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Originally Posted by Zgringo
Other than reading to gain all your knowledge, what type of degrees do you hold and what research have you performed to even suggest engineers should read your unfounded statements.
Your defination of a particle is something above atomic level.
As for not understanding the difference between heat and tempature, I understand both quite well, and if you don't think applying 2,000F creates heat then you have the problem. I could care less what heat does to a peltier element, were not trying to creat electric power, were talking turbo's.
As for the guys at STS to read old theory, their putting fact into play and rewriting old theory with new technology. There proof is on the dyno, not some unfounded theory.
Don't critize their itelligance, correct you ignorance.
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As I have mentioned before (in another thread) I'm currently a college student and I have now about one and a half year left intil I have a masters in mechanical engineering.
Applying 2000 degF doesn't mean that there is much heat involved. For example heating a piece of say 1 kg nickel to 1093 degC (all from 0 degC, 2000 degF = 1093 degC) does involve less heat than for example heating 10 kg nickel to 150 degC or 2 kg water to 60 degC.
You and the guys at STS are both ignoring the thermodynamics of turbines, and try covering that fact up with statements about old theories and unfounded theories is just silly.
Let's say that we have an exhaust turbine with the following conditions:
m = 0.5kg/s
P1 = 3bar
P2 = 1bar
Cp = 1.0 kJ/(kg*K) (specific heat)
γ = 1.4 (ratio of specific heats)
During test 1 the turbine inlet temperature (T1) was 1200K and during test 2 900K. The output for the idealized turbine will then be as follow:
Test 1
T2 = T1*(P2/P1)^[(γ-1)/γ] = 1200*(1/3)^((1.4-1)/1,4) = 877K
P = m*Cp*(T1-T2) = 0.5*1,0*(1200-877) = 161.5 kW
Test 2
T2 = T1*(P2/P1)^[(γ-1)/γ] = 900*(1/3)^((1.4-1)/1,4) = 658K
P = m*Cp*(T1-T2) = 0.5*1,0*(900-658) = 121 kW
NOTE: (relations between heat, impulse caused by fluidparticles and turbine power)
M = F*r
M = m(r1*c1-r2*c2)
Y = P/m = u2*c2 - u1*c2
P = Δh*m but also P = m*Y and P = q*V*Y and therefore
Δh = Y = Cp*(T1-T2)
Where u = blade velocity, c = tangential component of blade velocity.
With a radial turbine one can approximate by Δh = U1*C1 as blade velocity is greatest at the inlet and the outlet flow is mainly axial.
Δh and Y are defined as "specific theoretical bladework", the amount of work extracted per massflow.
the specific bladework can also be written as:
Y = 1/2[(c2^2-c1^2)+(u2^2-u1^2)+(w1^2-w2^2)]
In the case of a reaction turbine, where the fluid undergoes a change in static pressure, the decreaseing pressure will result in an increase in velocity.
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The turbine which offers the most excess power (turbine power minus compressor power consumption) first will offer the best spool up. The time until a high turbine expansion ratio (defined as P2/P1) is reached is also important.
Turbines, pistons or a peltier element and an electric motor.... they all convert heat energy into kinetic energy.
EDIT: In the "test" equations I used [] instead of () as the latter was turned into smilies...