Airship Lift By Hot Air From Its Engines

Introduction

Airships use helium. To go up the bag (sometimes called the envelope) is filled. To go down some helium is compressed into small bags, called ballonets. Helium is getting scarce and expensive, and there are problems associated with the use of other gases. Hot air can be used instead of helium and such ships use a gas burner to heat the air. Here it is shown that airships can use air which is heated by the engines.

To ensure that the values used are realistic an example airship is used, currently made by HAV, the Airlander 10. Specifications are given in the appendix.

The engines of this airship are diesel, and a general rule of thumb is that the energy consumed by such engines is distributed thus: 35% power output, 35% exhaust & 30% water jacket. The power output of the example engines (four are used) is 1MW. This means that the heat from the water jacket is 895kW. This heat is available in the air leaving the engine radiators. Allowing for the inaccuracy of the rule of thumb, and for losses, assume that only 400kW is available.

Figure 1
Figure 1

Figure 1 shows how the air from the radiators can be directed into the bag, to raise the temperature of the air in the bag. The pressure in the bag is maintained at ambient pressure by a pressure relief valve releasing air from the top of the bag, so the density of that air is decreased, and the bag becomes buoyant.

Ambient air is drawn through the radiator as in normal practice. From the radiator the air, instead of escaping to atmosphere as normal, enters a duct. In the duct is a valve which can take any position between its extents A & B. At A all the air is directed into the bag. At B, the air is vented to the atmosphere. Any position between A & B can can be selected to control the rate of ascent.

To descend, the valve at the top of the bag can be opened and a valve at the bottom of the bag can be opened, thus letting ambient (cooler) air into the bag.

ICAO, for the ambient air in the troposphere
PslStandard pressure at sea level = 101325Pa
TslStandard temperature at sea level = 288.15K
ρslStandard density at sea level = 1.225kg/m3
HtHeight of tropopause = 11000m
LTemperature lapse rate = (Tsl -Tt) / Ht = 0.0065K/m
PtPressure at tropopause = 22633Pa
TtTemperature at tropopause = 216.65K
g9.80665m/s2
RUniversal gas constant = 8.31447J/mol.K
MMolar mass of dry air = 0.0289644kg/mol
RsSpecific gas constant = R / M = 287.048J/kg.K

T = Tsl - LH
P = Psl (1 -
LH
Tsl
)
g
RsL
ρ =
PM
RT
=
P
RsT

In the lower stratosphere (constant T): Hs = Height above HT

P = Pt e
( g
(Ht - Hs)
RsTs
)
aambient
bair in bag
ffinal
iingoing air
oout
mmass = V.ρ kg
VbVolume of bag
VsVolume of ship, say = Vb
WbWeight of air in bag = Vbb.g
WsWeight of (rest of ) ship
WtWb + Ws
ρsDensity of ship = Wt / Vb
Cspecific heat, kJ/kg.K
Ttemperature, K
Vvolume

Buoyancy

Upthrust = ρa g Vb

Acting against this is Wt

So total buoyancy force = ρa g Vb - Wt

Maintenance of Constant Height

For constant height: ρa g Vb = Wt

Pa g Vb
Rs Ta
= Wb + Ws = Vb ρb g + Ws =
Vb Pb g
Rs Tb
+ Ws
Vb Pb g
Rs Tb
=
Pa g Vb
Rs Ta
- Ws
Vb Pb g =
Rs Tb Pa g Vb
Rs Ta
- Rs Tb Ws
= Tb (
Pa g Vb
Ta - Rs Ws
)
Tb = (
Vb Pb g
(
Pa g Vb
Ta - Rs Ws
)
)

As stated above, Pb = Pa.

Example: required height = 500. From HAV's Airlander 10, Vb = 38000 & Ws = 20000.

