course Phy 242 Please see my note about the problem in assignment 8. ՍSؾR|崆assignment #006
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15:52:11 query introset change in pressure from diameter change given original vel and diameter
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RESPONSE --> We are given r1, r2, P1, v1, and asked to find the p2. Using Bernoulli's Equation we know that for P1 to remain constant P2 must decrease to compensate for the increase in velocity. We know that the velocity must increase because the diameter fo the hole has become smaller and since water is not compressible the velocity must increase. First, we find the area proportion: area proportion = 'pi * (r1/r2)^2 v2 = v1 * area proportion Then we take the v2 and plug it into: 0.5 * 'rho * v^2 = P Now we have P2 - P1.
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15:52:21 ** The ratio of velocities is the inverse ratio of cross-sectional areas. Cross-sectional area is proportional to square of diameter. So velocity is inversely proportional to cross-sectional area: v2 / v1 = (A1 / A2) = (d1 / d2)^2 so v2 = (d1/d2)^2 * v1. Since h presumably remains constant we have P1 + .5 rho v1^2 = P2 + .5 rho v2^2 so (P2 - P1) = 0.5 *rho (v1^2 - v2^2) . **
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RESPONSE --> Understood.
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15:56:15 query video experiment 4 terminal velocity of sphere in fluid. What is the evidence from this experiment that the drag force increases with velocity?
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RESPONSE --> As the velocity increases it takes more & more weight to cause it to be pulled up. This shows that for the velocity to increase a larger force has to be acting on the sphere in the fluid.
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15:56:31 ** When weights were repetitively added the velocity of the sphere repetively increased. As the velocities started to aproach 0.1254 m/sec the added weights had less and less effect on increasing the velocity. We conclude that as the velocity increased so did the drag force of the water. **
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RESPONSE --> Understood.
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16:14:05 query univ phy problem 14.85 (14.89 10th edition) half-area constriction then open to outflow at dist h1 below reservoir level, tube from lower reservoir into constricted area, same fluid in both. Find ht h2 to which fluid in lower tube rises.
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RESPONSE --> First we take Bernoulli's Equation. P1 = 0.5*'rho*v1^2 = P2 + 0.5*'rho*v2^2 Then substituting the conitnuity equation for v2 we get: P1-P2 = 0.5*'rho*v1^2[(A1^2/A2^2)-1] We know that the pressure difference P1-P2 = 'rho*g*h. So when we substitute we get: 'rho*g*h = 0.5*'rho*v1^2[(A1^2/A2^2)-1] When we solve this for h we get: h = {0.5*'rho*v1^2[(A1^2/A2^2)-1]}/('rho * g) This h is the difference in the heights so now all we have to do is substract the h from h1 and we will get h2. h1 - h = h2
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16:15:02 ** The fluid exits the narrowed part of the tube at atmospheric pressure. The widened part at the end of the tube is irrelevant--it won't be filled with fluid. So Bernoulli's Equation will tell you that the fluid velocity in this part is vExit such that .5 rho vExit^2 = rho g h1. Now compare three points: point 1, in the narrowed tube; point 2 at the top of the fluid in the lower tube; and point 3 at the level of the fluid surface in the lower container. At point 1 the pressure is atmospheric and velocity is vExit. Pressure is atmospheric because there is no pressure loss within the tube, since friction and viscosity are both assumed negligible. At point 2 fluid velocity is zero and since there is no fluid between the narrowed tube and this point there is no net rho g h contribution to Bernoulli's equation. So we have .5 rho v1^2 + P1 = .5 rho v2^2 + P2. P1 is atmospheric and v1 is vExit from above, while v2 = 0 so P2 = atmospheric pressure + .5 rho vExit^2 = atmospheric pressure + rho g h1. Now comparing point 2 with point 3 we see that there is a difference h in the fluid altitude, with velocity 0 at both points and atmospheric pressure at point 3. Thus P2 + rho g h2 = P3 + rho g h3, or (atmospheric pressure + rho g h1) + rho g h2 = atmospheric pressure + rho g h3. Thus rho g (h3 - h2) = rho g h1 and h3 - h2, which is the height of the fluid in the lower tube, is just equal to h1. If we assume that somehow the fluid manages to expand on escaping the narrowed tube so that it fills the once-again-widened tube, and exits with vExit as above, then the velocity in the narrowed tube will be 2 * vExit. This leads to the conclusion that pressure change from small to large tube is .5 rho (2 vExit)^2 - .5 rho vExit^2 = .5 rho (3 vExit^2). Since pressure is atmospheric in the large tube, pressure in the small tube is atmospheric pressure + 3 ( .5 rho vExit^2). If we use this pressure for point 1 and follow the steps given above we conclude that h3 - h2, the height of the fluid column in the lower tube, is 3 h1. This is the book's answer. Again I don't have the problem in front of me and I might have missed something, but the idea of the fluid expanding to refill the larger pipe doesn't seem consistent with the behavior of even ideal liquids, which are pretty much incompressible at ordinary pressures. However note that I am sometimes wrong when I disagree with the textbook's solution. **
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RESPONSE --> I believe that I made this problem much more complex and in doing so I may have got it wrong. I will make sure to print this out and study this answer.
