Summary Gasturbine Performance lecture week 1 - 6

Gas turbine Performance lecture 1
Introducing the GE H80 turboprop engine (goals week 3 + 4)
By incorporating sophisticated technologies like bladed disc rotor design, improved
compressor and turbine aerodynamics, and high temperature capability materials, GE is
delivering more power and better efficiency for aerial application and utility aircraft
missions.

The principle of propulsion
1. Air is being accelerated.
2. Air experiences force from
balloon.
3. Balloon experiences same force in
opposite direction from air.

𝐹 =𝑚∙𝑎

𝑎𝑐𝑡𝑖𝑜𝑛 = − 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛

Newtons second law

𝐹 =𝑚∙𝑎

𝐹 = 𝑚 ∙ ∆𝑣

m = massastroom

Velocities air and aircraft

𝐶2 (𝑎𝑖𝑟𝑐𝑟𝑎𝑓𝑡) = 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑖𝑟𝑐𝑟𝑎𝑓𝑡 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑡𝑜 𝑡ℎ𝑒 𝑤𝑖𝑛𝑑 (𝑇𝐴𝑆)

𝐶? (𝑖𝑛) = 𝐶2 ∗ = 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑖𝑟 𝑎𝑡 𝑖𝑛𝑙𝑒𝑡 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑡𝑜 𝑡ℎ𝑒 𝑒𝑛𝑔𝑖𝑛𝑒

𝐶B (𝑗𝑒𝑡) = 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑖𝑟 𝑎𝑡 𝑜𝑢𝑡𝑙𝑒𝑡 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑡𝑜 𝑡ℎ𝑒 𝑒𝑛𝑔𝑖𝑛𝑒
*Alleen bij hoge snelheden (cruise)

Thrust

𝐹 = 𝑚 ∙ (𝐶B − 𝐶? )

Losses
Power from engine:
1. Power towards aircraft (𝜂F ) propulsive efficiency
2. Acceleration air

, Definition 𝜂F :

𝑃HIJKHLL 𝑃2?OPO2KM 𝑝𝑜𝑤𝑒𝑟 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑 𝑡𝑜 𝑎𝑖𝑟𝑐𝑟𝑎𝑓𝑡
𝜂F = = =
𝑃MNM2L 𝑃2?OPO2KM + 𝑃2?O 𝑝𝑜𝑤𝑒𝑟 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑 𝑏𝑦 𝑒𝑛𝑔𝑖𝑛𝑒

Power towards aircraft:

𝑃2?OPO2KM = 𝐹 ∙ 𝑣

𝑃2?OPO2KM = 𝑚 ∙ (𝐶B − 𝐶? ) ∙ 𝐶2

𝑃2?OPO2KM = 𝑚 ∙ (𝐶B − 𝐶? ) ∙ 𝐶?

Power towards air:

𝑃2?O = ∆𝐸U,2?O

1 1
𝑃2?O = 𝑚 ∙ (𝑣NHM )Y − 𝑚 ∙ (𝑣?Z )Y
2 2


1 Y 1
𝑃2?O = 𝑚 ∙ [𝐶B − 𝐶? \ − 𝑚 ∙ 0Y
2 2

1 Y
𝑃2?O = 𝑚 ∙ [𝐶B − 𝐶? \
2

Conclusion:

𝑃2?OPO2KM
𝜂F =
𝑃2?OPO2KM + 𝑃2?O

𝐶? 1 Y
𝜂F = 𝑚 ∙ [𝐶B − 𝐶? \ ∙ ∙ [𝐶B − 𝐶? \ ∙ 𝐶? + 𝑚 ∙ [𝐶B − 𝐶? \
𝑚 2

2
𝜂F =
𝐶B
1+𝐶
?


Efficiency versus thrust
We want:
1. A strong engine.
2. An efficient engine.

^_
^`
problem... Damn, but... 𝐹 = 𝑚 ∙ [𝐶B − 𝐶? \

So, solution: a large m (massastroom). Preferably SMALL acceleration of LARGE mass.