How to quickly convert between cubic meters and cubes

This post was last edited by ylb913 on 2016-9-5 at 19:52. Using the formula P0V0/T0 = PV/T, where P0 = 1 standard atmosphere (1 standard atmosphere = 101.325 KPa) and T0 = 273.15 K, we have: V0 = 273.15PV/T. Conversely, V = V0T/273.15P. V0 represents the flow rate under standard conditions, while V represents the flow rate under operating conditions; P should be used as absolute pressure rather than gauge pressure.

Here’s a simple algorithm for you. Cubic volume ÷ 22.4 × average molecular weight ÷ density. :)

This post was last edited by ylb913 on 2016-9-5 20:01. First of all, this is a one-way calculation, converting from standard units to cubic meters under actual operating conditions; no formula for converting cubic meters under actual operating conditions back to standard units is provided. Second, how is the average molecular weight determined? How is the density determined under actual operating conditions? In reality, the calculation process is more complex. A compositional analysis of the gas is required to calculate its average molecular weight, while the density of the gas still needs to be calculated using the equation PV=nRT.

This post was last edited by an instrument repairman on 2016-9-5 at 21:41. The Clausius-Clapeyron equation is PV=nRT, where R is a constant; therefore, for the same gas, nR remains constant. By rearranging this equation, we obtain P0V0/T0=PV/T. With P0=101.325 KPa and T0=273.15 K, it follows that V0=273.15PV/101.325T. Conversely, V=101.325V0T/273.15P

V0=273.15PV/101.325T, and inversely, V=101.325V0T/273.15P

This post was last edited by ylb913 on 2016-9-5 at 22:18. The original formula for PV=nRT was added in my reply on page 4; since R is a constant and n remains unchanged, it follows that P0V0/T0 = PV/T. Moreover, the derivation formula that follows is also quite common and can be applied directly. For those who can ask such a question, there’s no need to explain too much; if they know that P0V0/T0 = PV/T, then they should also understand that PV = nRT… as well as the unit conversions for pressure. What remains is the question of how to calculate absolute pressure... To put it plainly, the pressure under standard conditions is already specified, and there’s no dispute regarding that. But how is the absolute pressure calculated under actual operating conditions? ——In reality, the pressure that is measured is mostly gauge pressure. The atmospheric pressure varies from place to place, and even at the same location it changes throughout different seasons or at different times of day. Using the formula \"101.325 kPa + gauge pressure (with units in kPa)\\" is often unreasonable. However, I have never seen anyone use the actual atmospheric pressure value when converting cubic meters in standard conditions to those under actual operating conditions. I believe in the practice of introducing the local annual average atmospheric pressure in different regions.

Thank you for the advice. Generally, the material balance is determined first; as a result, the density and average molecular weight data are available, making it simple to use the aforementioned method. Without these data, your method is relatively reliable.

The density under condition x for a cube / the density under standard conditions is the standard cube density

;P;P;P;P;P;P