Technology and Operations
This forum is for discussing technological & operational matters pertaining to U-boats.
Re: Torpedo Launching
Posted by:
TomW
()
Date: November 03, 2003 03:58AM
I did an estimate of the pressurizable volume of a Type VIIC U-Boat:
Per "uboat.net", the submerged displacement of a Type VIIc is 871 tons.
Since submerged ship displaces its volume in water, I assume this is given in metric tons (2200 lbs) or long-tons
(2240 lbs) and equates to approximately 1,916,200 lbs of seawater.
From standard seawater density of 64 lbs/cu.ft, this gives a pressure hull volume of approximately 29,940 cubic feet.
As a check, the PH overall length is listed as 50.5 m = 165.68'
and the beam (or diameter) is listed as 4.7 m = 15.42'
Assuming cylindrical PH with hemispherical ends, the calculated volume
per these dimensions is 29,981 cubic feet.
This confirms displacement volume above.
As a round number, I will use 30,000 cubic feet (empty PH).
Undetermined is how much of this volume is pressurizable?
Obviously, some of this volume is taken-up by tankage (trim tanks, negative
tank, etc). Crew is not pressurizable. Most equipment (electrical panels) is pressurizable, as is stowage lockers,
tool boxes, etc. Even stuffing the hull with provisions for patrol, I believe it would be difficlt to reduce
the pressurizable volume to 50% of the total pressure hull volume of 30,000 cu.ft.
My gut feel on this pressurizable volume is 60% to 75% of total pressure hull volume. For the first estimate, I've
decided to split the diffence and use 2/3rds, or 66% of total PH volume which is 20,000 cu ft. If anyone has a
more accurate number, I'll use it.
Now, regarded torpedo ejection, the two pieces of information provided govern this process.
If the torpedo is accelerated at an average acceleration (a), for a period of time (t), and distance (d),
it will attain a final velocity (v). Since torpedo tube length (d) is given as 25 feet, and final velocity
(v) is 33 ft/sec, "t" and "a" can be calculated. The eject time (t) is calculated as 1.52 seconds, and the
average acceleration (a) is 21.7 ft/sec2. These values also pass the "sanity check" and compare well with my data
for other "vehicles" launched from a submerged sub.
Since the 1.52 second eject time is "in the ballpark", I believe the average acceleration to be valid as well.
Now, four (4) basic factors govern the air pressure used to eject the torpedo. The pressure applied to the piston must:
#1 Overcome the external sea pressure at the torpedo tube muzzle at Periscope Depth (P.D.)
#2 Overcome the sliding friction between the torpedo and the tube.
#3 Overcome the hydrodynamic drag of the 21" diameter torpedo moving thru the water.
#4 Accelerate the torpedo at an average of 21.7 ft/sec2 to achieve at final velocity of 33 ft/sec.
Given P.D. or 14 meters and PH diameter of 4.7 meters, I have assumed the lower torpedo tubes to be at a depth of
12 meters and have a sea pressure of 17.5 psi at the torpedo tube muzzle.
Regarding the sliding friction, I've used a coefficient of friction (u) of approx 0.5; that is, the sliding friction
is 1/2 the weight of the torpedo/piston or approximately 1800 lbs. Given the piston area of 346.36 sq.in., an air
pressure of 5.2 psi is required to overcome the sliding friction of a "greased" torpedo.
I am still working on the calcs for hydrodynamic drag; however, it is low at small velocity and increases with the
square of the velocity.
Given a torpedo mass of approximately 109.75 lbs-m, the force required to achieve and acceleration of 21.7 ft/sec2
is 2382 lbs, Again, given a piston area of 346.36 sq.in., the required air pressure is only 6.9 psid across the piston.
Adding the pressures, an average air pressure of 30 psi behind the piston is required to eject the torpedo at P.D.
The full pressure of 24 atmospheres was not used; this would have imparted an acceleration of over 1000 ft/sec2!!
and the torpedo would have left the tube at a velocity well above 33 ft/sec. Unfortunately, I am not an expert
on Uboat systems, but I am assuming that the 24 atmosphere header was used to charge one (or more) eject air flasks
which would then be "dumped" directly into the torpedo tube to eject the torpedo. In another assumption, I'm
estimating an eject flask volume of approx. 8 cu ft to 10 cu ft. By prepressurizing the eject flask(s) to 24 atmospheres,
then "dumping" this volume directly into a torpedo tube, an average torpedo tube pressure of 30 - 50 psi can be achieved,
ovecoming sea pressure, and friction and imparting a 33 ft/sec eject velocity to the torpedo. I suspect that one
(or more) torpedo eject flasks were used to achieve the air volume required; a pressure regualating valve would
likely be too large to maintain the required outlet pressure.
Can any of the Uboat experts confirm that torpedo eject air flasks existed?
More assumptions:
If this is true, my prelim calcs indicate that when fired torpedo tube was vented inboard, the pressure inside
the boat increased by approximately 0.3 psi per tube. Therefore, if four (4) torpedos were fired, boat pressure
increased to 1.2 psi. THe force of this pressure on a 600 mm hatch is over 525 lbs! This certainly seems like
enough to lift the first person thru the hatch!
