(Solution Manual) Ballistics Theory and Design of Guns and Ammunition

Carlucci,Jacobson, 3

Social Science

Solution Manual

3rd Edition

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Solution Manual for Ballistics Theory and Design of Guns and Ammunition, The downloadable file consists of 549 pages, and the total size of the file is 4.46 MB. Purchase now only for highly discounted price.

Author(s): Carlucci, Jacobson

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Ballistics: The Theory and Design of Ammunition and Guns 3rd Edition Solutions Manual Part 0 Donald E. Carlucci Sidney S. Jacobson 2.1 The Ideal Gas Law Problem 1 - Assume we have a quantity of 10 grams of 11.1% nitrated nitrocellulose (C6H8N2O9) and it is heated to a temperature of 1000K and changes to gas somehow without changing chemical composition. If the process takes place in an expulsion cup with a volume of 10 in3, assuming ideal gas behavior, what will the final pressure be in psi? Answe r  =2inlbf292p Solution: This problem is fairly straight -forward except for the units. We shall write our ideal gas law and let the units fall out directly. The easiest form to start with is equation (IG -4) RTm pg=V (IG-4) Rearranging, we have VRTmpg= Here we go () ( ) ( ) ()( ) ()3ONHC in10K 1000ftin12kJlbfft6.737kgkgmol 2521 K kgmolkJ314.8gkg 10001g10 9286   −              =p  =2inlbf292p You will notice that the units are all screwy – but that’s half the battle when working these problems! Please note that this result is un likely to happen. If the chemical composition were reacted we would have to balance the reaction equation and would have to use Dalton’s law for the partial pressures of the gases as follows. First, assuming no air in the vessel we write the decompositio n reaction. ()sC N CO5OH4 ONHC2 2 9 2 8 6 + + + → Then for each constituent (we ignore solid carbon) we have VTNpi i= So we can write () ( ) ( ) ()  ()      −                      = inft 121 lbfftkJ 6.7371in10gkg 000,11g10kgkgmol 2521K 1000K- kgmolkJ314.8kgmolkgmol 4 3ONHCONHC ONHC ONHCONHC ONHCOH OH92869286 9286 92869286 92862 2p  =2 OHinlbf168,1 2p () ( ) ( ) ()  ()      −                      = inft 121 lbfftkJ 6.7371in10gkg 000,11g10kgkgmol 2521K 1000K- kgmolkJ314.8kgmolkgmol5 3ONHCONHC ONHC ONHCONHC ONHCCO CO92869286 9286 92869286 9286p  =2 COinlbf460,1 p () ( ) ( ) ()  ()      −                      = inft 121 lbfftkJ 6.7371in10gkg 000,11g10kgkgmol 2521K 1000K- kgmolkJ314.8kgmolkgmol 1 3ONHCONHC ONHC ONHCONHC ONHCN N92869286 9286 92869286 92862 2p  =2 Ninlbf292 2p Then the total pressure is 2 2 N CO OH p p pp + + =  = + + =2 2 2 2inlbf920,2 inlbf292 inlbf460,1 inlbf168,1p 2.2 Other Gas Laws Problem 2 - Perform the s ame calculation as in problem 1 but use the Noble -Abel equation of state and assume the covolume to be 32.0 in3/lbm Answer:  =2inlbf2.314p Solution: This problem is again straight -forward except for those pesky units – but we’ve done this before. We start with equation (VW -2) ( ) RTm cb pg= −V (VW -2) Rearranging, we have cbRTmpg −=V Here we go () ( ) ( ) ()( ) ()() () ()          −   −              = lbmin0.32kglbm2.2gkg 10001g10 in10K 1000ftin12kJlbfft6.737kgkgmol 2521 K kgmolkJ314.8gkg 10001g10 3 3ONHC9286p  =2inlbf2.314p So you can see that the real gas behavior is somewhat different than ideal gas behavior at this low pressure – it makes more of a difference at the greater pressures. Agai n please note that this result is unlikely to happen. If the chemical composition were reacted we would have to balance the reaction equation and would again