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16:15:33 query univ phy problem 14.85 (14.89 10th edition) half-area constriction then open to outflow at dist h1 below reservoir level, tube from lower reservoir into constricted area, same fluid in both. Find ht h2 to which fluid in lower tube rises.
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RESPONSE --> This is the same question as before is it not.
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16:15:52 ** The fluid exits the narrowed part of the tube at atmospheric pressure. The widened part at the end of the tube is irrelevant--it won't be filled with fluid. So Bernoulli's Equation will tell you that the fluid velocity in this part is vExit such that .5 rho vExit^2 = rho g h1. Now compare three points: point 1, in the narrowed tube; point 2 at the top of the fluid in the lower tube; and point 3 at the level of the fluid surface in the lower container. At point 1 the pressure is atmospheric and velocity is vExit. Pressure is atmospheric because there is no pressure loss within the tube, since friction and viscosity are both assumed negligible. At point 2 fluid velocity is zero and since there is no fluid between the narrowed tube and this point there is no net rho g h contribution to Bernoulli's equation. So we have .5 rho v1^2 + P1 = .5 rho v2^2 + P2. P1 is atmospheric and v1 is vExit from above, while v2 = 0 so P2 = atmospheric pressure + .5 rho vExit^2 = atmospheric pressure + rho g h1. Now comparing point 2 with point 3 we see that there is a difference h in the fluid altitude, with velocity 0 at both points and atmospheric pressure at point 3. Thus P2 + rho g h2 = P3 + rho g h3, or (atmospheric pressure + rho g h1) + rho g h2 = atmospheric pressure + rho g h3. Thus rho g (h3 - h2) = rho g h1 and h3 - h2, which is the height of the fluid in the lower tube, is just equal to h1. If we assume that somehow the fluid manages to expand on escaping the narrowed tube so that it fills the once-again-widened tube, and exits with vExit as above, then the velocity in the narrowed tube will be 2 * vExit. This leads to the conclusion that pressure change from small to large tube is .5 rho (2 vExit)^2 - .5 rho vExit^2 = .5 rho (3 vExit^2). Since pressure is atmospheric in the large tube, pressure in the small tube is atmospheric pressure + 3 ( .5 rho vExit^2). If we use this pressure for point 1 and follow the steps given above we conclude that h3 - h2, the height of the fluid column in the lower tube, is 3 h1. This is the book's answer. Again I don't have the problem in front of me and I might have missed something, but the idea of the fluid expanding to refill the larger pipe doesn't seem consistent with the behavior of liquids, which are pretty much incompressible at ordinary pressures. However note that I am sometimes wrong when I disagree with the textbook's solution. **
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RESPONSE --> ok
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͵|`Rҳ assignment #007 { Physics II 09-23-2006
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16:25:13 Query set 5 problems 16-20 explain how we calculate the average force exerted by a particle of mass m moving at speed v which collides at a time interval `dt elastically and at a right angle with a solid wall
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RESPONSE --> First, we must know the average mass of each particle (pm). When a particle collides it has a change in velocity. 'dv = 2 * (v1) To get the momentum change of all the particles: 'dv * total mass of the particles = momentum change Then we use the Impulse-Momentum Theorem to determine the average force: avg.F = momentum change/time interval
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16:25:23 ** The impulse exerted on a particle in a collision is the change in the momentum of that particle during a collision. The impulse-momentum theorem says that the change in momentum in a collision is equal to the impulse, the average force * the time interval between collisions. The average force is thus change in momentum / time interval; the time interval is the round-trip distance divided by the velocity, or 2L / v so the average force is -2 m v / ( 2L / v) = m v^2 / L If there were N such particles the total average force would be N * m v^2 / L If the directions are random we distribute the force equally over the 3 dimensions of space and for one direction we get get 1/3 the force found above, or 1/3 N * m v^2 / L. This 3-way distribution of force is related to the fact that for the average velocity vector we have v^2 = vx^2 + vy^2 + vz^2, where v is average magnitude of velocity and vx, vy and vz the x, y and z components of the velocity (more specifically the rms averages--the square root of the average of the squared components). **
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RESPONSE --> Understood.
