- #1
RagincajunLA
- 19
- 0
I have a homework problem that I completed but I am not sure if it is correct. I was wondering if you guys could let me know if my thought process is correct of if I need to modify my solution. The problem is:
An adiabatic and rigid open vessel contains 4 kg of air at 300 K and has a volume of 1m^3. Air at 450 K and 400 kPa is entering the vessel at a rate of 0.10 kg/s. Assume the air is calorimetrically perfect ideal gas (CPIG) with constant specific heat ratio (k) of 1.40. What is the instantaneous rate of extensive energy increase in the vessel?
I started off by completing a foundational energy balance of
\frac{dE}{dt}=\frac{dU}{dt}+\frac{dKE}{dt}+\frac{dPE}{dt}=\frac{dU}{dt}=\dot{m}_{in}T_{in}
Since this is a CPIG, its enthalpy can be simplified to h_{in}in=c_{p}T_{in} and the constant pressure specific heat can be simplified to c_{p}=\frac{kR}{k-1}. By using all this, I get a final expression of
\frac{dE}{dt}=\frac{dU}{dt}=\frac{\dot{m}kRT_{in}}{k-1}
where Tin is the temperature of the incoming air.
For some reason I felt like I left some stuff out of the problem. For instance, I wasn't sure if I needed to use the pressure of the incoming air, the mass of the air in the control volume, or the volume of the control volume. Please let me know If I completed this problem correctly or steer me in the right direction. Thank you
An adiabatic and rigid open vessel contains 4 kg of air at 300 K and has a volume of 1m^3. Air at 450 K and 400 kPa is entering the vessel at a rate of 0.10 kg/s. Assume the air is calorimetrically perfect ideal gas (CPIG) with constant specific heat ratio (k) of 1.40. What is the instantaneous rate of extensive energy increase in the vessel?
I started off by completing a foundational energy balance of
\frac{dE}{dt}=\frac{dU}{dt}+\frac{dKE}{dt}+\frac{dPE}{dt}=\frac{dU}{dt}=\dot{m}_{in}T_{in}
Since this is a CPIG, its enthalpy can be simplified to h_{in}in=c_{p}T_{in} and the constant pressure specific heat can be simplified to c_{p}=\frac{kR}{k-1}. By using all this, I get a final expression of
\frac{dE}{dt}=\frac{dU}{dt}=\frac{\dot{m}kRT_{in}}{k-1}
where Tin is the temperature of the incoming air.
For some reason I felt like I left some stuff out of the problem. For instance, I wasn't sure if I needed to use the pressure of the incoming air, the mass of the air in the control volume, or the volume of the control volume. Please let me know If I completed this problem correctly or steer me in the right direction. Thank you
Last edited:
