Sunday, April 15, 2007

HW #5, P2 - Adiabatic Gas Turbine - 5 pts

Argon gas enters an adiabatic turbine at 900 kPa and 450oC with a velocity of 80 m/s and leaves at 150 kPa and a velocity of 150 m/s. The inlet cross-sectional area is 60 cm2. If the power output of the turbine is 250 kW, determine the exit temperature of the argon. The process operates at steady-state and argon behaves as an ideal gas.

21 comments:

Anonymous said...

i could not get the volume flow rate or volume head so i don't know how to get the mass flow rate.

Dr. B said...

problem:
Volumetric flow rate is velocity times cross-sectional area. You know both v1 and A1, so you can calculate m1 and m2 = m1.

Anonymous said...

I found the change in enthalpy but I can't figure out what the initial enthalpy is because the initial temperature is 450C but the NIST site only lists enthalpies of argon up to 427C.

Anonymous said...

Actually, I just realized I probably don't need the enthalpies.

Anonymous said...

Hmm... I still can't figure out what the initial enthalpy is.

Anonymous said...

How are we supposed to get the initial enthalpy -- NIST doesn't have data for 450 degrees C

Anonymous said...

I think the answer to your question is that we should first of all assume that Argon behaves as an ideal gas. Then we can use the Cp given in class (0.5203 kJ/kg-K), multiplied by delta T, and that should be delta H. In math, this is H2 - H1 = Cp * (T2-T1), which can be rearranged to yeild T2, given that you have already solved for H2-H1 and are given T1 and Cp...

Anonymous said...

I don't understand how you can get the mass flow rate. There's no way of getting the density or specific volume of argon from NIST

sparticus said...

You can get V hat by using PV = nRT and rearranging it into PVhat = RT/MW. Once you have Vhat then use the volumetric flow rate to find the mass flow rate.

Dr. B said...

Anon 3:17 PM:
Use Cp = 0.5203 kJ/kg and calculate deltaH = Cp * deltaT because Cp can be considered to be a constant.

Dr. B said...

Anon 3:18:
You don't really need the enthalpy, but you do need the change in the enthalpy.

deltaH = Cp * deltaT

Dr. B said...

questioning 7:53 PM:
Thank you. You are exactly right.

Dr. B said...

Anon 8:07 & Sparticus:
Thank you Sparticus. You are completely correct. Or, try the IG EOS in the following form:
P * Vdot = mdot * R * T / MW

Anonymous said...

I keep getting 283C for the answer instead of 267C. Anyone else having this problem?

Anonymous said...

Yeah I also get 283C instead of 267C...

sparticus said...

I got 267C, it is possible!

Anonymous said...

sparticus and others-

how did you get 267 ? I too keep getting 283

Anonymous said...

I bet it has something to do with the molar mass.

Anonymous said...

Need to account for the kinetic energy. The given answer is correct.

Dr. B said...

To everyone getting 283 degC:
Anon Coward gave you the right answer. Changes in kinetic energy ARE important. You should guess this when inlet and outlet velocities are given.

Anonymous said...

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes
are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Argon is an ideal gas with
constant specific heats.
Properties The gas constant of Ar is R = 0.2081 kPa.m3/kg.K. The constant pressure specific heat of Ar is
cp = 0.5203 kJ/kg·°C (Table A-2a)
Analysis There is only one inlet and one exit, and thus m&1 = m& 2 = m& . The inlet specific volume of argon
and its mass flow rate are
( )( ) 0.167 m /kg
900 kPa
0.2081 kPa m /kg K 723 K 3
3
1
1
1 =
⋅ ⋅
= =
P
RT v
ARGON
A1 = 60 cm2
P1 = 900 kPa
T1 = 450°C
V1 = 80 m/s
P2 = 150 kPa
V2 = 150 m/s
250 kW
Thus,
(0.006 m )(80 m/s) 2.874 kg s
0.167m /kg
1 1 2
1 1 3
1
m = AV = = /
v
&
We take the turbine as the system, which is a control volume since
mass crosses the boundary. The energy balance for this steady-flow
system can be expressed in the rate form as
& & &
& &
E E E
E E
in out
in out
− =
=
Rate of net energy transfer
by heat, work, and mass
system
(steady)
Rate of change in internal, kinetic,
potential, etc. energies
14243 144424443
Δ 􀃊0 = 0
⎟ ⎟


⎜ ⎜

⎛ −
= − − +
+ = + ≅ Δ ≅
2
( / 2) ( + /2) (since pe 0)
2
1
2
2
out 2 1
2
out 2 2
2
1 1
W m h h V V
m h V W m h V Q
& &
& & & &
Substituting,
⎥ ⎥


⎢ ⎢


⎟ ⎟


⎜ ⎜

− ⎛
= − ⋅ − +
2 2
2 2
2 1000 m /s
1 kJ/kg
2
(150 m/s) (80 m/s)
250 kJ/s (2.874 kg/s) (0.5203 kJ/kg o C)(T 450o C)
It yields