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16:28:28 Summarize the relationship between the thermal energy that goes into the system during a cycle, the work done by the system during a cycle, and the thermal energy removed or dissipated during the cycle.
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RESPONSE --> The thermal energy that goes into a system during a cycle is used to expand air that will raise and as it cools lower a certain number of pistons. While this process is occuring a certian amoung of the thermal energy is lost as heat. This is an example of a heat engine.
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16:29:36 ** Work-energy is conserved within an isolated system. So the thermal energy that goes into the system must equal the total of the work done by the system and the thermal energy removed from the system. What goes in must come out, either in the form of work or thermal energy. **
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RESPONSE --> I understand that this is a combination of Newton's 1st and 2nd Laws of Thermodynamics.
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16:31:22 If you know the work done by a thermodynamic system during a cycle and the thermal energy removed or dissipated during the cycle, how would you calculate the efficiency of the cycle?
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RESPONSE --> The efficiency of a cylce (e) is directly proportional to the work done and inversly proportional to the heat that is lost. e = W/Q e is efficiency W is work Q is heat lost
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16:31:37 ** STUDENT SOLUTION: Efficiency is work done / energy input. Add the amount thermal energy removed to the amount of work done to get the input. Then, divide work by the energy input. **
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RESPONSE --> ok
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16:31:41 prin phy and gen phy problem 15.2, cylinder with light frictionless piston atm pressure, 1400 kcal added, volume increases slowly from 12.0 m^3 to 18.2 m^3. Find work and chagne in internal energy.
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RESPONSE -->
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16:31:43 Work done at constant pressure is P `dV, so the work done in this situation is `dW = P `dV = 1 atm * (18.2 m^3 - 12 m^3) = (101.3 * 10^3 N/m^2) * (6.2 m^3) = 630 * 10^3 N * m = 6.3 * 10^5 J. A total of 1400 kcal = 1400 * 4200 J = 5.9 * 10^6 J of thermal energy is added to the system, the change in internal energy is `dU = `dQ - `dW = 5.9*10^6 J - 6.3 * 10^5 J = 5.9 * 10^6 J - .63 * 10^6 J = 5.3 * 10^6 J. It is worth thinking about the P vs. V graph of this process. The pressure P remains constant at 101.3 * 10^3 J as the volume changes from 12 m^3 to 18.2 m^3, so the graph will be a straight line segment from the point (12 m^3, 101.3 * 10^3 J) to the point (18.2 m^3, 101.3 * 10^3 J). This line segment is horizontaland the region above the horizontal axis and beneath the segment is a rectangle whose width is 6.2 * 10^3 m^3 and whose altitude is 101.3 * 10^3 N/m^2; its area is therefore the product of its altitude and width, which is 6.3 * 10^5 N m, or 6.3 * 10^5 J, the same as the word we calculated above.
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RESPONSE -->
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16:31:45 prin phy and gen phy problem 15.5, 1.0 L at 4.5 atm isothermally expanded until pressure is 1 atm then compressed at const pressure to init volume, final heated to return to original volume. Sketch and label graph.
