Overview
With the aircraft advancing and beign at home for a few weeks over Christmas it was decided to firm up the cooling of the cowl.
Why?
Simple, I would like to close up the hole under the nose gear but this is currently the primary means of cooling and engine air other components will be required.
The radiator is the primary concern as it size is fixed and options are few. Let get this straight it has worked in the factory aircraft for 300 plus hours and there are others on the way but one effect of closing a cowl is the internal temperatures climb and air has to have a routed in/out.
The first blog simply says you need ducts so that easy just build an internal cowl. The original design showed a connection to the fake exhaust - nice but not practical as to look correct they have to reside on the fuselage and the connection would be a bear.
What was needed was a means of extracting the air while maintaining the cowls look and this restricted the ideas to one - louvers. After a lot of work I have settled on a set of automotive louvers 5'' x 6.5 '' overall for a WRX.
Proposed duct - 550 mm overall |
The inlet in light orange will be installed as the airflow is underrated in climb remembering that engine and oil cooling have to be handled by the radiator and the calcs show the radiator and water flow underrated this mode. The heat exchanger for the oil will be incorporated into the engine shroud for a number of reasons but one is to allow the exchanger to have an air blast to help offset load on the radiator.
The front duct will decelerate the incoming air raising the dynamic pressure with the duct on the reverse side accelerating the air reversing the process and then raising the pressure at the exit again.
A wicker is just an additional trip at the front vent
The airflow behind the vent would be interesting
The airflow behind the vent would be interesting
RV12 Radiator / oil cooler duct |
Another vent will be placed on the port side to vent 12 kw of heat created by the cylinders and radiant heat from the exhausts, radiator and other items.
Note: This will only be effective with a sealed cowl
The fins are cooled by a downdraft pendulum used on the RV12, this will be covered in a later blog but in the heat exchanger plates located on its underside will be installed into the shroud inlet using incoming air to cool the cylinders taking some load off the radiator as the 64 mm diameter inlet has an excess of air available.
If door are installed they will have two sets of louvers fitted to allow the radiated heat to be ejected on the underbelly [see note above]
The final item examined was the supercharger air supply as this is also part of this equation. A 75 mm od supply was used for a simple reason, its what is available for the aftermarket automotive market. The inlet and filter are both carbon fibre and very light, cost effective with all connection being achieved using 75 / 64 mm od SCATT hose.
Comment
The rough estimate of the workload outlined below shows that there is a potential load at full power of 48 kw on the radiator with the water pump providing 50 Kw of water with 65 Kw of air available in level flight. If correct it means the system has to be very efficient in climb and when full power is beign used as every cubic cm of air must be harnessed along with the cooling water and there is a need to verify the capacity of the radiator in the real world carefully!.
Wishing I had paid a lot more attention when in sitting in the classes - always wise after the event so below is some rusty thermodynamics.
The initials calcs were based on a article for the Europa by Jans
Apologize for any error in advance
Maximum
power output Rotax Supercharged - 100 kW
Consumption
at maximum power 40 l/h, i.e. 11.1 ml/s.
Gasoline
represents 35 kJ of energy per ml.
Power
consumption is therefore 11.1 x 35 = 388 kW
Efficiency
is 100 / 388 = 26%.
Maximum
cruise power output = 79 kW
Consumption
at maximum cruise power is 29 l/h, i.e. 8.0 ml/s.
Power
consumption is 280 kW
Efficiency
is 28%.
75%
cruise power output 55.1 kW
Consumption
at 75% cruise power is 20.4 l/h, i.e. 5.67 ml/s.
Power
consumption is 198 kW
Efficiency
is 27.8%.
Total
heat production to be removed estimated :
388
– (85 * 1.13) – (12.5 * 1.13) = 278 kW.
Assumptions
Rotax 914 heat removal prescription for maximum power operation (86 kW)
has
to reject 45 kW as follows:
30
kW through cooling radiator
9
kW through oil radiator
6
kW from cylinder barrel fins
Supercharger
factor = 912S/914 = 105/86 = 1.22
Under
the cowling.
As the
engine and exhaust system heat up they radiate more heat.
Stefan-Boltzmann
says: about 5 / 10^11 x T^4 kW/m^2, T in Kelvin.
Table -
T in Kelvin ('F) and corresponding radiation flux in kW/m^2 - :
700
(800'F) 12.0
800
(980'F) 20.5 - supercharger runs about 200 degrees cooler than turbo
900
(1160'F) 32.8
1000 (1340'F) 50.0 - turbo calcs
1100
(1520'F) 73.2
Estimate
of 800K area (4 primary exhaust pipes): 0.10 m2
Radiated
heat: 2 kW
Estimate
of 800K area (muffler, turbo): 0.15 m2
Radiated
heat: 3 kW
Say
another 5 kW to be removed from under the cowling.
Total (39 * 1.22) + 5 = 53 kW.
Estimated
exhaust gas energy 278 – 53= 225 Kw - that why a turbo is best for raw power
Assumptions
Similar
efficiency for direct and indirect - via liquid - air cooling:
Inlet
areas :
Cooling
inlet radiator 210 cm^2 (estimated)
Cylinder
barrel fins 20 cm^2 (estimate on diameter 5.0 cm)
Gear
door louvers 18 * 7.5 * 1 = 135 cm^2 (18 louvers total)
11 ml
gasoline per second with mass 11 * 0.7 g/s = 8.7 g/s
requires
about 14 * 8.7 = 121 g/s of air per second
Assumptions
Air velocity into ducts is 40% of
aircraft velocity
Specific
Gravity 1.25kg/m3 at sea level pressure
Climb best rate =
120 km/hr (33 m/s) worst case air flow
Inlet radiator
duct 210 cm/sq
Inlet supercharger
44 cm/sq
Specific heat
capacity air = 1005 Kj/kg/deg K
Specific heat
capacity water – 4.2 Kj/kg/deg K
Flow water = 73
l/min or 1.2 l/sec
Calculations:
Air flow supercharger
Air flow supercharger
Using Q = (a*v*0.4) * s.g. = g/sec
44 * 33 * 0.4 * 1.25 = 726 g/s
NACA 3'' od Duct [44 sq cms] will supply enough air
by a 6:1 margin
Air flow radiator inlet
210 * (33 * 0.4) =
2370 l/s climb
210 * (60 * 0.4) =
5040 l/s cruse
Assume required radiator
delta t = 10 deg /c (283 deg /K)
The
Specific Heat formula is: c
= ΔQ / (m × ΔT)
Where:
c: Specific Heat , in J/(kg.K)
ΔQ: Heat required for the temperature change, in J
ΔT: Temperature change, in K
m: Mass of the object, in kg
Where:
c: Specific Heat , in J/(kg.K)
ΔQ: Heat required for the temperature change, in J
ΔT: Temperature change, in K
m: Mass of the object, in kg
Theoretical capacity air
1.005 = Q/ (2.4 * 1.25) * 10
Q = 30 KW climb
dT = (Q / (Heat output turbo x Supercharger factor)) * target radiator
= (30 / 39 x 1.22) * 10 = 6.2 deg C
= (30 / 39 x 1.22) * 10 = 6.2 deg C
1.005 = Q/ (5.0 x 1.25) * 10
Q = 63 kW cruse
dTmax = 63 / 48 x 10 = 13.1 deg C
Theoretical capacity water
4.2 = Q/ (1.2 * 1) * 10
Q = 50 kW
Q = 50 kW
No comments:
Post a Comment