Just using the surface area of our cylinder (From Appendix 3a), we have 117 m x 72 m + 2 x 414 (ends) = 9,252 m^2 at 440 gsm = 4,071 kg.
Appendix 3ii-d) The Running Rigging 298 kg.
The Command gondola must be able to transmit signals to the tail surfaces, and it must also be able to vent gas from each gas bag at will. Assuming steel wire at 10m /kg . Assume two cables per fin. Four fins (upper and lower rudder, left and right elevator) for 8 wires. Assume the controls have to travel 170 m, to give 1,360 m.
Gas vents. Assume 19 vents (there should be one per gas bag, and one gas bag per 9 m), there is one cable per valve, and an average distance from the control gondola of 85 m, to give 1,615 m.
Running rigging is 2,975 m or 298 kg.
Appendix 3ii-e) The Gondolas 1,050kg
I have no idea. Gondolas need to be able to carry the engines and people servicing them, plus the command gondola. Woodhouse [p.209] allocates about 150 kg each. We have decided on 6 engines and a command gondola, so, 7 x 150 = 1050kg.
Appendix 3ii-f) Engines 1,560 kg
Woodhouse [p.209] allocates 260 kg per 100 hp engine, for engine, mufflers, radiator, water, propeller, and hub. Hot-bulb engines aren't known for their power to weight ratios, we'll assume 260 kg for every complete 40 HP engine. Our airship needs 104 HP for 15 mps as a cruise speed. We want some spare power for emergencies, so we'll assume 6x 40 hp = 240 hp. 1,560 kg (6.5 kg/ hp). Giving a top speed in still air of 16 – 19 mps (37.9 – 42.7 mph)-the lower value represents performance with only 71% power being delivered as thrust. The six engines means the Sao Martinho can provide the required cruise power of 126 HP even with two of the six engines out for repairs and maintenance.
Appendix 3ii-g) Trim ballast and trim pumps and pipes 1,260 kg.
An airship flying level is an airship flying economically. If there is an angle of attack-using the hull to gain lift-more drag is experienced, and more power (thus more fuel) is needed for a given forward velocity. To maintain level trim an airship has trim ballast. This is usually water that can be pumped to various holding tanks to balance the ship (The crew can double as emergency trim ballast.). Normal ballast, which is dumped to reduce weight when trying to gain height (averaging 4% according to Brooks [p.186]), is not trim ballast.
Woodhouse says a navy patrol blimp has 90 lbs of trim ballast on a 5,275 lb gross lift airship (1.7%). Using this value, the Sao Martinho would have trim ballast of ~873 kg (1).
We now have to be able to move it to tanks forward and aft. We'll assume 5/8" inch copper tube at about 0.5kg/m, and we’ll want about 340 m-170 kg (2).
Hand operated pumps at, call it 25 kg (3)
Copper tanks to hold 1,500 liters. 6 x 275 liter drums at 16kg = 96 kg (4).
(We need some empty space so we can move water from place to place, hence available volume exceeds actual volume of trim ballast.)
We also need water tanks for the ~2,097 kg of ballast water in say, 6 x 300 liter tanks at about 16kg each = 96 kg (5).
Totaling 1,260 kg.
Appendix 3ii-h) Fuel tanks, fuel piping, and pumps 810 kg.
We have 5,897 kg of fuel. Assume it is petrol, and we have ~6,700 liters. We want to spread them along the length of the keel, so 12 x 600 liter tanks at 20 kg each = 240 kg (1)..
Plumbing for the fuel, 85 m x 10, plus two runs of 170 m to connect all tanks = 1,020 m at 0.5 kg per m 510 kg (2).
Pumps per tank 12 x 5 kg = 60 kg (3)
Total 810 kg.
Appendix 4: Conversion rates
All currency conversions are based on an exchange rate of one guilder = USE$40.
One Guilder = 36 kruezer)
One Guilder = USE$40
One Thaler = 90 kruezer
One Thaler = USE$100
One Spanish Ducat = 110 kruezer
One Spanish Ducat = USE$122
One Spanish Reale (a "piece of eight") = 10 kruezer
One Spanish Reale = USE$11
One Venetian Ducat = 88 kruezer
One Venetian Ducat = USE$98
One English Pound = 400 kruezer
One English pound = USE$444
One French Livre = 36 kruezer
One French Livre = USE$40
One Florin = 60 kruezer
One Florin = USE$67
One Gulden = 60 kruezer
One Gulden = USE$67
One Rixdaler (Swedish) = 90 kruezer
One Rixdaler = USE$100
One Rigsdaler (Danish) = 90 kruezer
One Rigsdaler = USE$100
(Note: there are 3 kruezer to the groschen, and 20 groschen to the thaler.)
One cubic meter (m^3) = 35.315 cubic feet (ft^3)
One kilogram (kg) = 2.204 pounds (lbs)
One meter per second (mps) = 2.237 miles per hour (mph)
One KW (1000 watts) = 1.34 horse power (HP)
One Horse Power (HP) = 746 watts
Appendix 5: Calculating energy to produce one m^3 of hydrogen by blowing steam over red-hot iron filings:
The chemical equilibrium equation is:
3Fe + 4H2O ‹=› Fe3O4 + 4H2
This tells us that for every water molecule input, we get a H2 molecule.
At STP one mol of a gas occupies 22.41 lt. One m^3 of H2 at STP needs 1,000/22.41 lt/mol = 44.623 moles.
Atomic weight H2O = 18.01528 g
44.623 mol x 18.01528 g = 803.92g (0.80392 kg)
To heat 1kg water from say, 10 degrees C to 100 degrees C
Specific heat = 1 cal/ gram/ degree C. 1000 g x (100-10) = 90,000 cal
Latent heat of vaporization (convert water to steam at 100 degrees C) = 540 cal/g @100 degrees C. 1000 g x 540 = 540,000 cal
Total energy to get 1 kg of water from 10 degrees C to steam is: 90k + 540 k = 630,000 cal/ kg
@ 4.1813 j/cal = 2,634,219 j/kg
Energy to make steam to produce 1 m^3 of H2 = 2,634,219 x 0.80392 = 2,117,701 j.
At 2,117,701 j per m^3, we are replacing all the hydrogen vented to cover burned fuel (5,897 kg, / 1.09 = 5,410 m^3) = 11,456,762 kj.
Charcoal at 29,600 kj/ kg = 387 kg at 4.2 thaler / 1000 kg = 1.63 thaler
Coal at 27,000kj/kg = 424 kg at 2.28 thaler / 1000kg = 0.97 thaler
Dry wood at 15,000 kj/kg = 764 kg at 2.64 thaler/ 1000kg = 2.02 thaler
A very big, permanent double-boiler system might produce 284 m^3 (10,000 ft^3) per hour. Portable systems, similar in size to the wagon mounted systems used by the Union in the ACW might have a capacity of up to 60 m^3 per hour. Refilling airships is a slow process.
Bibliography
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