| Basic Thermodynamics ~ J. Pohl © | www.THERMOspokenhere.com (204-E232) |
On strafing runs WWII P-51 Mustangs often fired bursts of 1000 rounds from its six - 50 caliber machine guns. Those streams of "fired" lead changed the air speed of the Mustang. The sketch includes data of one such event.
Calculate the P-51's air speed after it fires a burst of 1000 rounds.
♦ We take the system to be the mass of the P-51 and the 1000 rounds of 50 cal bullets. The mass of such fighters was 5400 pounds and its speed 300 miles per hour (440 ft/s) when firing commenced. Each bullet (mass = 0.1 lbm) left its machine gun muzzle with a speed of about 3000 feet per second. A sketch of before and after is helpful.
We take the airplane and the bullets it carries (then fires in the event) as our system. Since there is "plane" momentum and "bullets" momentum, we place a summation, left of equality, on momentum.
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(1)
System consists of all of the remaining ammunition and the fighter. |
Initial Condition: An event is a change from some initial condition to a final condition. Initially the plane and bullets move horizontally at same constant speed. The state is called uniform motion. Propeller thrust causes the velocity, the drag (the opposing force of air as the plane moves through it) equals the thrust force. So velocity and momentum are constant for all times prior to firing (t < 0+). Stated by equation, this is:
| (2) Momentum is constant and the force of the prop equals that of aerodynamic drag. |
The above equation tells that prior to firing the system momentum is zero in two ways. Left-of equality says the bullets and aircraft speeds are constant. Right-of-equality says "thrust equals drag."
Trigger Time: The six guns are triggered in a burst. To use the momentum equation we write it in general and consider it at the time: t = 0+
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(3) System mass remains constant as bullets speed ahead of the fighter. |
All Forces of the event are notated. You might wonder about the forces as the guns recoil. But the recoil effects are not forces since they do not occur at the system boundary. The bullets. powder and aircraft are all part of the system. Forces of the momentum equation are effects of the surroundings. To proceed toward solution of the first order differential equation, we separate variables then apply the integration operator:
| (4) Same procedure as always. Separate variables and integrate. Let the math take the lead. |
We like to do easy parts first. The integrand, right of equality (Fthrust - Fpropeller) was zero before change started, when time was t < 0+. We suppose the firing happens very fast so that either "dt" is zero or (Fthrust - Fpropeller) remains what it was at the start ~ zero. Use either approximation to set "right of equality" equal to zero for the event. Left of equality integrates immediately.
| (5) Integration of an exact differential yields the difference of its quantities. |
The summation is required - our system has two components. Now we expand the sum.
 | (6) There are two components to our system. |
The direction of all vectors is to the right, so we scalar multiply or "dot" multiply ("• I") the equation then put numbers into it.
 | (7) As always, care with units is required. |
This amounts to a worrisome, 14% decrease in speed. We suppose 14% is the least reduction. In events of "squandered energy," approximations tend to be "rosey."