have to use Dalton’s law for the partial pressures of the gases. Again, assuming no air in the v essel we write the decomposition reaction. ()sC N CO5OH4 ONHC2 2 9 2 8 6 + + + → Then for each constituent (again ignoring solid carbon) we have ( )cbTNpi i-V= So we can write () ( ) ( ) ()  ()() () ()      −                −                    = inft 121 lbfftkJ 6.7371 lbmin0.32kglbm2.2gkg 10001g10 in10gkg 000,11g10kgkgmol 2521K 1000K- kgmolkJ314.8kgmolkgmol 4 3 3ONHCONHC ONHC ONHCONHC ONHCOH OH92869286 9286 92869286 92862 2p  =2 OHinlbf257,1 2p () ( ) ( ) ()  ()() () ()      −                −                    = inft 121 lbfftkJ 6.7371 lbmin0.32kglbm2.2gkg 10001g10 in10gkg 000,11g10kgkgmol 2521K 1000K- kgmolkJ314.8kgmolkgmol5 3 3ONHCONHC ONHC ONHCONHC ONHCCO CO92869286 9286 92869286 9286p  =2 COinlbf571,1 p () ( ) ( ) ()  ()() () ()      −                −                    = inft 121 lbfftkJ 6.7371 lbmin0.32kglbm2.2gkg 10001g10 in10gkg 000,11g10kgkgmol 2521K 1000K- kgmolkJ314.8kgmolkgmol 1 3 3ONHCONHC ONHC ONHCONHC ONHCN N92869286 9286 92869286 92862 2p  =2 Ninlbf314 2p Then the total pressure is 2 2 N CO OH p p pp + + =  = + + =2 2 2 2inlbf142,3 inlbf314 inlbf571,1 inlbf257,1p Problem 3 – A hypothetical “air mortar” is to be made out of a tennis ball can using a tennis ball as the proje ctile. The can has a 2 -1/2” inside diameter and is 8” long. If a tennis ball of the same diameter weighs 2 oz. and initially rests against the rear of the can, to what air pressure must one pressurize the can to in order to achieve a 30 ft/s launch of the tennis ball? Assume that the tennis ball can be held against this pressure until released, that it perfectly obturates and also assume an isentropic process and ideal gas behavior with  = 1.4 for air. Solution: First we need to get some parameters i n order. We start with a chamber volume. Since the tennis ball rests against the bottom of the can, the chamber volume is the volume of a cylinder of 2 -1/2” diameter by 1 -1/4” length minus the volume of half a sphere of 2 -1/2” diameter. The empty chambe r has a volume of ()() 3 2 2in136.6 in25.1in5.24U = = () 3 3 3 3 sphere in181.8 in5.26 6V = = = d 2VU Vsphere 0 −= ( )( ) 3 3 3 0 in045.2 in2181.8in 136.6 V = − = There are several ways to solve this problem – the simplest one is to follow the method I described in the notes. We can determine our f ake chamber length through Al=0V Al0V= ( ) ()in417.0 in5.24in 045.2 2 23 = =l We shall assume muzzle exit of the projectile occurs as the equator of the tennis ball passes through the plane of the end of the can. Then the travel o f the projectile is the length of the can less half the diameter of the tennis ball ()() in75.6 in25.2in8 = − =L The formula for muzzle velocity given an isentropic expansion of air was provided in the notes as equation (IG -28) () ()( )() ()   − −− − +−=1 11 12l LlmlRTmV pi g m (IG-28) If we assume ideal gad behavior we recognize that the term mgRTi is the initial pressure times the initial chamber volume thus we can write i g i RTm p =0V Making this substitution we have () ( )( )() ()   − −− − +−=1 11 0 1V2l LlmlpV pi m Rearranging and solving for the pressure allows us to write ()() ( )() ()    − − −− +−=1 1 1 021 V2 l Llm lVpp m i We now have a direct substitution with the data provided and calculated. () () ()( )( )( )() ( ) ( )( )( )( ) ()  −−− +−        = − −− 20.4- 4.11 4.110.4 14.1 322 2 s lbfft lbm2.32 in 417.0 75.6 417.04.11ozlbm 161oz2 in 417.0in 045.22ftin12 sft30 ip  =2inlbf033.6ip 2.4 Thermodynamics Problem 4 - The M898 SADARM projectile weighs 102 .5 lb. The projectile was fired from a 56 caliber, 155mm weapon and a pressure -time trace was obtained. The area under the pressure -time curve was (after converting the time to distance) calculated to be 231,482 psi -m. Calculate the muzzle energy of the projectile in MegaJoules. Assume the bore area to be 29.83 in 2. Answer MJ7.30=E Solution: As given in the problem statement W p 102.5 lb = L 56 0.155  m = thus L8.68m = g 32.2 lb ft  lbf s 2  = and A 29.83 in 2 = If we were not given the average pressure we would now have to calculate it from the pressure -time trace, but since the individual who wrote the problem was such a nice guy and gave us the average pressure we shall use it directly to calculate th e muzzle velocity. We shall call the area under the curve Apx A px 231482 psi m  = The average pressure is then the value of Apx divided by the length of travel PApx L= P 2.667 104 psi = 1() The muzzle velocity is then found from Vm2gPAL Wp= Vm 1.15 103m s= 2() The muzzle energy is calcula ted from E W p 2 g  V m 2  = E 3.072 107 J = 3() To convert these units to Joules we used 1.356 Joules per foot pound force Problem 5 - An 8” Mk. 14 Mod. 2 Navy cannon is used at NSWC Dahlgren, VA for “canister” firings. These firings are used to gun harden electronics which are carried in an 8” projectile. The projectile used weighs 260 lb. The measured muzzle velocity is around 2800 ft/s. Calculate the muzzle energy of the projectile in MegaJoules. Assume the bore area to be 51.53 in 2. The rifled leng th of the tube (distance of projectile travel) is 373.65 in. Answer MJ43E Solution: As given in the problem statement lbm260=pW in65.373=L and   −−=2s lbfft lbm2.32cg 2in53.51=A The m uzzle energy is calculated directly from 2 2m cpVgWE= () ()( )     −−=22 2 2sft2800 s lbfft lbm2.322lbm 260E  ( ) MJ43lbfftJ356.1ft lbf000,652,31   −− =E 2.5 Combustion Problem 6 - Calculate the A -F ratio for the combustion of the following fuels. Calculate the ratio with both theoreti cal air and 10% excess air. a.) Benzene – C6H6 b.) n-Butane – C4H10 c.) Ethyl Alcohol – C2H5OH Answer a.) 13.24 and 14.56, b.) 15.42 and 12.5.96, c.) 8.98 and 9.88 Solution : Benzene C6H6 + 7.50 O 2 + (7.5)(3.76)N 2 →6CO 2 + 3H 2O + 28.20N 2 1 Mole C 6H6 = 78.11 lbm Mass of Air = (7.5)(4.76)(28.97) = 1034.23 lbm A-F = 11.7823. 1034 = 13.24 With 10% Excess Air: A-F = 11.78)1.1)(23. 1034( = 14.56 n – Butane C4H10 + 6.50 O 2 + (6.5)(3.76)N 2 →4CO 2 + 5H2O + 24.44N 2 1 Mole C 4H10 = 58.12 lbm Mass of Air = (6.5)(4.76)(28.97) = 896.33 lbm A-F = 12.5833.896 = 15.42 With 10% Excess Air: A-F = 12.58)1.1)(33.896( = 16.96 Ethyl Alchohol C2H5OH + 3O 2 + (3)(3.76)N 2 →2CO 2 + 3H 2O + 11.28N 2 1 Mole C 2H5OH = 46.07 lbm Mass of Air = (3)(4.76)(28.97) = 413.69 lbm A-F = 07.4669.413 = 8.98 With 10% Excess Air: A-F = 07.46)1.1)(69.413( = 9.88 Problem 7 – Let us examine a pressure vessel identical to the e xample problem in the text containing 0.001 kg of methane (CH 4) and 0.002 kg of air. The enthalpy of formation for the methane is -74,850 kJ/kgmol and its molecular weight is 16.04 kg/kgmol. The reaction will begin at 298 K and we shall remove enough hea t from the vessel that the final temperature becomes 1,000 K. a.) determine much heat is given off. b.) compare the result in a.) above with the example problem in this chapter. Answer a.) kJ386.0−=Q , b.) This situation removes 2.984 kJ more energy than the example. Solution: We need to balance the chemical reaction on a molar basis so we shall determine how many moles of methane and air we have in the container. This is