mathematical question related to Centrifugal Pumps question?
818883
Es = L*r*w (w=2piN) , N=600 rpm=10rps , L=9 \nEs = L*r*2pi*N) , N=10, L=9 \nEs = 9*0.18*2*3.14*10\nEs (Shaft Energy)= 101.734 Newton/second which is not a unit of power.\n\nPlease specify the units of r,Es,L=95 or 9 ?\nWhat is the unit of Ee? HP or KW?\nfurther you have given following data which are contradictory.\nPump Shaft Speed =1200 ?\nEe (Electrical Energy) = 67.2\nT = 1.62\nw (Rotational Speed)= 62.8
805064
lol \nYou have homework that you don't know the answer to.  \n\nWhy would someone ask these here out of interest? Like you have these specific parameters and conditions that you are just raring to want to know just out of the blue.\n\n Quit wasting time writing and typing questions out waiting here for answers when you could be the next engineering graduate who actuallly UNDERSTANDS this stuff by doing it yourself.
375635
Kp=Kc(RT)n\na.Kp=Kc(RT)-1/2\nb.Kp=KcRT\nc.Kp=KcRT\nd.Kp=Kc(RT)2
816469
well F=mv/t given: m=78kg, v=24m/s, t=60s\n\nF=78(24)/60=31.2N
944167
PV=nRT\nV = nRT/P\nV = (114g/44g/mol)(0.08206L-atm/mol-K)(28+273K)/0.987atm\nV = 64.84L or 0.0648 m^3
512055
2,800\n\nYour original solution is one tenth of a mL, so multiply that by ten to get the end volume.  Do the same thing with the number of particles: 280x10=2,800.
903092
1) b)\n\n2) [H2O]^2 / 2[H2] [O2]\n  \n   [NO]^2 / [N2] [O2]\n\n3) b)\n\n4) b)
829328
Water pumps can  be reciprocal , geared or centrifugal.  A reciprocal pump works like an automobile cylinder.  water in the cylinder is forced out when the piston rises.  A valve closes to hold the water there.  On the downstroke, a valve opens to admit more water to the cylinder.  On the upstroke that valve closes, and the discharge valve opens.\n\nIn a geared pump, two meshing gears rotate in a very tight enclosure.  Water is admitted on one side.  The spaces between teeth carry water around the case to the discharge side.\n\nCentrifugal pumps spin water with a paddlewheel-like arrangement.  This gives the water energy in the form of velocity.  When this water is forced into a pipe, the energy converts from velocity to pressure.
514815
OK, the barrel´s mass is 40Kg (notice: its mass is 40 Kg, not its weight). \n\na) So, your formula is Weight = mass(m)*gravity´s acceleration(g)= 40Kg*10m/s^2 = 400 KG*m/s^2 (=400 Newton)\n\nb) Work = force * distance\nSince work is calculated only for the vertical, and the force equals the weight, you have:\nWork = 400N*5m = 2000 Joule\n\nc) Eff = 250N*5m/2000J*100 = 0.625*100 = 62.5%\n(You got 12.5% because you forgot to multiply 250N for 5 meters. Remember that efficiency is a relation of works, and work is force * DISTANCE).\n;o)
883912
Let the banking angle be '$'.\nThe reaction force will be perpendicular to the banking surface.\nLet the reaction force be 'R'.\nLet 'm' be the mass of the car.\nvelocity of car (v)=27 mt/s\nradius(r)=60 mt.\n\nThe centrifugal force will be in horizontal direction. And this force will be balanced by a component of the reaction force 'R'.\nR sin$= mv^2/r ---------------------------(1) \n\nThe weight of the car will be vertically downward. This weight will be balanced by Rcos$.\nRcos$=mg  ----------------(2)\n\nNow dividing eqn(1) by eqn(2)\n\nRsin$/Rcos$=mv^2/r /mg\n=>tan$=v^2/rg = 27 x 27/60 x 9.8 = 729/588 = 1.239\n\n$= tan(inverse)(1.239)\n\nsolve above tan(inverse)  function. and find the value of the banking angle
798267
This is a Hess' Law problem.\n\nThe strategy I like to teach for these is to get the biggest thing you need first then get rid of the biggest thing you don't need next.  So we want to get the 2 N2O5 first then get rid of the resulting 4 HNO3.\n\nUse -2 times the first eq to get your 2 N2O5, -4 times the third to get rid of your 4 HNO3, finally -2 times the second to get rid of the water:\n\n2 N2O5 + 2H2O > 4 HNO3   -152.4 kJ\n4 HNO3              >  2 N2 + 6 O2 + 2 H2  +696.0 kJ\nO2 + 2H2           >   2 H2O  -572 kJ\n\nEverything cancels properly to get the desired equation, and delta H is -28.4 kJ.