I will continue to refine my calcs, but any technical input(s) are welcomed!
TomW
Per "uboat.net", the submerged displacement of a Type VIIc is 871 tons.
Since submerged ship displaces its volume in water, I assume this is given in metric tons (2200 lbs) or long-tons
(2240 lbs) and equates to approximately 1,916,200 lbs of seawater.
From standard seawater density of 64 lbs/cu.ft, this gives a pressure hull volume of approximately 29,940 cubic feet.
As a check, the PH overall length is listed as 50.5 m = 165.68'
and the beam (or diameter) is listed as 4.7 m = 15.42'
Assuming cylindrical PH with hemispherical ends, the calculated volume
per these dimensions is 29,981 cubic feet.
This confirms displacement volume above.
As a round number, I will use 30,000 cubic feet (empty PH).
Undetermined is how much of this volume is pressurizable?
Obviously, some of this volume is taken-up by tankage (trim tanks, negative
tank, etc). Crew is not pressurizable. Most equipment (electrical panels) is pressurizable, as is stowage lockers,
tool boxes, etc. Even stuffing the hull with provisions for patrol, I believe it would be difficlt to reduce
the pressurizable volume to 50% of the total pressure hull volume of 30,000 cu.ft.
My gut feel on this pressurizable volume is 60% to 75% of total pressure hull volume. For the first estimate, I've
decided to split the diffence and use 2/3rds, or 66% of total PH volume which is 20,000 cu ft. If anyone has a
more accurate number, I'll use it.
Now, regarded torpedo ejection, the two pieces of information provided govern this process.
If the torpedo is accelerated at an average acceleration (a), for a period of time (t), and distance (d),
it will attain a final velocity (v). Since torpedo tube length (d) is given as 25 feet, and final velocity
(v) is 33 ft/sec, "t" and "a" can be calculated. The eject time (t) is calculated as 1.52 seconds, and the
average acceleration (a) is 21.7 ft/sec2. These values also pass the "sanity check" and compare well with my data
for other "vehicles" launched from a submerged sub.
Since the 1.52 second eject time is "in the ballpark", I believe the average acceleration to be valid as well.
Now, four (4) basic factors govern the air pressure used to eject the torpedo. The pressure applied to the piston must:
#1 Overcome the external sea pressure at the torpedo tube muzzle at Periscope Depth (P.D.)
#2 Overcome the sliding friction between the torpedo and the tube.
#3 Overcome the hydrodynamic drag of the 21" diameter torpedo moving thru the water.
#4 Accelerate the torpedo at an average of 21.7 ft/sec2 to achieve at final velocity of 33 ft/sec.
Given P.D. or 14 meters and PH diameter of 4.7 meters, I have assumed the lower torpedo tubes to be at a depth of
12 meters and have a sea pressure of 17.5 psi at the torpedo tube muzzle.
Regarding the sliding friction, I've used a coefficient of friction (u) of approx 0.5; that is, the sliding friction
is 1/2 the weight of the torpedo/piston or approximately 1800 lbs. Given the piston area of 346.36 sq.in., an air
pressure of 5.2 psi is required to overcome the sliding friction of a "greased" torpedo.
I am still working on the calcs for hydrodynamic drag; however, it is low at small velocity and increases with the
square of the velocity.
Given a torpedo mass of approximately 109.75 lbs-m, the force required to achieve and acceleration of 21.7 ft/sec2
is 2382 lbs, Again, given a piston area of 346.36 sq.in., the required air pressure is only 6.9 psid across the piston.
Adding the pressures, an average air pressure of 30 psi behind the piston is required to eject the torpedo at P.D.
The full pressure of 24 atmospheres was not used; this would have imparted an acceleration of over 1000 ft/sec2!!
and the torpedo would have left the tube at a velocity well above 33 ft/sec. Unfortunately, I am not an expert
on Uboat systems, but I am assuming that the 24 atmosphere header was used to charge one (or more) eject air flasks
which would then be "dumped" directly into the torpedo tube to eject the torpedo. In another assumption, I'm
estimating an eject flask volume of approx. 8 cu ft to 10 cu ft. By prepressurizing the eject flask(s) to 24 atmospheres,
then "dumping" this volume directly into a torpedo tube, an average torpedo tube pressure of 30 - 50 psi can be achieved,
ovecoming sea pressure, and friction and imparting a 33 ft/sec eject velocity to the torpedo. I suspect that one
(or more) torpedo eject flasks were used to achieve the air volume required; a pressure regualating valve would
likely be too large to maintain the required outlet pressure.
Can any of the Uboat experts confirm that torpedo eject air flasks existed?
More assumptions:
If this is true, my prelim calcs indicate that when fired torpedo tube was vented inboard, the pressure inside
the boat increased by approximately 0.3 psi per tube. Therefore, if four (4) torpedos were fired, boat pressure
increased to 1.2 psi. THe force of this pressure on a 600 mm hatch is over 525 lbs! This certainly seems like
enough to lift the first person thru the hatch!
I will continue to refine my calcs, but any technical input(s) are welcomed!
TomW
