THERMODYNAMICS... piston-cylinder problem... PLEASE help?
410173
This formula might help\n\nchange in S=(m1)(c1)(ln(Tf/Tc))+(m2)(c2)(ln(Tf/Tc))\nm1 & m2=substance mass in this case oxygyn gass (O2)=15.999(2)\nc1 & c2 are spcific heats\nTc =cold temp\nTh = Hot Temp\nTf=((m1)(c1)(Tc)+(m2)(c2)(Th))/((m1)(c1)+(m2)(c2))\n\nI'm assuming m^3/kg is the pressure (Volume/Mass) ???\n\nhope this helps!!
908729
(a) ω = ωo + αt\n0 = 150 + α * 132\nα = -150/132\n≈ -1.136 rev/s²\n\n(b) θ = ωot + ½αt²\n= 150 *132 - ½*1.136*132²\n≈ 9903 revolutions (nearest rev)\n\n(c) See (b)\n\n(d) 75 rpm = 75*2π rad/min\nα = rω²\n= 50 *(75*2π)²\n≈ 11103300 cm/min²\n= 111033 m/min²\n= 111033/3600 m/s²\n≈ 30.84 m/s²\n\n(e) ???? I do not understand what it is you are seeking.
485796
Follow the steps with reference to the webpage ...\n\nStep 1 Characteristics of air\n\nLet's fit a rubber to a can like Figure 1. You can understand easily that the rubber expands when the can is heated (Figure 2), and the rubber contracts when the can is cooled (Figure 3). It is caused that a pressure of the air in the can works to the rubber when the air is heated, shown in allows of Figure 2. Of course, you cannot see the pressure by your eyes.\n\n \nStep 2 What is a displacer piston?\n\nNext, let's put in a piston into the can like Figure 4. A diameter of the piston must be somewhat smaller than that of the can, because the piston works to move (displace) the air up and down in the can. And, please heat the bottom side of the can and cool the upper side of the can. After it has enough temperature difference, move the piston up and down by your hand. When the piston is moved up, the rubber expands because there is a lot of hot air in the can (Figure 5). It corresponds to Figure 2. When the piston is moved down, the rubber contracts because there is a lot of cold air in the can. It corresponds to Figure 3.\nIn the case of the Stirling engine, this piston, which moves (displaces) the air and make the pressure changes is called a displacer piston.\n\n \nStep 3 Crank mechanism ~ from piston motion to rotation ~\n\nDo you understand about the characteristics of the air and the work of the displacer piston? They are very important to understand how does the Stirling engine work.\nFirst, connect from the piston and a bent wire with a thread like Figure 6. When the bent wire is rotated, the piston is moved up and down. This is called a crank mechanism.\nPlease heat the bottom side of the can and cool the upper side of the can, similar to above description. When your hand rotates the bent wire, the piston is moved up and down, and the rubber expands and contracts repeatedly (Figure 7).\n\n \nStep 4 Power piston ~ function of the rubber ~\n\nThe Stirling engine converts from the motion of the rubber to the rotation of the bent wire. Please connect from the rubber to the bent wire with a rod. In this time, a force of the rubber (expansion and contraction) has to be the direction, which rotates the bent wire. In short, you must bend the bent wire the just right angle (90 degrees) from the piston like Figure 8 and 9.\n\n \nStep 5 Flywheel ~ for the smooth rotation ~\n\nThis engine has not worked yet. If you try to work this engine (Figure 8 and 9), the rubber keeps the most expansion or contraction. In order to rotate smoothly and repeatedly, you must fit a rotated mass with the bent wire. It is called a flywheel.\nGenerally, the flywheel is circular like Figure 10. But at this point, please bend the end of the bent wire, and fit a mass at the edge of the wire like Figure 11. The mass works as the flywheel and to be balanced to the weight of the piston. So, you must fit the mass against the piston.\n\nThis page of diagrams courtesy of Koichi Hirata.\n\nPlease click the link for the diagrams ...\nhttp://www.hotairengines.com/WorkingDescription/Work.htm\n\nWith Love\nsridhar
317109
Take table for R134a refrigerant:\n\na) delta s = entropy change\n--inlet for evaporator assuming 50%moisture= interpolate to get S \n@200kpa for R134a\n0       Sg   \n0.5     S\n1       Sf\nget Sg and Sf from table at 200kpa for R134a. \n \n--out for evaporator = Sg at 200Kpa (get from table) since it is sat vap.\n\nthe difference of both Sfg and Sg is the answer.\n\nb)  delta S = Q / T  where Q = 120 KJ and T = -5+273 = 268K\n\nc) add both a) and b) is the total entropy
728936
1. obtain formulas for surface area and volume of cylinder\n2. let volume equal 1000 and solve this volume equation for h\n3. substitute the expression from previous step into surface area    equation to get equation with only r's\n4. differentiate surface area formula with respect to r set this equal to 0 and solve for r i got   r=10/3rd root(pi)\n5. substitute value for r into volume equation solve for h  i got h=10/(pi)^(2/3)\n\nincidentally, i used 2nd derivative test to conclude r is minimum\n\n\n\ni'll get back to u with more
613974
You need to use the combined gas law:\nP1V1/T1 = P2V2/T2, but your temps need to be in Kelvins, not Celsius.\n\nIt also seems that the volume is being held constant, so we can cancel volume out of the problem.\n\nSo, convertine the temperature first, we get:\n10C + 273 = 383K for our temp1\n\nSo, subbing into the formula we get:\n100mmHg/283K = 900mmHg/x\n\nSolve for x and get your answer.  The temperature will be in Kelvins.
