Post by finiteparts on Jul 27, 2024 12:03:14 GMT -5
I thought that I would share a few thoughts on cycle performance to help those new to the hobby understand the impact of losses to the overall cycle performance.
Additionally, I will provide a quick approach to calculating and visualizing some of the losses that impact performance, via the use of Temperature verses Entropy plots.
To get started, lets get familiar with the T-s diagram. As shown below, the x-axis is the entropy. Entropy is a state variable, which means that if we know two other state conditions, such as temperature and pressure, then there is a fixed entropy value that corresponds to these conditions. Entropy can be thought of, for our purposes, as a measure of the amount of energy unavailable to do work in a system. Thus, more entropy is bad, because it is basically a loss in the work the system can do and it is manifested in our engines by such things as mixing losses, friction losses (flow and mechancial), vorticity created by wakes in the flow path, etc, etc.
The lines in the x-y space are called isobaric lines, where the "iso-" prefix means "the same" and "baric" is from the Greek "baros" meaning weight, which is combined to mean constant pressure (since pressure is essentially a weight on an area). These lines of constant pressure are plotted for 1 bar to 5 bar as labelled in the upper right corner.
As can be seen, the isobaric lines diverge as the temperature increases and it is this divergence which makes the cycle work. If they stayed parallel, then no amount of heat could be added to overcome the losses introduced into the cycle due to the internal entropy producing processes. Luckily for us, they do diverge and thus as the amount of internal losses increase, we can overcome them by increasing the overall system temperature.
As an example, below is shown a simple gas turbine cycle. As can be seen, the ideal or lossless amount of spec work across a component is a vertical line from the starting point to the next point's isobar, i.e. a vertical line. A vertical line means that there was no entropy change across that component.
The light blue line on the lower left shows the smallest amount of specific work that would be needed to compress the air from 1 bar to 4 bar, which is roughly 142 kJ/kg. But in reality, our compressors will require more power than that because there are all kinds of losses within the compressor stage of our engines. So this is easy to visualize because we still have to end up at 4 bar, but we know there will be lost energy, which is entropy. So the line goes up an to the right.
Since isentropic energy is defined as:
compressor isentropic efficiency = (ideal del h) / (actual del h)
where h is the enthalpy, which, if we assume that we are using an ideal gas, can be calculated as:
h = Cp * T
Due to the assumptions used to find these equations, the correct Cp to use is an average between the Cp at the high temperature and the Cp at the low temperature. I attached a Cp vs Temperature plot for dry air further down, to help reader's estimate their average Cp for whichever component they are working with.
Since the proper Cp to use is the average Cp, the change in enthalpy (del h) can be calculated as:
del h = Cp *(T2 - T1)
and thus the isentropic efficiency can be recast as:
compressor isentropic efficiency = Cp*(T2ideal - T1) / Cp*(T2actual - T1) = (T2ideal - T1) / (T2actual - T1)
Now, if we shuffle things around we can get:
(T2actual-T1) = (T2ideal -T1) / compressor isentropic efficiency
So the actual temperature rise is just a scaled amount of the ideal heat rise in the compressor, scaled by the inverse of the isentropic efficiency.
Similarly for a turbine:
(T4actual - T3) = turbine isentropic efficiency * (T4ideal - T3)
Which is a direct scale...so the temperature drop across a 65% efficient turbine will be 65% of the ideal temperature drop.
It should be noted that the amount of specific work (work per mass) that is achieved over a component is given by:
Ws = del h = Cp * (T2-T1) [kJ/kg]
So, on the plot, the specific work can be understood to be the vertical distance between the two operating points of the component in question. You can "see" the amount of work available. This is a powerful tool for understanding how things change due to cycle variations!
And the actual work can be found by multiplying by the mass flow to get:
W = mdot * Cp *(T2 -T1) [kJ/s]
So now, if we focus more closely at the turbine side of this cycle as shown below:
If I start with a 65% efficient turbine (shown with the blue ideal line) and take a 177 C temperature drop across the stage I can get the 195 kJ/kg spec work to meet the compressor plus mechanical losses needed to keep the engine running at this steady state point. But if somehow, I loose 8% efficiency due to increased clearances (this is referenced to John's recent change as an example), then I can no longer get enough of a drop in temperature across the 4 bar to 1 bar pressure ratio and this inlet temperature.
The reduction in efficiency will mean that the change in entropy (del s) will be larger and thus there is less temperature drop available across this stage. The solution to get to the needed power demand is to increase the turbine inlet temperature, in this case by 68.7 C. This also raises the exhaust temperature because the whole right side of the cycle has shifted upward on the isobars.
Hopefully, this helps to understand how to use the T-s diagram to estimate cycle performance. I will try to expand on these concept in the future so show how to use this method to show why low pressure cycles cannot achieve self-sustaining operation with our typical efficiencies, the impact of combustor pressure drop, the impact of adding an exhaust nozzle, etc....
