Thermodynamics. Turbine, hwo to use enthelpy to find mass flow?

AI Thread Summary
To determine the mass flow of air through a high-speed turbine operating adiabatically, the first law of thermodynamics is applied, leading to the equation dW/dt = m(h_i - h_e). The enthalpy change is crucial, and while the use of steam tables is suggested for finding h1 and h2, an alternative approach using the ideal gas properties is also discussed. The change in internal energy alone is insufficient, as it does not account for expansion work against atmospheric pressure, which must be included in the total work calculation. The discussion emphasizes the importance of accurately determining enthalpy changes to solve for mass flow effectively.
eventob
Messages
32
Reaction score
0

Homework Statement



The problem (tried my best to translate it):
A small high speed turbine is operating on compressed air. It deliveres dW/dT=100 W. At the inlet, the pressure is 400 kPa and the temperature 50*C.

At the exit, the pressure is 150 kPa and the temperature -30*C.

Neglect the velocity and assume an adiabatic process. Find the necessary mass flow of air through the turbine.


Homework Equations


First law for control volume
Definition ofg enthalpy

The Attempt at a Solution


I have derived the first law for a control volume:
dE/dt=(dQ/dt)-(dW/dt)+∑m_i (h_i+0.5v_i^2+gz_i)-∑m_e (h_e+0.5v_e^2+gz_e)

Where t is time, and m_i og m_e is rate of change of mass flow at the inlet and exit, respecitively.
Assumed steady state: dE/dt=0.
Adibatic dQ/dt=0.
Also m_e´=m_i´=m´

By neglecting kinetic and potential energy associated with gravity, i end up with:

dW/dt=m(h_i-h_e) <=> m=(dW/dt) / (h_i-h_e)

So far so good, but now I need to find the change of enthalphy. We were supposed to solve this task without the use of steam tables. I have tried to use the definition of constant volume heat capacity, but no luck so far. Any input?

Thanks in advance. :)
 
Physics news on Phys.org
I think all you have to do is apply the first law to an adiabatic process (ΔQ=0) so:

ΔU = -W => dU/dt = -dW/dt

So start by determining the internal energy change of 1 mole of air going from 323K to 243K. [This is not a reversible process so the change in pressure does not help you determine the work done]. Assume air is an ideal diatomic gas with a Cv of 5R/2.

AM
 
Let me ask, are you using temperature/pressure tables for air?

If so then with the known temp/pressure at the inlet (T1/P1) and the known temp/pressure at the outlet (T2/P2) then you can refer to the tables to find the inlet and outlet enthalpies, h1 and h2 respectively.

Thus Q = m(dot)*(h1 - h2)

Solve for m(dot)

Not sure if this is what your looking or but that is my two cents
 
LabGuy330 said:
Let me ask, are you using temperature/pressure tables for air?

If so then with the known temp/pressure at the inlet (T1/P1) and the known temp/pressure at the outlet (T2/P2) then you can refer to the tables to find the inlet and outlet enthalpies, h1 and h2 respectively.

Thus Q = m(dot)*(h1 - h2)

Solve for m(dot)

Not sure if this is what your looking or but that is my two cents
My suggestion to use change in internal energy is not correct because the work done on the turbine by the gas is not the total work that is done. Since the volume of the air doubles there is also the expansion work done against the atmosphere. This is taken into account in the change in enthalpy but not with the change in internal energy.

Since atmospheric pressure is constant you can determine the work done on the atmosphere: Watm = PatmΔV

By the first law,

ΔQ = 0 = ΔU + W where W is the total work done by the air = Wturbine + Watm

ΔU = -Wturbine + -Watm → Wturbine = - (ΔU + Watm) = -(ΔU + PatmΔV)

AM
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
I was thinking using 2 purple mattress samples, and taping them together, I do want other ideas though, the main guidelines are; Must have a volume LESS than 1600 cubic centimeters, and CAN'T exceed 25 cm in ANY direction. Must be LESS than 1 kg. NO parachutes. NO glue or Tape can touch the egg. MUST be able to take egg out in less than 1 minute. Grade A large eggs will be used.
Back
Top