231576
That means either the piston rings or cylinders is worn or the valve guide seals are bad. Since it is a 99 I would be looking at the piston rings or cylinders. This basically intails rebuilding the engine which will be costly. You may be better off adding oil treatment and 40W oil.
957415
Use PV = n RT, RT = constant. Therefore n = PV * constant in the three cases.\nWe have \nH2 :  P = 372 torr V = 2 liter  n_H2=  744 * constant\nN2:  P = p torr V = 1 liter  n_N2 = p * constant\nCombined: P = 303 torr V=3 liter  n_combined = 909 * constant\n\nTherefore  744 + p = 909 => p = 165 torr
64647
2x (G2 + M2 ---> 2GM)      DH=-280.0*2= -560.0 kJ\n2x (L + 2M2 ---> LM4)      DH=-350.0*2= -700.0 kJ\n   (L2G4 ---> 2L + 2 G2)   DH=122.0*(-1)=-122.0 kJ (change sign when flip over)\n\nAdd them together, we'll have:\n2G2 + 2L + 6M2 + L2G4 ---> 4GM + 2LM4 + 2L + 2G2\nthen cancel 2G2 and 2L on both side, we end up with:\nL2G4 + 6M2 ---> 2LM4 + 4GM\nthe enthalpy of this reaction is, therefore the sum of each enthalpies calculated above.\nDH=(-560.0) + (-700.0) + (-122.0)= -1382.0 kJ (exothermic process)\nhope that'll help!
617970
I am terribly afraid of the equilibrium problem.. the first one on the free response!!! well, good luck!
1010084
Propane  = C3H8. The molar mass is\n(3 x 12 + 8 x 1) gram = 44 gram.\nYou give 44.1 g. That a little bit better.\n\nThe volume of the cylinder is\nAh = pi r2 h = 3.14 x (0.050 m)^2 x 1.30 = 0.212 m3.\n\nBefore: pV = nRT\n1.4×10^6 Pa x 0.212 m3 = n x 8.31 J/(mol x K) x (273 + 23)K\nSo n(begin) = 119 mol.\n\nAfter: pV = nRT\n2.4×10^5 Pa x 0.212 m3 = n x 8.31 J/(mol x K) x (273 + 23)K\nSo n(end) = 20 mol.\n\nUsed is (119 - 20) mol propane, being 99 x 44.1 gram = 4.4 kg
934101
what's the height?  you need the height to finish this problem\nwhat type of shape is it?  a cylinder? cone?\nbe specific or no one can help you
329557
Try this....\n\nhttp://www.efunda.com/formulae/solid_mechanics/mat_mechanics/pressure_vessel.cfm
918429
use the formulas your given\nto find the volume of the cylinder\nthen pretend the semi hemi on  the top is a sphere find the volume and multiply by 1/2\n\nadd two totals to find volume
1121436
let's see. The car in on large piston, right?\n\nso the pressure increases by 90/250 units.\n\nthe additional weight needed to compensate from from the small piston side is (90/250)*10=90/25 kg\n\nthis weight must be provided by a weight of oil going up in the small piston, it's volume being 90/25/560 m3, \n\nso the large piston must go down to provide this volume by\n90/25/560/.025=0.2571 m  (slightly under a foot)\n(m2 = 10,000cm2, so 250cm2=0.025m2)\n\nSo I guess 0.2571 m  is the answer. \nDo not remember any of this physics, but this seems to be the only way to link the numbers.\nLemme know if it is right :)
174163
thermodynamics?  heat transfer?  physics?
651714
Homework?\nOK, only a hint. Consider a cylinder made of the circle and with a height dh. What is the volume of this cylinder? \n\nNow displace the cylinder along the path over dh, etc.