exactly the same procedure as in the chapter example problem. For the methane we have: ( )  ( )  44 4 CH5 CH CH kgmol 1023.6 kgmolkg04.16kg 001.0− =  = N For the air we have ( ) ( ) air5 air air kgmol 1090.6 kgmolkg97.28kg 002.0− =  =N Our balanced reaction is then ( ) ( )() () ( ) ( ) ( )( )()( )25- 5- 25- 25-2 2-5 4-5 N 1045.5sC1023.6 H 1056.9OH 109.2N79.0 O 0.21 10 6.90 CH10 6.23  +  +  + → +  +  We shall examine the reactants first. For the methane we have ( )( )  ( ) ()      −+ −  = − +− → K 298K kgmolkJ314.80kgmolkJ850,74 kgmol 1023.65 CH 2980 CH4 4TR h h Nu T f ( ) kJ82.4 4 4 CH 2980 CH −= − +→ TR h h Nu T f For the oxygen and nitrogen we have ( )()( )  ( ) ()      −+  = − +− → K 298K kgmolkJ314.800 kgmol 1090.621.05 O 2980 O2 2TR h hNu T f ( ) kJ036.0 2 2 O 2980 O −= − +→ TR h hNu T f ( )()( )  ( ) ()      −+  = − +− → K 298K kgmolkJ314.800 kgmol 1090.679.05 N 2980 N2 2TR h hNu T f ( ) kJ13.0 2 2 N 2980 N −= − +→ TR h h Nu T f The enthalpies of the reactants are the same as in the example problem therefore ( )    kJ986.4 kJ13.0 kJ036.0 kJ82.4 −= − − −= −  reacu reac ii TR hN For the products we have (using the tables in the appendix) ( )( )  ( ) ( )      − + −  = − +− → K 000,1K kgmolkJ314.8 993,25 845,241 kgmol 109.25 OH 2980 OH2 2TR h h Nu T f ( ) kJ50.6OH 2980 OH2 2−= − +→ TR h h Nu T f ( )( )  ( ) ( )      − +  = − +− → K 000,1K kgmolkJ314.8 664,200 kgmol 1056.95 OH 2980 H2 2TR h hNu T f ( ) kJ18.1OH 2980 H2 2+= − +→ TR h hNu T f ( )()( )  ( ) ( )      − +  = − +− → K 000,1K kgmolkJ314.8kgmolkJ468,210 kgmol 1090.679.05 N 2980 N2 2TR h hNu T f ( ) kJ72.0 2 2 N 2980 N = − +→ TR h hNu T f ( )( )  ( ) ( )      − +  = − +− → K 000,1K kgmolkJ314.8kgmolkJ795,110 kgmol 1023.65 C 2980 C TR h hNu T f ( ) kJ217.0C 2980 C += − +→ TR h hNu T f The enthalpies of the products are then given by ( )     kJ38.4 kJ217.0 kJ72.0 kJ18.1 kJ50.6 −= + + + −= −  produ prod ii TR hN The heat given off by the reaction is then calculated as ( )( ) kJ606.0 kJ 986.4 kJ38.4 += −− −=Q The example in the text generated a heat output of 3.37 kJ so we have actually remov ed 2.764 kJ more energy by reducing the temperature of the products. Problem 8 – A really interesting person takes the tennis bal l mortar we built in problem 3 and modifies it – squirting in and igniting 0.003 oz. of acetylene gas (C 2H2(g)). If we assume the combustion kinetics are fast enough such that the energy release occurs before the ball can move we want to determine the muzzle velocity of the tennis ball. Proceed along the following steps: a.) Balance the stoichiometric reaction equation for ace tylene. b.) Balance the actual equation neglecting the volume the acetylene occupies in the chamber. Assume the air initially in the chamber is at 14.7 psia and 77 ºF. c.) Determine the increase in internal energy of the gas as we have done in class d.) Assuming the gas is calorically perfect ( U = mgcvT) and that cv = 0.33 BTU/lbm -ºR for the mixture, determine the increase in temperature of the gas. NOTE: you will have to do c.) and d.) by iteration, first assuming a final reaction temperature, c arrying out the calculation for u and seeing if the T you get matches – if not iterate again – once you get an answer within say 10% that is good enough. e.) Based on the result of d.) above, determine the initial pressure on the tennis ball assuming t he specific gas constant of the products is R = 80 ft -lbf/lbm -ºR. f.) Use the result of e.) and possibly your results from problem 2.) to determine the muzzle velocity of the tennis ball. Assume  = 1.4 g.) Determine the temperature