Pb = Pa = Psl (1 -
LH
Tsl
)
g
RsL
= 101325 (1 -
0.0065 x 500
288
)
9.80665
287 x 0.0065
= 95458Pa

Ta = Tsl - LH = 288 - 0.0065 x 500
= 285K

Tb = (
38000 x 95458 x 9.80665
(
95458 x 9.80665 x 38000
285 - 287 x 20000
)
)
= 299K

If the bag is internally lined with Aerofoam or survival blanket material, then the thermal conductivity, λ, of the bag will be 0.044W/m2.K:

Heat loss = Surface area . δT . λ

≃ 15050 x 11 x 0.044

= 7.3kW

Example: required height = 6000.

Pb = Pa = Psl (1 -
LH
Tsl
)
g
RsL
= 101325 (1 -
0.0065 x 6000
288
)
9.80665
287 x 0.0065
= 47154Pa

Ta = Tsl - LH = 288 - 0.0065 x 6000
= 249K

Tb = (
38000 x 47154 x 9.80665
(
47154 x 9.80665 x 38000
249 - 287 x 20000
)
)
= 271K

Heat loss = 11.3kW

Example: required height = 11000.

Tb = (
38000 x 22633 x 9.80665
(
22633 x 9.80665 x 38000
217 - 287 x 20000
)
)
= 255K

Heat loss = 25.2kW

As shown above, 400kW is available. This is ample.

Rate of ascent

Regarding the air from the radiator: Cair ≃ 1.

Vb = 38000.

At sea level: ρa = 1.225, mb = 46550.

So the air from the radiator can raise the bag temperature by

400000
46550
 = 8.6K/s

At a temperature of 288 + 8.6,

ρ =
P
RsT
 = 1.19
H′ =
Tsl
L
( 1 - (
ρRsT
Psl
)
RsL
g
)
=
288
0.0065
( 1 - (
1.19 x 287 x 296.6
101325
)
287 x 0.0065
9.80665
)
= 2.26

In other words, a climb rate of 136m/min.

At H = 500: Pa = 95458 & Ta = 285

ρa =
P
RsT
=
95458
287 x 285
= 1.167

mb = 44346.

So the air from the radiator can raise the bag temperature by

400000
44346
 = 9K/s

At a temperature of 285 + 9,

ρa =
P
RsT
 = 1.13
H′ =
285
0.0065
( 1 - (
1.13 x 287 x 294
95458
)
287 x 0.0065
9.80665
)
= 9.7

In other words, a climb rate of 582m/min.

At H = 11000

Pa = 22663

Ta = 217

ρa =
22663
287 x 217
 = 0.364

mb = 13828.

So the air from the radiator can raise the bag temperature by

400000
13828
 = 29K/s

At a temperature of 217 + 29,

ρa =
P
RsT
 = 0.32
H′ =
217
0.0065
( 1 - (
0.32 x 287 x 246
22663
)
287 x 0.0065
9.80665
)
= 19.7

In other words, a climb rate of 1184m/min.

The engine power will decrease with altitude, but power reduction is not considered here.

Figure 2
Figure 2

So far the heat from the water jacket has been considered, but more heat is available from the exhaust. It is not suggested that the exhaust is vented into the bag, which would cause contamination of the bag material and internal components. Instead, the heat from the exhaust can be used as shown in Figure 2.

The exhaust pipe is part of a duct with a valve which works as explained in Figure 1. With the valve in position B the exhaust is vented to atmosphere. When the exhaust is directed into the bag the flow is then through a heat exchanger and then is vented to atmosphere.

First Conclusion

The use of heat from diesel engines is a practical proposition.

The Future

As airship speeds increase the use of jet engines will become practicable.

The Turboprop

Figure 3
Figure 3

An example of a current engine of the required power output is the General Electric CT7. This has an air flow rate of 4.5kg/s, and the exhaust temperature will be ≃ 1073K. With four engines, and letting C = 1, the heat power available is 19.3kW. From Maintenance of Constant Height it can be seen that this is sufficient for the ceiling of the Airlander, 6000m. Figure 3 shows how to achieve the heat exchange.

