5
$\begingroup$

A synthesis gas mixture ($\text{H}_2:\text{CO}$ with molar proportion 2:1) is needed for methanol synthesis. The final mixture must be available in a gas cylinder of 50 L, say cylinder $M$, which contains, initially, pure $\text{H}_2$ at 5 bar. The carbon monoxide which will compose the final mixture is available in a smaller cylinder (20 L), say cylinder $m$ in a volumetric proportion of 9:1, at 120 bar. Both cylinders are, initially, at 303 K. A 1/8” steel tube must be used with a globe valve placed 1 m from the small cylinder $m$ and 4 m from the cylinder $M$.

How can one calculate the pressure and temperature of both cylinders in the process where the gas transfer is done opening totally and suddenly the globe valve, after 15 s? The dynamics of the problem seem to be difficult to implement.

$\endgroup$
1
  • $\begingroup$ This seems like a homework question. What have you tried? A globe valve doesn't suddenly open. It takes quite a few rotations of the handle to fully separate the plug from the seat. Assuming you have a valve that can be suddenly opened fully, try assuming that your final temp and pressure equalize, then figure out the volume of each species now in each cylinder. $\endgroup$
    – morristtu
    May 2, 2016 at 8:49

1 Answer 1

1
$\begingroup$

You essentially have a fluid mechanics problem here that can be approximated fairly well without heat transfer during the process of mixing. One approach (not too difficult) is to simply discretize the problem into steps of a fraction of a second performing the following computations:

  1. Compute the mass flow rate from the upstream pressure, valve flow coefficient (Cv), and downstream pressure. Because this is a fairly high pressure problem, there will be some pressure drop from the tubing itself (but much smaller than the valve). You can choose whether to incorporate that loss into this step. See valve sizing equations from Crane Technical Paper 410, among many other references, for further details. You'll need to split the C and O into separate masses and assume they are evenly mixed for your own sanity (so you deduct 90% of the mass flow from C, 10% from O at each step).
  2. Deduct the mass flow that would exit the first tank and add it to the second tank (taking into account your time step).
  3. Recompute your upstream and downstream pressure based on the new masses.
  4. Repeat until you reach steady state (i.e. your pressures aren't changing anymore). Note that in a real world scenario the valve may have a minimum pressure to result in flow ("cracking pressure" for check valves, for ex) - so steady state may mean P1 != P2. You can then back out the temperature in each cylinder from the ideal gas law.

If you need to deal with heat exchange during the process, this turns into a fully-fledged 1D compressible flow solver.

I suspect the problem was meant to be constructed as some sort of Bernoulli Equation example where you handwave and say the completely open valve acts exactly as a pipe, rather than the dominant factor (since they give you pipe dimensions but no valve details).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.