of the gases at sh ot exit. For acetylene n = 26.038 lbm/lbmol, h0f = +97,477 BTU/lbmol Solution: We begin by balancing the stoichiometric equation. () ()()2 2 2 2 2 2 2 N4.9OH CO2 N5.276.3 O5.2 HC + + → + +g As in the homework we will first need to determine the amount of air initially in the chamber. The volume of the chamber was determined in problem 2.) as ( )( ) 3 3 3 0 in045.2 in2181.8in 136.6 V = − = The density of the air can be found from the ideal gas equation of state as RTpRT pv = → =  The air weighs 28.97 lbm/lbmol and the density of air is calculated from () ( ) ( ) ()() =     −−    =32 inlbm0000428.0 R 537ftin12R lbmollbfft1545lbmollbm97.28 inlbf7.14  So the amount of air we actually have is ( ) ( ) lbm 0000875.0 in 045.2 inlbm0000428.0 V3 3 0 = = =airm The amount of fuel was given in ounces ( )( ) lbm 0001875.0ozlbm0625.0oz 003.0 = =fuelm For the actual combustion we need to use our mass information from before and convert it to molar values. For the fuel we have ( ) ( )  lbmol 10 201.7 lbmollbm038.261lbm 0001875.06− =  = = fuelfuel fuelnmN ( ) ( )  lbmol 10 020.3 lbmollbm97.281lbm 0000875.06− =  = = airair airnmN for each lbmol of air we know that 1/4.76 lb -moles of it is oxygen so we have ( )    lbmol 10 344.6 lbmol 10 020.376.417 6 O2− − =  =N ( )    lbmol 10 385.2 lbmol 10 020.376.476.36 6 N2− − =  = N Now we can write our actual equation as ( ) ()( ) ( )→  +  + − − 26 27 2 2-6N 10 385.2 O 10 344.6 HC 10 201.7 g ( ) ( )( ) ( )26 26 5 26N 10 385.2 H 10 932.5C 10 440.1OH 10 269.1− − − − +  +  +  Now for the reaction calculation. Since the combustion occurs over a constant volume there is no work done on the projectile so our first law of thermodynamics is provided as U U UP R + = Since we are dealing with internal energies and not enthalpies we need to calculate RuT for the initial and final state of the gases. Unfortunately we do not know the temperature of the products after the reaction takes place let’s assume it is 2000 ºR. Fo r the reactants we have ( ) ()() −=    −−=lbmollbfin980,955,9ftin12R 537R lbmollbfft1545 TRu I used units of inches because everything else we have in the problem statement is in inches. Let’s look at the internal energies for each of the reactants Reactant Enthalpy of formation (BTU/lbmo l) Enthalpy of formation (in -lbf/lbmol) C2H2(g) +97,477 +910,232,428 O2 0 0 N2 0 0 The conversion used here is as follows () ( ) () −→   −   lbmollbfin9.337,9ftin12BTUlbfft16.778lbmolBTUx x For the products we have Product Enthalpy of formation (BTU/ lbmol) Enthalpy of formation (in -lbf/lbmol) H2O (g) -104,040 -971,515,116 N2 0 0 C(s) 0 0 H2 0 0 Since the reactants will be invariant here let’s deal with them first. We calculate UR first ( ) ( ) ( )TRh hNTRh hNTRh h N UufufufR −+ + −+ + −+ =   2 2 22 N O HC Plugging in the numbers we have we get ( ) ( ) ( ) ( ) ( ) ( ) −−+  + −−+  + −−+ +  = −−− lbmollbfin980,955,900 lbmol 10 385.2 lbmollbfin980,955,900 lbmol 10 344.6 lbmollbfin980,955,90 428,232,910 lbmol 10 201.7 676 RU  lbfin453,6 − +=RU At 2000 ºR we have the following values for RuT and the products – we will neglect the effect of the solid carbon other than the effect with RuT. ( ) ( )() −=    −−=lbmollbfin000,080,37ftin12R 2000R lbmollbfft1545 TRu Produc t Enthalpy at temperature (BTU/lbmol) Enthalpy at temperature (in -lbf/lbmol) H2O (g) 13,183 123,102,854 N2 10,804 100,887,752 C(s) 0 0 H2 10,337 96,526,906 We calculate Up in at 2000 ºR as ( ) ( ) ( )( )TRh hNTRh hNTRh hNTRh h N Uufufufufp −+ + −+ + −+ + −+ =    C N H OH2 2 2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) −−+  + −− +  + −− +  + −− + −  = −−−− lbmollbfin000,080,3700 lbmol 10 440.1 lbmollbfin000,080,37 752,887,1000 lbmol 10 385.2 lbmollbfin000,080,37 906,526,960 lbmol 10 932.5 lbmollbfin000,080,37 854,102,123 116,515,971 lbmol 10 269.1 5666 pU  lbfin153,1 − −=pU Then P RU UU − = ( ) ( )   lbfin606,7 lbfin 153,1 lbfin 453,6 − = − −− − =U Comparing this to the expression given in the problem statement Tcm Uvg = ( )() ( ) ()( )R 537 2000ftin12BTUlbfft16.778R lbmBTU33.0 lbm 0001875.0 0000875.0 −   −   + =U  lbfin240,1 − =U Based on this let’s try a final temperature of 4000 ºR. ( ) ( )() −=    −−=lbmollbfin000,160,74ftin12R 000,4R lbmollbfft1545 TRu Product Enthalpy at temperature (BTU/lbmol) Enthalpy at temperature (in -lbf/lbmol) H2O (g) 36,251 338,508,213 N2 27,587 257,604,647 C(s) 0 0 H2 26,071 243,448,391 We ca lculate Up in at 4000 ºR as ( ) ( ) ( )( )TRh hNTRh hNTRh hNTRh h N Uufufufufp −+ + −+ + −+ + −+ =    C N H OH2 2 2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) −−+  + −− +  + −− +  + −− + −  = −−−− lbmollbfin000,160,7400 lbmol 10 440.1 lbmollbfin000,160,74 647,604,2570 lbmol 10 385.2 lbmollbfin000,160,74 391,448,2430 lbmol 10 932.5 lbmollbfin000,160,74 213,508,338 116,515,971 lbmol 10 269.1 5666 pU  lbfin5.523 − −=pU Then P RU UU − = ( ) ( )   lbfin976,6 lbfin5.523 lbfin 453,6 − = − −− − =U Comparing this to the expression given in the problem statement Tcm Uvg = ( )() ( ) ()( )R 537 4000ftin12BTUlbfft16.778R lbmBTU33.0 lbm 0001875.0 0000875.0 −   −   + =U  lbfin934,2 − =U Now let’s try a final temperature of 7000 ºR. ( ) ( )() −=    −−=lbmollbfin000,780,129ftin12R 000,7R lbmollbfft1545 TRu Product Enthalpy at temperature (BTU/lbmol) Enthalpy at temperature (in -lbf/lbmol) H2O (g) 76,146 711,043,733 N2 54,109 491,901,896 C(s) 0 0 H2 52,678 505,264,431 We calculate Up in at 7000 ºR as ( ) ( ) ( )( )TRh hNTRh hNTRh hNTRh h N Uufufufufp −+ + −+ + −+ + −+ =    C N H OH2 2 2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) −−+  + −− +  + −− +  + −− + −  = −−−− lbmollbfin000,780,12900 lbmol 10 440.1 lbmollbfin000,780,129 896,901,4910 lbmol 10 385.2 lbmollbfin000,780,129 431,264,5050 lbmol 10 932.5 lbmollbfin000,780,129 733,043,711 116,515,971 lbmol 10 269.1 5666 pU  lbfin6.679 − =pU Then P RU UU − = ( ) ( )   lbfin773,5 lbfin6.679 lbfin 453,6 − = − − − =U Comparing this to the expression given i n the problem statement Tcm Uvg = ( )() ( ) ()( )R 537 7000ftin12BTUlbfft16.778R lbmBTU33.0 lbm 0001875.0 0000875.0 −   −   + =U  lbfin477,5 − =U This answer is close to 5% so we’re good to go. The temperature is now 7,000 ºR – pretty hot! – that’s why acetylene works so good as a cutting torch fuel. Now the initial pressure on the tennis ball comes through the ideal gas equation of state. RT pi= ( ) ( )() ( )()    −− +=ftin12R 000,7R lbmlbfft80 in 045.2lbm 0001875.0 0000875.0 3 ip  =2inlbf5.903ip The muzzle velocity of the tennis ball is then () ( )( )() ()   − −− − +−=1 11 0 1V2l LlmlpV pi m ()( ) ( )( )( ) ()() ( )( )( )( )( ) ()  −−− + −       =− −− 20.4- 4.11 4.110.4 14.1 3 2 s lbfft lbm2.32 in 417.0 75.6 417.0 4.11ozlbm 161oz2ftin12in 417.0in 045.2 inlbf5.9032 mV  =sft7.64mV For the temperature of the gases when muzzle exit occurs we can use equation (IG -19) ()1 VV−   = c iTT (IG-19) Assuming muzzle exit occurs as we have stated earlier we can write 2VU Vsphere can e − = ()() 3 2 2 2 can in27.39 in8in5.24 4U = = = Ld then we have ( )( )33 3 e in179.352in 8.181in27.39 V = − = Then the temperature at muzzle exit is given by ( )( ) ( )( )14.1 33 in 35.179in 2.045R 000,7−    =T  F 783,1 R 243,2 = =T