558084
Start with the ideal gas law: PV=nRT.  Since no gas enters or leaves the system, n is constant, and R is the universal gas constant, the product nR has the same value at the beginning and as at the end.  So, collect it to one side, and move the other factors to the other side.\n\nP1 * V1 / T1 = nR = P2 * V2 / T2\n\nSolve for T2\n\nT2 = T1 * (P2 * V2) / (P1 * V1)\n= 300e4 K (.0003e15 atm * .046e-15 cc)/(50e3 atm * .0004e-7 cc)\n=2.07 * 10^7 K
806029
1st link \n\nb) F = 40 N\n\nc) F = 60 N\n\n2nd link\n\nb) linear velocity = 0.59 m/s\n\nc) Taking g = 9.8 m/s^2\n\nv = 0.83 m/s\n\nd) It would not effect the answer as the expression for v which is sqrt(mu * r * g) is independent of the mass. \n\n3rd link\n\na) Rotate the setup, note the reading on the balance. Find the time required for 1 circuit.\n\nb) No work is done\n\nc) T = 5.48 N\n\nd) 5.9 %\n\ne) If the cord is exactly horizontal, it won't touch the rod and hence cannot be rotated.
347099
I have used the Bernoulli equation to calculate pressure drop through piping and to size pumps (determine required head pressure).  You can't get much more real than that.
601468
The answer is (d) if this is the question:\n\nInitially the tank is full.  The outlet pipe is opened.  When the water in the tank is 3/4th full, both inlet pipes are opened.  Calculate the fractional level of the water in the tank after 2:05 hr.\n\nSince inlet A and outlet C cancel each other out at 50L/hr, only need to consider inlet B at 25L/hr.  Since 2:05 hr is the same as 2.083 hr (5 min = 0.083 hr), the amount of water added by inlet B is 25L/hr * 2.083 hr = 52.08 L.  Will use 52 L in the calculation.\nThe total amount of water in the tank after 2:05 hr is the initial 750 L + 52 L = 802 L.\nThe fraction level of the water is 802 L / 1000 L = 0.802\nSince 4/5 = 0.8 the answer is (d).
714549
a)\nConsider the energy of the sled:\nIt loses potential energy of m*g*R*(1-cos(a))\na is the angle that has opened from vertical\n\nThe kinetic energy is\n1/2*m*v^2\nso v=sqrt(2*g*R*(1-cos(a)))\n\nfor b\nthe normal force is\nm*g*cos(a)\nthe centrifugal force is\nm*v^2/R\nv=sqrt(R*g*cos(a))\n\nfor c\nusing the maximum velocity for departure as\nv^2=R*g*cos(a)\nplug into\nv^2=2*g*R*(1-cos(a))\nR*g*cos(a)=2*g*R*(1-cos(a))\ncos(a)=2(1-cos(a))\ncos(a)=2/3\nabout 48 degrees\n\n\nj
822266
Conservation of energy:  KE=PE\n1/2*Mb*Vb^2= (Mb+Mw)*g*h\nVb^2=2*(Mb+Mw)*g*h/Mb\nVb^2=2*(0.01+2)*10*0.1/0.01\nVb^2=402\nVb=20.05 m/s