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RESPONSE -->
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16:31:47 When a confined ideal gas is expanded isothermally its pressure and volume change, while the number of moled and the temperature remain constant. Since PV = n R T, it follows that P V remains constant. In the initial state P = 4.5 atm and V = 1 liter, so P V = 4.5 atm * 1 liter = 4.5 atm * liter (this could be expressed in standard units since 1 atm = 101.3 kPa = 101.3 * 10^3 N/m^2 and 1 liters = .001 m^3, but it's more convenient to first sketch and label the graph in units of atm and liters). During the isothermal expansion, therefore, since P V remains constant we have P V = 4.5 atm liters. At a pressure of 1 atm, therefore, the volume will be V = 4.5 atm liter / P = 4.5 atm liter / (1 atm) = 4.5 liters. The graph follows a curved path from (1 liter, 4.5 atm) to (4.5 liters, 1 atm). At the gas is compressed at constant pressure back to its initial 1 liter volume, the pressure remains constant so the graph follows a horizontal line from (4.5liters, 1 atm) to (1 liter, 1 atm). Note that this compression is accomplished by cooling the gas, or allowing it to cool. Finally the gas is heated at constant volume until its pressure returns to 4.5 atm. The constant volume dictates that the graph follow a vertical line from (1 liter, 1 atm) back to (4.5 liters, 1 atm). The graph could easily be relabeled to usestandard metric units. 1 atm = 101.3 kPa = 101.3 * 10^3 Pa = 101.3 * 10^3 N/m^2, so 4.5 atm = 4.5 * 101.3 * 10^3 Pa = 4.6 * 10^3 Pa = 4.6 * 10^3 N/m^2. 1 liter = .001 m^3 so 4.5 liters = 4.5 m^3. Since P V = 4.5 atm liters, P = 4.5 atm liters / V. This is of the form P = c / V, with c a constant. For positive values of V, this curve descendsfrom a vertical asymptote with the vertical axis (the V axis) through the point (1, c) then approaches a horizontal asymptote with the horizontal axis. For c = 4.5 atm liters, the curve therefore passes through the point (1 liter, 4.5 atm). As we have seen it also passes through (4.5 liters, 1 atm).
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RESPONSE -->
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16:31:49 gen phy problem 15.12, a-c curved path `dW = -35 J, `dQ = -63 J; a-b-c `dW = - 48 J gen phy how much thermal energy goes into the system along path a-b-c and why?
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RESPONSE -->
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16:31:50 ** I'll need to look at the graph in the text to give a reliably correct answer to this question. However the gist of the argument goes something like this: `dQ is the energy transferred to the system, `dW the work done by the system along the path. Along the curved path the system does -35 J of work and -63 J of thermal energy is added--meaning that 35 J of work are done on the system and the system loses 63 J of thermal energy. If a system gains 35 J of energy by having work done on it while losing 63 J of thermal energy, its internal energy goes down by 28 J (losing thermal energy take internal energy from the system, doing work would take energy from the system so doing negative work adds energy to the system). So between a and c along the curved path the system loses 28 J of internal energy. In terms of the equation, `dU = `dQ - `dW = -63 J -(-35 J) = -28 J. It follows that at point c, the internal energy of the system is 28 J less than at point a, and this will be the case no matter what path is followed from a to c. Along the path a-b-c we have -48 J of work done by the system, which means that the system tends to gain 48 J in the process, while as just observed the internal energy goes down by 28 Joules. The system therefore have `dQ = `dU + `dW = -28 J + (-48 J) = -76 J, and 76 J of internal energy must be removed from the system.**
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RESPONSE -->
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16:31:52 gen phy How are the work done by the system, the thermal energy added to the system and the change in the internal energy of the system related, and what is this relationship have to do with conservation of energy?
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RESPONSE -->
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16:31:54 ** If a system does work it tends to reduce internal energy, so `dW tends to decrease `dU. If thermal energy is added to the system `dQ tends to increase `dU. This leads to the conclusion that `dU = `dQ - `dW. Thus for example if `dW = -48 J and `dU = -28 J, `dQ = `dU + `dW = -28 J + -48 J = -76 J. **
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RESPONSE -->
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16:31:56 gen phy How does the halving of pressure caused a halving of the magnitude of the work, and why is the work positive instead of negative as it was in the process a-b-c?