Problem 9 – A potato is stuffed into the 3” diameter exhaust pipe of a car that is not running t oo well. 1 gram of incompletely combusted combustion products (assume gaseous Heptane) mixes with a stoichiometric amount of air behind the potato and ignites. If the potato is wedged 4” into the exhaust (i.e. it has 4” of travel) and weighs 0.25 lbm, an d assuming the combustion takes place before the potato moves, determine the theoretical maximum “muzzle” velocity of the potato. Also calculate the muzzle velocity assuming an isentropic expansion. For gas expansion purposes you can assume the volume av ailable initially behind the potato is equal to the volume between the point of obturation and the end of the exhaust and assume a “smeared” specific heat ratio of 1.3 for the product gases. Also assume the combustion begins at 500 K and completes at 1,500 K. Assume the total enthalpy at 500 K for n-Heptane (C 7H16) is -120,000 kJ/kgmol. Solution: We know that the combustion is occurring under stoichiometric conditions so we can write the balance reaction equation as follows: () ()()2 2 2 2 2 16 7 N36.41OH8 CO7 N1176.3 O11 HC + + → + +g This is on a one mole of fuel basis. We need to write it for what we actually have which is 1 gram of fuel. First we determine the molecular weight of the Heptane to be 100.2 kg/kgmol. Therefore the number of moles of Heptane we have is () ( )  kgmol 101 kgmolkg2.100gkg 10001g1 5 HC167−=      = N So the balanced equation becomes ( ) ()( ) ( ) ( ) ( ) ( )2-5 2-5 2-5 2-5 2-5 16 7-5N 1036.14 OH 108 CO 107 N 1036.14 O 1011 HC 101  + +  →  +  +  g Now let’s deal with the thermochemistry. We start with our second law of thermodynamics equation, simplified by the fact that there is no heat transfer and no shaft work. Th en the energy of the fuel -air mixture equals the work done on the projectile plus the energy of the products of combustion plus the work done on the gas. (Note the difference here than in the text where we lumped the work done on the gas with the losses) P P R W H H + = Let’s look at the internal energies for each of the reactants Reactant Total Enthalpy (kJ/kgmol) C7H16(g) -120,000 Enthalpy of formation (kJ/kgmol) Δh(500) (kJ/kgmol) O2 0 6,097 N2 0 5,920 For the products we have Product Enthalpy of formation (kJ/kgmol) Δh(1,500) (kJ/kgmol) H2O (g) -241,845 48,121 CO 2 (g) -393,546 61,681 N2 0 38,404 We will rearrange our second law equation as follows p R p H H W − = We calculate HR first ( ) ( ) ( )h hNh hNh h N H f f fR + ++ ++ =   2 2 167 N O HC Plugging in the numbers we have we get ( ) ( ) ( ) ( ) ( ) ( ) +  + +  + − = −− − kgmolkJ920,50 kgmol 1036.41 kgmolkJ097,60 kgmol 1011kgmolkJ000,120 kgmol 101 55 5 RH kJ919.1=RH We calculate Hp in a similar manner ( ) ( ) ( )h hNh h Nh h N H f f fp + ++ ++ =   2 2 2 N CO OH ( ) ( ) ( ) ( ) ( ) ( ) +  + + − + + − = −− − kgmolkJ404,380 kgmol 1036.41 kgmolkJ681,61 546,393 kgmol 107kgmolkJ121,48 845,241 kgmol 108 55 5 pH kJ8.22−=pH Then the work done on the projectile is ( ) kJ8.24 kJ8.22 kJ919.1 = −− =pW Now we need to recall that this work equals the muzzle energy of the projectile kJ8.24212= =mV Wp Therefore the theoretical maximum velocity for this reaction is ()() ()()( ) () =      =    = =sm661 lbmkg 2.21lbm25.0smkg800,242 lbmkg 2.21lbm25.0kJ8.242 222 mWVp This is over twice the speed of sound! To get a more realistic answer let’s look at the ideal gas equation of state. If the projectile doesn’t move, the volume behind it is ()()( ) 3 4 33 3 22 m 106.4 inm0254.0in4in43V− = = =  Ali The volume of the exhau st pipe when the potato