COOLING DRAG

Cylinder heads, oil coolers, and other heat exchangers require a flow of air through them for purposes of cooling. Usually, the source of this cooling air is the free stream, possibly alimented to some extent by a propeller slipstream or bleed air from the compressor section of a turbojet. As the air flows through the baffling, it experiences a loss in total pressure, Дp, thus extracting energy from the flow. At the same time, however, heat is added to

the flow. If the rate at which the heat is being added to the flow is less than the rate at which energy is being extracted from the flow, the energy and momentum flux in the exiting flow after it has expanded to the free-stream ambient pressure will be less than that of the entering flow. The result is a drag force known as cooling drag.

It is a matter of “bookkeeping” as to whether to penalize the airframe or the engine for this drag. Some manufacturers prefer to estimate the net power lost to the flow and subtract this from the engine power. Thus, no drag increment is added to the airplane. Typically, for a piston engine, the engine power is reduced by as much as approximately 6% in order to account for the cooling losses.

Because of the complexity of the internal flow through a typical engine installation, current methods for estimating cooling losses are semiempirical in nature, as exemplified by the Lycoming installation manual (Ref. 4.12). Before considering an example from that manual, let us examine the basic fundamentals of the problem. A cowling installation is schematically pictured in Figure 4.30. Far ahead of the cowling, free-stream conditions exist. Just ahead of the baffle, the flow is slowed so that the static pressure, PB, and the temperature, TB, are both higher than their free-stream values. As the flow passes through the baffle, PB drops by an amount Дp because of the friction in the restricted passages. At the same time, heat is added at the rate Q, which increases TB by the amount AT. The flow then exits with a velocity of VE and a pressure of PE, where PE is determined by the flow external to the cowling. The exit area, AE, and the pressure, pE, can both be controlled by the use of cowl flaps, as pictured in Figure 4.31. As the cowl flaps are opened, the amount of cooling flow increases rapidly because of the decreased pressure and the increased area. Downstream of the exit, the flow continues to accelerate (or decelerate) until the free-stream static pressure is reached. Corresponding to this state, the cooling air attains an ultimate velocity denoted by V„.

If m represents the mass flow rate through the system, the cooling drag,

Figure 4.30 Schematic flow through a heat exchanger.

Figure 4.31 The use of cowl flaps to control engine cooling air. (a) Flow of cooling air around an air-cooled engine, (b) Typical cowl flaps on horizontally-opposed engine.

Dc, will be given by the momentum theorem as

Dc=tn(Vo-V„) (4.45)

The rate, Д W, at which work is extracted from the flow is

AW = i2m(V02-Vj) (4.46)

Observe that the cooling’drag is obtained by dividing the increment in the

energy by a velocity that is the average between V„ and V»; that is,

For a fixed-baffle geometry, the pressure drop through the baffle will be proportional to the dynamic pressure ahead of the baffle.

Ap <x pVB (4.48)

Vb can be related to the mass flow, m, through the baffle and an average flow area through the baffle.

m = pABVB (4.49)

Substitution of Equation 4.49 into Equation 4.48 results in

2

Др a — (4.50)

P

In the preceding, for a fixed baffle, the fictitious area, AB, has been absorbed into the constant of proportionality.

The rate at which heat is conducted away from the baffles by the cooling air must depend on the difference between the temperature of the baffles and that of the entering cooling air. For a piston engine, the baffle temperature is the cylinder head temperature (CHT). Denoting the cooling air temperature just ahead of the baffles by TB, the rate of heat rejection by the engine can be written as

Q « CHT – TB (4.51)

In Equations 4.50 and 4.51, the constants of proportionality will depend on the particular engine geometry and power. These constants or appropriate graphs must be obtained from the manufacturer. One is tempted to scale Q in direct proportion to the engine power, P, and to scale AB with the two-thirds power of the engine power, in which case Equations 4.50 and 4.51 become

Др oc (4.52)

Q « (CHT – TB)P (4.53)

However, these are speculative relationships on my part; they are not substantiated by data and therefore should be used with caution.

If sufficient information is provided by the engine manufacturer to relate Ap, m, and p, and to estimate Q for a given temperature difference between CHT and the cooling air, then one is in a position to calculate the cooling drag. The details of this are best illustrated by means of an example given in Reference 4.12. In accordance with the reference, the English system of units will be used for this example.

This example will consider a horizontally opposed engine delivering 340 bhp operating at a 25,000 ft pressure altitude and a true airspeed of 275 mph. The outside air temperature is taken to be 20 °F higher than staifdard for this altitude, which means an OAT of -10°F. Lycoming recommends 435 °F as a maximum continuous cylinder head temperature for maximum engine life. This example is to represent a cruise operation with cowl flaps closed. (This does not mean that the exit is closed; see Figure 4.31.) It is therefore assumed that the exit static pressure is equal to the ambient static pressure. Given the foregoing, the problem is to size the area of the exit and to calculate the cooling drag.