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RESPONSE -->
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16:31:57 ** Work is the area under the pressure vs. volume curve. If you have half the pressure between two volumes the graph has half the altitude, which leads to half the area. The 'width' of a region is final volume - initial volume. If the direction of the process is such that final volume is less than initial volume (i.e., going 'backwards', in the negative x direction) then with 'width' is negative and the area is negative. **
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RESPONSE -->
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16:45:57 query univ phy problem 19.56 (17.40 10th edition) compressed air engine, input pressure 1.6 * 10^6 Pa, output 2.8 * 10^5 Pa, assume adiabatic.
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RESPONSE --> We can use the ideal-gas equation: PV=nRT We know that V,n, and R are all constant so we throw them out and get. P1/T1 = P2/T2 Since the exhaust cannot have any frost the temperature can go no lower than 1 degree Celsius. T1 = (P1*T2)/P2 = (1.60*10^6 Pa * 1 degree Celsius)/2.80*10^5 Pa = 5.71 degrees Celsius
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16:46:26 ** For an adiabatic process in an ideal gas you know that PV = nRT and PV^`gamma is constant. You are given P1 and P2, and you want T2 > 273 K to prevent formation of frost. Assume T2 = P2 V2 / (n R) = 273 K and n R = (P2 V2) / 273 K . Then T1 = P1 V1 / (n R) = P1 V1 * 273 K / (P2 V2) = (P1 / P2) * (V1 / V2) * 273 K. Since PV^`gamma = constant it follows that V1 / V2 = (P2 / P1)^(1/`gamma) = (P1 / P2)^(-1/`gamma). Thus T1 = (P1 / P2) ( P1 / P2)^(-1/`gamma) * 273 K = (P1 / P2)^(1 - 1/`gamma) = (P1 / P2)^(1-1/1.4) * 273 K = (P1 / P2)^.29 * 273 K = 5.6^.29 * 273 K = 443 K, approx. **
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RESPONSE --> Understood
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17:21:18 query univ 19.62 (17.46 10th edition) .25 mol oxygen 240 kPa 355 K. Isobaric to double vol, isothermal back, isochoric to original pressure.
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RESPONSE --> Given: n = 0.250 moles P = 2.40*10^5 Pa T = 355 K Using PV = nRT, we can solve for the initial V. V = (nRT)/P = (0.250 * 8.314 * 355)/2.40*10^5 = 3.07*10^-3 L Given the initial Volume we know that during the iobaric process the volume is doubled and the pressure is constant. This means that: V1/T1 = V2/T2 T2 = (6.14*10^3L * 355 K) / 3.07*10^-3 L = 710 K During the isothermal process the V2 goes back to V1 and the T is constant. This means that: P1V1=P2V2 P2 = (2.40*10^5 Pa * 6.14*10^-3 L) / 3.07*10^-3 L = 4.80*10^6 Pa During the isochoric process the pressure returns to normal and the V is constant. This means that: P1/T1 = P2/T2 T2 = (2.40*10^5 Pa * 710 K)/4.80*10^6 Pa = 35.5 K The work done in the isobaric process is: W = P (V2-V1) = 2.40*10^5 Pa(6.40*10^-3 L - 6.07*10^-3 L) = 736.8 No work is done during the iosthermal process because there is no introduction of heat. Also, during the iochoric process no work is done because the volume is constant and no work is done on its surroundings. Total work done is 736.8 (Pa*L) ? I will email you the graph.
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17:22:34 ** .25 mol oxygen at 240 kPa occupies about V = n R T / P = .25 mol * 8.31 J / (mol K) * 355 K / (2.4 * 10^5 N/m^2) = .003 m^3, very approximately. Doubling volume, `dV = 2 * V - V = V = .003 m^2 and P = 2.4 * 10^5 Pa so P `dv = 700 J, very approximately. During isothermal compression we have n = const and T = const so P = n R T / V. Compressing to half the volume, since PV = const, gets us to double the pressure, so max pressure is 2 * 240 kPA = 480 kPa. To get work we integrate P dV. Integral of P dV is calculated from antiderivative n R T ln | V |; integrating between V1 and V2 we have n R T ln | V2 | - n R T ln | V1 | = n R T ln | V2 / V1 |. In this case V2 = V and V1 = 2 V so V2 / V1 = 1/2 and we have `dW = n R T ln(1/2) = .25 mol * 8.31 J/(mol K) * 710 K * (-.7) = -1000 J, approx. So net work is about 700 J - 1000 J = -300 J **
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RESPONSE --> Understood.
