heat is numerically equal to the change in internal energy since non-flow work (W_n) is zero.
  
• Change in Enthalpy (Change in H)
  
• Change in Entropy (Change in S)
  Since we are under the isometric process, we will be using the specific heat capacity of the substance at constant volume.
  
• Reversible steady flow work (W_s)
  

So we now learnt the basic concept and formulas that are very important in dealing with isometric processes. So let us test ourselves with these sample review problem.

A perfect gas has a value of R = 58.8 ft-lb_f per lb_m per unit degree Rankine and k = 1.26. If 20 BTU are added to 5 lb_m of this gas at constant volume when the initial temperature is 90 degree Fahrenheit, find the following:

• Final Temperature (T₂)
• Change in Enthalpy (Change in H)
• Change in Entropy (Change in S)
• Change in Internal Energy (Change in U)
• Work for a Non-flow Process (W_n)

So the solution goes like this. We were provided with the gas constant of the perfect gas which is R = 58.8 ft-lb_f per lb_m per unit degree Rankine and an adiabatic index of k = 1.26. In order to have a straightforward calculation the first thing we need to do is obtain the specific heat capacity of the gas at constant volume (C_v) and at constant pressure (C_p). And here are the calculations:

Thus we have C_v and C_p of 0.29 and 0.37, respectively. Both in units of BTU per unit lb_m per unit degree Rankine.

• Final Temperature (T₂)
  So for the final temperature (T₂), since we are provided with the amount of heat being added to the gas at 5 lb_m. And we just obtained the specific heat capacity of the gas at constant volume, so we can now obtain the T2 by equating the formula for the change in internal energy which is numerically equal to the heat being added to the gas. And here is the calculation;
  

  
  And we have our T₂ at 563.8 degree Rankine.

• Change in Enthalpy
  Since we now have the T₂ and we previously obtained the specific heat capacity of the perfect gas (C_p), we can easily substitute the values to the formula. And here is the calculation;
  

  
  And we have our change in enthalpy which is at 25.53 BTU.

• Change in Entropy
  Since we are in an isometric process, we will be using the specific heat capacity of the perfect gas at constant volume. And the calculation goes like this;
  

  
  And we obtained an entropy of 0.036 BTU per unit degree Rankine.

• Change in Internal Energy
  As you have noticed in the above problem, it has stated that there is a 20 BTU of energy being added to the perfect gas. And since heat added is numerically equal to the internal energy, the change in internal energy is equal to 20 BTU.
  

  

• Work for a Non-flow Process
  For a non-flow process, work for isometric process is always equal to zero.
  

  

So that ends my discussion and sample problem regarding the thermodynamic process - isometric to be exact.

Thank you for spending your time reading this blog of mine.

Much love and respect.

Ace | @josephace135

Reference:

1. Hipolito B. Sta. Maria, Thermodynamics