Steam at 5 MPa and 400°C enters a nozzle steadily with a velocity of 80 m/s,
and it leaves at 2 MPa and 300°C. The inlet area of the nozzle is 50 cm2,
and heat is being lost at a rate of 120 kJ/s. Determine...
a.) The mass flow rate of the steam
b.) The exit velocity of the steam
c.) The exit area of the nozzle
12 comments:
Can anyone verify the answer to part b. I'm getting delta v = 560 and v2 = 640. Seems fishy that I get the solution for delta v, not v2.
Verify 11:01 PM
Something is wrong, but I cannot tell what from the info you have given. deltaEkin = (v2^2-v1^2)/2gc
NOT (v2-v1)^2/2gc
That would be my best guess about what went wrong.
Your value for deltaV is amazingly close to the correct value for v2 !
can you give me any help on part b? I tried both the 1st law eqn for nozzles and the general 1st law SISO, double-checked units, but I still can't get an answer anywhere near 560 m/s for the outlet flow velocity.
For part a I am using the equation mass flow rate = density * velocity * area and for density I am using the inverse of V from the superheated vapor table for water. I get very close to 15. What am I doing wrong?
kobe-
Are you in the right units? you may have kg/mol or something and not m3/kg. That looks like the right formula and i got the right answer.
Could someone please explain why we use gc in calculating v2? In the thermoCD examples we don't use it and those seem like the same type of problems.
We use gc because the potential energy term in the 1st law equation is delta(E^), energy per unit mass. we divide delta(v^2) by gc to get J/kg, which is what we want. So we have to include gc in our calculations. I know this because it's where I messed up in the previously stated post (I apparently needed to check my units a third time.)
KINETIC energy term, not potential energy term, sorry.
kevin 5:58 PM
The only things that are common mistakes in a problem like this one are:
1- messing up the units.
2- Using the wrong form of the 1st Law. This is a flow system, so you should use the form of the 1st law that includes shaft work and deltaH.
Kobe & DKress
The formula Kobe gives is correct. DKress has a good suggestion about units. They are the most common source of errors. For example, 50 cm^2 = 0.005 m^2.
DKress 11:09 PM
I always use gc, even in SI units. It is just a good habit to get into. In SI, gc = 1 kg*m/(N*s^2). So it looks just like a unit conversion to you.
Truth be told, in SI, as in this problem, you do not need to use gc.
Kevin 12:04 AM
Thank you making a clear statement about how to use gc and why units are SO important.
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