The exhaust is ducted. The duct has two interconnected valves. With the valves at B the exhaust is vented to atmosphere. With the valves at A the exhaust heat is directed into the bag, the flow then being through heat exchanger and then to atmosphere.

Figure 4
Figure 4

Another way of getting heat from a jet engine, turboprop, turbofan or turbojet, is to take air from the compressor, as shown in Figure 4.

Some air is bled from the compressor outlet into a duct, 2, containing a valve, 3, which can move between the extents A & B as before. With the valve at B the air is redirected back to the engine, 4. With the valve at A the air is directed to the bag, 5.

The Compressor

The compression is isentropic and a CR of 4 is typical of current engines. γ for air = 1.4.

Pa
Po
= (
Vo
Va
)
γ
So Po = Pa(4
γ
)
To
Ta
= (
Po
Pa
)
(1 -
1
γ
)
So To = Ta (
Po
Pa
)
(1 -
1
γ
)
ρo =
Po
RsTo

At sea level: Ta = 288 & Pa = 101325.

At 500m: Ta = 285 & Pa = 95458.

For an ascent between these heights let Ta & Pa be the median values, ie, 286.5 & 98392.

From the compressor: Po = 685238, To = 499 & ρo = 4.78

From Maintenance of Constant Height:

Tb = (
Vb Pb g
(
Pa g Vb
Ta - Rs Ws
)
)  = 299

For a mixture of two masses of air at different temperatures:

Tf =
m1 C1 T1 + m2 C2 T2
m1 C1 + m2 C2
C1 ≃ C2 ≃ 1, so Tf =
m1 T1 + m2 T2
m1 + m2

Once the masses are mixed the composition will be:

( Va - Vi ) @ 288K + Vi @ 499K

So Tf =
288 x 1.225 ( Va - Vi ) + 499 x 4.78Vi
1.225 ( Va - Vi ) + 4.78Vi
=
288 x 1.225Va - 288 x 1.225Vi + 499 x 4.78Vi
1.225Va - 1.225Vi + 4.78Vi
=
288 x 1.225Va + 2032Vi
1.225Va + 3.56Vi
Tf ( 1.225Va + 3.56Vi ) = 288 x 1.225Va + 2032Vi
1.225VaTf + 3.56ViTf = 288 x 1.225Va + 2032Vi
2032Vi - 3.56ViTf = 1.225VaTf - 288 x 1.225Va
Vi ( 2032 - 3.56Tf ) = 1.225Va ( Tf - 288 )
Vi =
1.225Va ( Tf - 288 )
2032 - 3.56Tf
Vi =
1.225Va x 38000 ( 299 - 288 )
2032 - 3.56 x 299
 = 529
The work required = γRs (
To - Ti
γ - 1
)
= 1.4 x 287 (
499 - 285
0.4
)
 = 214963 J/m3

To match the rate of ascent accomplished by the radiator air of the diesel-powered ship, Vi must be supplied in 221s.

So the compressor power = 214963 x
529
221
= 515kW
 = 139kW / engine

(presuming four engines). Such an engine can be made.

Second Conclusion

The use of heat from a jet engine is a practical proposition.

Comparison With Helium

For maintenance of height there is no difference. For rate of climb, the density of helium at STP is 0.179kg/m3, so helium is far superior. The other advantage of helium becomes apparent when considering a cold start. When using air, first the engines must be warmed up and then the air in the bag must be heated. There is a patent covering the use of engines to heat the gas in the bag, but I suspect that the cold start is the reason that hot air is not used.

Appendix

HAV Airlander 10
Bag volume38000
Length92
Width43.5
Height26
Endurance5 days manned
Altitude6100
Engines4 x 261kW, 4 litre, turbocharged diesels
Total weight20000
Payload10000
Speedcruise 41, loiter 10

GE CT7 turboprop engine
SHP1100kW
Air flow4.5kg/s




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