Your gc is wrong.
gc is functionally unitless. However, gc in imperial units is 32.2 (lbm*ft)/(lbf*s^2). gc in metric is 1 kg*m/N*s^2.
F = m*a, right? No, F = m*a/gc.
In metric, this means 9.81N = 1 kg * 9.81 m/s2
However, in Imperial units, 1lbm * 32.2 ft/s^2 = 32.2 lbm*ft/s^2. For whatever...
Because density is a function of molecular weight of the gas flowing through the meter as well, which is the crux of your issue. Density is a function of both pressure and molecular weight. Ideal gas: rho = P*(M.W.)/ (R*T).
Please explain your reasoning. Coriolis meters are good for gas flow, and are perfect for your application, as they can be used to directly measure volumetric flow rate. E&H E300, a middle-of-the-line model, has a +/- 0.5% error on gas mass flow, and a decent range on allowable gas velocity. The...
This is likely due to how the GC program is set up. Elution times from the GC must be quite similar at the given setup, causing a unimodal or broad pead. Thus, the program was told that "x" to "y" time is considered a mixture of propane and butane since it does not have the resolution to...
Just an example for ethylene carbonate:
C3H4O3(s) + 2.5(O2)(g) -> 3(CO2)(g) + 2(H2O)(l)
Hcomb = (2*HfH2O+3*HfCO2) - (HfC3H4O3+HfO2)
Hcomb = (2*(-285 kJ/mol) + 3*(-393.5 kJ/mol))-(-581 kJ/mol + 0) = -1,169 kJ/mol
Note that this is the standard heat of formation, meaning it is taken at 298 K...
N8Cole,
Write out the combustion chemical formula and use the heats of formation (often found in NIST database) to determine the heat of combustion.
For those that NIST does not have the data (i.e. dimethyl carbonate), I found an online source that estimates the properties as well...
Shvet,
My point is that differently-shaped fittings are only an administrative control, not an engineering control. Unloading areas tend to keep a lot of double-ended fittings on-hand because truck drivers don't always arrive with the correct hose, and it is not uncommon to use that fitting...
Shvet,
I have found that locking unloading valves to be a better option than trying to supply different fitting for each and every unloading option. There is a particular case of an explosion (in a refinery, I think) where the truck driver went to the wrong unloading location only to find that...
Except it does. Take any milk jug or water bottle with a spout, fill it to the top, and turn it over, spout pointing downwards. Water will come out, the jug will flex inwards as pressure drops, and flow will temporarily stop as air enters the jug, temporarily reinflating it slightly. The cycle...
Are we simply ignoring the fact that due to the dynamics involved, air will gurgle back into the tank through the open valve at the bottom? The tank will eventually empty out, even if the top valve is closed. It simply will go through a constant cycle of spitting water out, drawing vacuum until...
IRStuff,
Normally, I agree. However, PE licensing is not a "Student" issue - it is a matter of a practicing engineer that is pushing for a professional license.
The only other possibly appropriate forum is "How to Improve Myself".
Georgeverghese,
This is a CSTR, not a plug flow reactor. Conversion at all points in the reactor is 65%.
OP, 0.65 mols of CH3CHO are converted, producing 1.3 mol of products. There are 0.35 mols remaining unconverted, for a total of 1.65 mols out. Vout = Vin * 1.65.
Again, one of the...
Spacetime is defined as reactor volume divided by inlet volumetric flow
The PE exam likes to throw a lot of extraneous information at you. Rate constant, molar density, etcs are all red herrings. The only relevant information is the decomposition equation and the spacetime.
For plug flow...
I haven't worked the second problem, but here's the first:
Set up a normal equilibrium expression (all values in concentrations, but, this being a gas, vol% can substitute for concentration:
k = NO/(N2^0.5 * O2^0.5)
x=NO
y=NO2
The concentration of N2 is the amount of initial N2 minus...
RomanKatz,
The difference is well explained by others above. Note that in heat transfer equations, Q = Q, meaning that the two equations must equal each other. You are getting two different values because you are not acknowledging the physical reality that the heat gained by the vessel's...
If this is continually sloped back to the top/side of the header, then there will be no significant change in measured pressure; the formed condensate from heat loss will just drain back down. Why put this in rather than direct mounting? You will be exposing your instrument to full steam...
Like this? Pretty common for steam applications.
It'll be mostly line pressure. A 1.5m "pig-tail" steam siphon is likely around 0.75m high. Perhaps half of that height is in the riser from the header, so you'll have, at most, approximately 0.25-0.4m or so of head loss from the condensate...
The other main question is more along the lines of mixing. It looks like your solvent is a higher boiler than water. However, since you have a tube rupture scenario, I assume you will have intimate mixing before the mixture dumps back into the tank.
In that case your release fluid will be a...
A few things come to mind:
First, and most importantly, this idea will not work. Since you have an weir overflow system, recirculating from Tank 2 back to Tank 1 will not serve to increase residence time. Your residence time is controlled solely via your wastewater inlet flow to Tank 1. If your...
Edward,
HCl(g) -> HCl(aq)
MgCO3(s)-> MgCO3(aq)
MgCO3(aq) + 2HCl(aq) -> H20(l)+CO2(g)+MgCl3(aq)
Use published values for heats of formation
MgCO3 Heat of Solution...