The mass density, p, can be calculated from the equation of state (Equation 2.1).

h

P Ps J.

where subscript s refers to standard values. From this, using Figure 2.3,

p = 0.00102 slugs/ft3

The free-stream dynamic pressure, q0, is thus

q0 = 83 psf

Neglecting any contribution from the propeller slipstream for the cruise condition, the reference, on the basis of experience, calculates the tem­perature at the engine face by assuming a 75% recovery of the dynamic pressure with a resulting adiabatic temperature rise. From Equations 2.1 and 2.30,

T

pHy-nivi ~ constant

or

= дсп Г786-3 + 0 75(83)1° L 786.3 J

= 460 °R

Thus, the temperature at the engine face is 0 °F.

With the assumed 25% loss in dynamic pressure resulting from the diffusion by the cowling, the static pressure at the engine face, assuming the flow there to have a negligible velocity, will be

pB = 848 psf

This corresponds to a pressure altitude of 23,200 ft. Next, Figure 4.32 (pro­vided by the manufacturer), is entered with the cooling air temperature at the

Щип 4.32 Required cooling airflow as a function of cooling air temperature at ehgine face.

engine face. For a CHT of 435 °F, this figure leads to a required cooling airflow of 2.55 lb/sec.

The baffle pressure drop is determined next from Figure 4.33 (also provided by the manufacturer). It would seem, in accordance with Equation 4.50, that this graph should be in terms of the density altitude instead of the pressure altitude. Nevertheless, it is presented here as taken from Reference

4.12. Entering this graph with an airflow of 2.55 lb/sec at a pressure altitude of 23,200 ft results in a pressure drop of 47 psf. Thus, downstream of the cylinders, the total pressure will be

pB – Ap = 801 psf

Instead of using a heat flow, Q, per se, the reference assumes (presum­ably based on experience) that the cooling air will experience a temperature rise of approximately 150 °F across the cylinders. For this particular example, this temperature rise can be expressed in terms of Q by using the specific heat at constant pressure.

Q = CpATm

For air Cp — 6000 ft/lb/slug/°R. Thus, for the airflow of 2.55 lb/sec,

Q = 6000(150)^|

= 71,273 ft-lb/sec = 129.6 hp

Notice that the above rejected heat amounts to 38% of the engine power. This is close to the 40 to 50% qupted in other sources.

The cooling air density just downstream of the cylinder heads after the temperature rise can be calculated from the equation of state.

Рв-Ьр

RT

801

1711(460+150)

= 0.000767 slugs/ft3

This flow is then assumed to expand adiabatically to the ambient static pressure of 786 psf. Thus the density at the exit, from Equation 2.30, is found to be

pE = 0.00076 slugs/ft3

The corresponding velocity is determined by the use of Equation 2.31.

’ 2y,	(pB — Ар	_ ‘ll/2 .2^1
у -1 ‘	V рв’	Pe) J
2(1.4)	/ 801	786
0.4	.000767	0.000757

)]

1/2

= 205 fps

This velocity, together with the density of the cooling flow at the exit and the required cooling flow, leads to a required exit area of 0.510 ft2. This number, as well as VE, differs slightly from Reference 4.12 as the result of calculating the flow state following the addition of heat in a manner somewhat different from the reference.

The resulting cooling drag for this example can be calculated from Equation 4.45.

Dc =— (403 – 205)

= 15.7 lb

This corresponds to an increment in the flat-plate area of 0.19 ft2. In terms of engine bhp, assuming a propeller efficiency of 85%, this represents a loss of

10.2 bhp, or 3% of the engine power.

For operating conditions other than cruise, it may be necessary to open the cowl flaps. Reference 4.12 states that a pressure coefficient at the cowl exit as low as —0.5 can be produced by opening the flaps to an angle of approximately 15°. By so doing, a relatively higher cooling flow can be generated at a lower speed, such as during a climb. Even though the engine power may be higher during climb, operating with open cowl flaps and with a richer fuel mixture can hold the CHT down to an acceptable value. Also, CHT values higher than the maximum continuous rating are allowed by the manufacturer for a limited period of time.

It will not be repeated here, but Reference 4.12 also performs a cal­culation similar to the foregoing but for climb conditions at 19,000 ft pressure altitude, 450 bhp, mixture rich, 130 mph true airspeed, and an OAT of 31 °R A CHT of 475 °F is allowed with the cowl flaps open to give an exit Cp of -0.5.

For this case, the required airflow is determined to be 1.95 lb/sec with an exit velocity of 73.8 fps. Thus, for this case,

Dc = 7.1 lb fc = 0.32 ft2

The equivalent power loss is only approximately 0.5% of the engine power for this case, even though the increment in equivalent flat-plate area is appreci­ably higher than that in cruise.