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17:26:27 univ phy describe your graph of P vs. V
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RESPONSE --> My graph shows a line decreasing from left to right. As the volume increases the pressure decreases.
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17:26:57 ** The graph proceeds horizontally to the right from original P and V to doubled V, then to the left along a curve that increases at an incr rate as we move to the left (equation P = 2 P0 V0 / V) until we're just above the starting point, then vertically down to the starting pt. **
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RESPONSE --> Understood.
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17:28:09 univ phy What is the temperature during the isothermal compression?
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RESPONSE --> 710 K This is the termperature that the isobaric process is at when it ends and the temperature is constant during isothermal processes.
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17:28:20 ** If vol doubles at const pressure then temp doubles to 710 K, from which isothermal compression commences. So the compression is at 710 K. **
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RESPONSE --> Understood.
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17:28:56 univ phy What is the max pressure?
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RESPONSE --> 4.8*10^6 Pa This is the highest pressure the gas reaches during the isothermal process.
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17:29:04 ** It starts the isothermal at the original 240 kPa and its volume is halved at const temp. So the pressure doubles to 480 kPa. **
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RESPONSE --> Understood.
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tJۇه{Cʣ assignment #008 { Physics II 09-23-2006
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17:30:05 prin phy and gen phy problem 15.19 What is the maximum efficiency of a heat engine operating between temperatures of 380 C and 580 C?
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RESPONSE -->
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17:30:07 The maximum possible efficiency is (T_h - T_c) / T_h, where T_h and T_c are the absolute max and min operating temperatures. T_h is (580 + 273)K = 853 K and T_c is (380 + 273) K = 653 K, so the maximum theoretical efficiency is max efficiency = (T_h - T_c) / T_h = (853 K - 653 K) / (853 K) = .23, approx. This means that the work done by this engine will be not greater than about 23% of the thermal energy that goes into it.
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17:30:11 query gen phy problem 15.26 source 550 C -> Carnot eff. 28%; source temp for Carnot eff. 35%?
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RESPONSE -->
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17:30:12 ** Carnot efficiency is eff = (Th - Tc) / Th. Solving this for Tc we multiply both sides by Th to get eff * Th = Th - Tc so that Tc = Th - eff * Th = Th ( 1 - eff). We note that all temperatures must be absolute so we need to work with the Kelvin scale (adding 273 C to the Celsius temperature to get the Kelvin temperature) If Th = 550 C = 823 K and efficiency is 30% then we have Tc =823 K * ( 1 - .28) = 592 K. Now we want Carnot efficiency to be 35% for this Tc. We solve eff = (Th - Tc) / Th for Th: Tc we multiply both sides by Th to get eff * Th = Th - Tc so that eff * Th - Th = -Tc and Tc = Th - eff * Th or Tc = Th ( 1 - eff) and Th = Tc / (1 - eff). If Tc = 576 K and eff = .35 we get Th = 592 K / ( 1 - .35 ) = 592 C / .6 = 912 K, approx. This is (912 - 273) C = 639 C. **
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17:45:15 univ phy problem 20.44 (18.40 10th edition) ocean thermal energy conversion 6 C to 27 C At 210 kW, what is the rate of extraction of thermal energy from the warm water and the rate of absorption by the cold water?
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RESPONSE --> I am not sure how to begin this problem. I think that the efficiency is 0.852 Because e = 1 - absolute value (Qc/Qh). e = 1- absolute value (-4/27) = 0.852
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17:48:26 ** work done / thermal energy required = .07 so thermal energy required = work done / .07. Translating directly to power, thermal energy must be extracted at rate 210 kW / .07 = 30,000 kW. The cold water absorbs what's left after the 210 kW go into work, or 29,790 kW. Each liter supplies 4186 J for every degree, or about 80 kJ for the 19 deg net temp change. Needing 30,000 kJ/sec this requires about 400 liters / sec, or well over a million liters / hour. Comment from student: To be honest, I was suprised the efficiency was so low. Efficiency is low but the energy is cheap and environmental impact in the deep ocean can be negligible so the process can be economical, if a bit ugly. **
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RESPONSE --> Where does the 0.07 come from? I understand the steps from there on.
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