Recent content by Chestermiller

What are the parameters in your equations, and what criterion are you using for failure?

Counter current heat exchangers, flash vaporizers, cooling towers, heat pumps, refrigeration units, mixing with ice, '''

Have you considered all possible scenarios. I think you should read up on the concept of exergy.

What do you mean by the term "cooling can be generated," and are you saying that this broad statement has no constraints? For example, do you only include just a gas, does this include cycles. Also, cooling of what and by what? How does this relate to cooling of a fluid stream in a heat...

If I understand you correctly, right for an adiabatic process.

Here's a big surprise: The surroundings doing adiabatic compression work on a gas causes its internal energy and temperature to increase. A gas doing adiabatic expansion work on its surrounding causes its internal energy and temperature to decrease. Isn't that what you are saying?

From a fundamental materials point of view, if the metal is isotropic and the sample in not constrained or loaded mechanically (essentially isotropic state of stress) in any way, then the sample experiences a homogenous isotropic strain, with, for small temperature changes, the strain being...

Cooling by expansion can be done with less energy consumption than by what? This seems to me to be basically just a standard reversible adiabatic expansion scenario, analogous to removing pebbles from a cylinder and piston oriented vertically. The increase in the potential energy of the...

Here are my calculations for scenario 1. Given: Initial Temperature = ##T_i=300.2K## Initial Pressure = ##P_i=101325\ Pa## Number of moles n =1.0 Calculations: Initial Volume$$V_i=\frac{nRT_i}{P_i}=\frac{(1)(8.314)(300.2)}{101325}=0.02463\ m^3$$ Final Volume$$V_f=3V_i=0.0739\ m^3$$ Final...

There is no net work done on the piston, and, thus, no net energy "spent by the piston." I show this below. From Newton's 2nd law of. motion applied to the piston, we have $$P_{gas}A+F-P_{atm}A=0\tag{1}$$The term ##P_{gas}A## represents the forward force of the gas on the inside face of the...

I totally disagree with your interpretation and stand by what I said previously. You should know that I have a lot of practical experience with thermodynamics (>60 years), so I am very confident in what I am saying. I can help you work your way through this if you are willing to work with me...

Yes. If you are pulling the piston outward, the net work is equal to the work to push back the atmosphere plus the work done to pull on the piston outward. For a reversible expansion, this is just equal to the work done by the gas. $$W_{gas}=P_{atm}\Delta V-\int{F_{piston}\frac{d V}{A}}$$...

In Post #17, I calculated the change in internal energy U in scenario 1. In this post, I will should that this matches the work done by the ideal gas in this case. The equation for the pressure in terms of the volume for both cases reads...

sorry;. If R=2 calories/mole-K, then 2.5R= 5 calories/mole-K. In our case,, R=8.314 Joules/mole-K, and 2.5 R = (2.5)(8.314) Joules/mole-K. See my corrected answer in Post#17.

A pressure of 1 kg/cm^2 corresponds to ##0.98\times10^5\ Pa## pressure. From the ideal gas law, for 1 mole gas at 300 K, the ideal gas law gives a volume of $$\frac{nRT}{P}=\frac{(0.08206)(300)}{1.0}=0.0246\ m^3$$ For an ideal gas, the volume of 1 mole of gas at 1 atm and 273.16 K comes out to...