is at the end is twice this value. The initial pressure is calculated from the ideal gas equation of state then using this volume VVTNRp TNR pu u = → = ( ) ( ) ( ) ( )kPa280,15 m 106.4K 500,1K kgmolkJ314.8 kgmol 1036.56 3 45 =    =−− p If we use the isentropic formulation for muzzle ve locity we can start from equation (LG - 180) () () ( ) −     ++= −   +x x cc c cdxlxlxpA xVxVcw 2 2 13 21 (LG-180) In our case we can safely assume the gas is massless and the initial velocity is zero so we have () () − −− + =   +=l cl c dxlx lpAdxllxpAl mV 0 02221   Integrating we have () ()()l cl c lx lpAdxlx lpAl mV 01 02 1221     −+= + =− −− −   Insert ing the limits of integration we obtain () ()     −−=    −−=− − − −    12 12 211 1 1 2 llpAl l lpAmVcc Rearranging we have   −−=−  12 21ll mpAVc Inserting our numbers we have ()()( ) ( ) ()()()in3.1144 2 lbmkg 2.21lbm25.0msmkg000,700,16 inm0254.04in323.11 2 2 33 32 2     −−         =− V  =sm601V Still way high but we neglected friction and LOTS of l osses. Problem 10 – Assume we have a quantity of 29 pounds of 11.1% nitrated nitrocellulose (C6H8N2O9) and it is placed in an empty chamber of a gun at 77°F and 14.7 psia. The chamber is 1160 cubic inches in volume. The propellant density is 0.060 lbm/i n3. If the air in the chamber is NOT neglected and assuming the volume is fixed: a.) Write the balanced equation for this combustion (assume the oxygen goes preferentially into CO 2 instead CO this time – you will find the difference later) b.) Using the tables in the textbook, estimate the adiabatic flame temperature of the resultant gas (Hint: recall the definition of adiabatic flame temperature) Solution: First we need to convert the units for the amount of propellant and air into moles. For the propellant it is exactly as found in the notes 0.115 lbmol. For the air in the chamber we have to play with a few equations. First, the volume the propellant occupies in the chamber is m p=V (1) () ( )3 3in483 inlbm060.0lbm29V =  =p (2) The volu me the air occupies is then p c air V-V V= (3)   3 3 3in677 in483 in 1160 V = − =air (4) The number of moles of air then can be found from the ideal gas equation of state as follows airuair airTRpNV= (5) Noting that the air temp erature must be in Rankine we have () () ( ) ()()R 537ftin12R lbmollbfft545,1in 677 inlbf14.73 2       =airN (6)   lbmol001.0=airN (7) Balancing the chemical equation we have ( ) ( ) ( ) ()( ) ()( )( )   ( )( )   ()( )( )  ()sC 00021.0 0.1155.3 N 00079.0 115.0 CO 00021.0 0.1155.3 OH 0.1154 N001.04.7676.3O4.761001.0 ONHC 0.115 22 2 2 2 9 2 8 6 − + + ++ + →   +  + (8) or ( ) ( ) ( ) ( ) ( ) ( ) ( )()sC400.0 N116.0 CO403.0 OH 0.460 N 00079.0 O 00021.0 ONHC 0.1152 2 2 2 2 9 2 8 6 + + + → + + (9) So this is our balanced react ion. To solve for the adiabatic flame temperature we need to write the first law of thermodynamics as in equation (CT -42) W H H QP R VC + = +.. (CT-42) The problem statement stated that the volume was fixed so W = 0 and there is no flow work. Since we are looking for the adiabatic flame temperature we assume that QC.V. = 0 also. Then we see that P RU U= (10) Looking at the notes we see that the enthalpy of formation for the propellant was given as 2,605.5 Btu/lbmol. The tables in the book are in kJ/kgmol so we have some unit conversions to do. ( ) ( ) () =     =kgmolkJ046,6kgmollbmol2.2BtukJ055.1lbmolBtu5.605,2 9286 ONHC0 fh (11) We can change this over to internal energy (because there was no flow work by using the definition of enthalpy, thus ( ) ( ) () =   − =kgmolkJ568,3 K 298K kgmolkJ314.8kgmolkJ046,6RU