- Chris
Additionally, I will provide a quick approach to calculating and visualizing some of the losses that impact performance, via the use of Temperature verses Entropy plots.
To get started, lets get familiar with the T-s diagram. As shown below, the x-axis is the entropy. Entropy is a state variable, which means that if we know two other state conditions, such as temperature and pressure, then there is a fixed entropy value that corresponds to these conditions. Entropy can be thought of, for our purposes, as a measure of the amount of energy unavailable to do work in a system. Thus, more entropy is bad, because it is basically a loss in the work the system can do and it is manifested in our engines by such things as mixing losses, friction losses (flow and mechancial), vorticity created by wakes in the flow path, etc, etc.
The lines in the x-y space are called isobaric lines, where the "iso-" prefix means "the same" and "baric" is from the Greek "baros" meaning weight, which is combined to mean constant pressure (since pressure is essentially a weight on an area). These lines of constant pressure are plotted for 1 bar to 5 bar as labelled in the upper right corner.
As can be seen, the isobaric lines diverge as the temperature increases and it is this divergence which makes the cycle work. If they stayed parallel, then no amount of heat could be added to overcome the losses introduced into the cycle due to the internal entropy producing processes. Luckily for us, they do diverge and thus as the amount of internal losses increase, we can overcome them by increasing the overall system temperature.
As an example, below is shown a simple gas turbine cycle. As can be seen, the ideal or lossless amount of spec work across a component is a vertical line from the starting point to the next point's isobar, i.e. a vertical line. A vertical line means that there was no entropy change across that component.
The light blue line on the lower left shows the smallest amount of specific work that would be needed to compress the air from 1 bar to 4 bar, which is roughly 142 kJ/kg. But in reality, our compressors will require more power than that because there are all kinds of losses within the compressor stage of our engines. So this is easy to visualize because we still have to end up at 4 bar, but we know there will be lost energy, which is entropy. So the line goes up an to the right.
Since isentropic energy is defined as:
compressor isentropic efficiency = (ideal del h) / (actual del h)
where h is the enthalpy, which, if we assume that we are using an ideal gas, can be calculated as:
h = Cp * T
Due to the assumptions used to find these equations, the correct Cp to use is an average between the Cp at the high temperature and the Cp at the low temperature. I attached a Cp vs Temperature plot for dry air further down, to help reader's estimate their average Cp for whichever component they are working with.
Since the proper Cp to use is the average Cp, the change in enthalpy (del h) can be calculated as:
del h = Cp *(T2 - T1)
and thus the isentropic efficiency can be recast as:
compressor isentropic efficiency = Cp*(T2ideal - T1) / Cp*(T2actual - T1) = (T2ideal - T1) / (T2actual - T1)
Now, if we shuffle things around we can get:
(T2actual-T1) = (T2ideal -T1) / compressor isentropic efficiency
So the actual temperature rise is just a scaled amount of the ideal heat rise in the compressor, scaled by the inverse of the isentropic efficiency.
Similarly for a turbine:
(T4actual - T3) = turbine isentropic efficiency * (T4ideal - T3)
Which is a direct scale...so the temperature drop across a 65% efficient turbine will be 65% of the ideal temperature drop.
It should be noted that the amount of specific work (work per mass) that is achieved over a component is given by:
Ws = del h = Cp * (T2-T1) [kJ/kg]
So, on the plot, the specific work can be understood to be the vertical distance between the two operating points of the component in question. You can "see" the amount of work available. This is a powerful tool for understanding how things change due to cycle variations!
And the actual work can be found by multiplying by the mass flow to get:
W = mdot * Cp *(T2 -T1) [kJ/s]
So now, if we focus more closely at the turbine side of this cycle as shown below:
If I start with a 65% efficient turbine (shown with the blue ideal line) and take a 177 C temperature drop across the stage I can get the 195 kJ/kg spec work to meet the compressor plus mechanical losses needed to keep the engine running at this steady state point. But if somehow, I loose 8% efficiency due to increased clearances (this is referenced to John's recent change as an example), then I can no longer get enough of a drop in temperature across the 4 bar to 1 bar pressure ratio and this inlet temperature.
The reduction in efficiency will mean that the change in entropy (del s) will be larger and thus there is less temperature drop available across this stage. The solution to get to the needed power demand is to increase the turbine inlet temperature, in this case by 68.7 C. This also raises the exhaust temperature because the whole right side of the cycle has shifted upward on the isobars.
Hopefully, this helps to understand how to use the T-s diagram to estimate cycle performance. I will try to expand on these concept in the future so show how to use this method to show why low pressure cycles cannot achieve self-sustaining operation with our typical efficiencies, the impact of combustor pressure drop, the impact of adding an exhaust nozzle, etc....
- Chris