adiabatic_ranger
Mechanical
- Aug 8, 2022
- 1
In calculating the work for a process involving an expanding gas in a piston/cylinder arrangement, it seems to me that I should be able to get the same answer using either the pressure/change in volume or the change in internal energy, but this is not the case. What am I missing? (Something, obviously.)
Here is a sample problem from my thermo textbook:
“A piston-cylinder assembly containing 0.1 lb of ammonia, initially a saturated vapor, is placed on a hot plate. Due to the surrounding atmospheric pressure and the weight of the piston, the pressure of the ammonia is held constant at 20 psi. Heating occurs slowly, and the ammonia expands at constant pressure until the temperature is 77 degrees F. What is the work for the process?”
Solution:
The initial volume can be found using the mass and the specific volume from the tables. At 20psi, ammonia vg=13.497 ft^3/lb. Therefore V1=0.1*13.497 = 1.35 ft^3.
The volume at state 2 can be found from the specific volume at 20 psi and 77 degrees. v2 = 16.7 ft^3/lb. So V2 is 0.1*16.7 = 1.67 ft^3.
Since the pressure is constant, W=p(V2-V1). W=(20 lb/in^2)(1.67-1.35 ft^3)(144in^2/ft^2) = 922 ft-lb, or 1.18 Btu. (This is the textbook answer)
All good so far. Now, from the energy balance, ΔU+ΔKE+ΔPE=Q-W. So if we assume that KE and PE changes are negligible (e.g. if the piston is weightless and all pressure comes from the surroundings), it should also be possible to calculate Q from ΔU.
Going to the tables, u_g for ammonia at 20psi = 564.3 Btu/lb. For superheated vapor at 77 degrees F, u = 594.2 Btu/lb. So:
Δu=594.2-564.3=29.9 Btu/lb.
ΔU=29.9 Btu/lb*0.1 lb=2.99 Btu
That is a very different number than 1.18 Btu. I know I’m missing something obvious, I just don’t know what..
Here is a sample problem from my thermo textbook:
“A piston-cylinder assembly containing 0.1 lb of ammonia, initially a saturated vapor, is placed on a hot plate. Due to the surrounding atmospheric pressure and the weight of the piston, the pressure of the ammonia is held constant at 20 psi. Heating occurs slowly, and the ammonia expands at constant pressure until the temperature is 77 degrees F. What is the work for the process?”
Solution:
The initial volume can be found using the mass and the specific volume from the tables. At 20psi, ammonia vg=13.497 ft^3/lb. Therefore V1=0.1*13.497 = 1.35 ft^3.
The volume at state 2 can be found from the specific volume at 20 psi and 77 degrees. v2 = 16.7 ft^3/lb. So V2 is 0.1*16.7 = 1.67 ft^3.
Since the pressure is constant, W=p(V2-V1). W=(20 lb/in^2)(1.67-1.35 ft^3)(144in^2/ft^2) = 922 ft-lb, or 1.18 Btu. (This is the textbook answer)
All good so far. Now, from the energy balance, ΔU+ΔKE+ΔPE=Q-W. So if we assume that KE and PE changes are negligible (e.g. if the piston is weightless and all pressure comes from the surroundings), it should also be possible to calculate Q from ΔU.
Going to the tables, u_g for ammonia at 20psi = 564.3 Btu/lb. For superheated vapor at 77 degrees F, u = 594.2 Btu/lb. So:
Δu=594.2-564.3=29.9 Btu/lb.
ΔU=29.9 Btu/lb*0.1 lb=2.99 Btu
That is a very different number than 1.18 Btu. I know I’m missing something obvious, I just don’t know what..
