Here are the solutions to the packet I handed out on Friday:
http://groups.google.com/group/bronx-ap-physics/web/Thermo+Review+Problem+Solutions.pdf
As requested, I will have a couple problems for you to play around with, but I expect most of what we do will come from you!
Sunday, December 21, 2008
Saturday, December 20, 2008
Bronx Tournament is ON!
The Bronx FLL Tournament is going on as scheduled today, Saturday, December 20, 2008.
Please leave yourself extra time to get to the event safely.
See you all there!
Please leave yourself extra time to get to the event safely.
See you all there!
Tuesday, December 16, 2008
Thermo-cycling in the Snow!
Handout answers:
1
a) T = 481 K
b) Wnet = +4000 J
c) For a full cycle, delta_U = 0. This means that for the entire cycle, Q = -W = -4000 J.
d) For any cycle, delta_U is zero, so it stays the same.
2.
a) AB,CD - isobaric, BC, DA - isochoric
b) 1.202 x 10^-4 moles (use P,V, and T at state D.)
c) Wab = -1.0 J, Wad = 0.5 J, all others zero
d) net work for cycle is the area inside = -0.5 J
e) Tb = 800 K, so delta_U = 0.750 J
f) P = Wnet/delta_t = 0.5 J/1.2 s = .417 W
3. b, c, a (remember that in c and a, the work is negative, and C has a work that is LESS negative.)
1
a) T = 481 K
b) Wnet = +4000 J
c) For a full cycle, delta_U = 0. This means that for the entire cycle, Q = -W = -4000 J.
d) For any cycle, delta_U is zero, so it stays the same.
2.
a) AB,CD - isobaric, BC, DA - isochoric
b) 1.202 x 10^-4 moles (use P,V, and T at state D.)
c) Wab = -1.0 J, Wad = 0.5 J, all others zero
d) net work for cycle is the area inside = -0.5 J
e) Tb = 800 K, so delta_U = 0.750 J
f) P = Wnet/delta_t = 0.5 J/1.2 s = .417 W
3. b, c, a (remember that in c and a, the work is negative, and C has a work that is LESS negative.)
Monday, December 15, 2008
In This House, we obey the laws of Thermodynamics!
Handout solutions:
1.
a) Tf = 400 K
b) The ratio of average kinetic energies becomes the ratio of the temperatures Tb/Ta = 3/4 since the Boltzmann constant divides out. (KE average = 3/2 kb*T)
c) W = -P*delta_V, but we have to find the pressure first. P = nRT/V = 8315 Pa.
W = -831.5 J
d) using 1st law, Q = delta_U - W = 1.5*(1 mole)(8.315 J/mol*K)(400 K - 300 K) - (-831.5 J) = 2,079 J
2.
a) 0.3 kg/(.004 kg/mol) = 75 moles
b) using 1st law, delta_U = Q + W. We can apply this to the entire process. During the first part, we can assume W = 0 and Q = +3100J. During the second, Q = 0, and W = +1000 J. Thus for the entire process, Q = 3100 J, and W = 1000 J. This means that delta_U = 4100 J.
c) Since delta_U = 4100 J = 1.5nRdelta_T, delta_T = 4.38 K. The final temperature is then 254.4 K.
3. It might help to draw a PV diagram on this problem. There are two segments, and three states to consider.
The initial temperature is 273 K, initial pressure = 101,300 Pa. We can use PV=nRT to find that the initial volume is 2.7 x 10-2 m^3.
At state 2, the volume is 2.7 x 10-2 m^3 (const. volume) and P = 303900 Pa. The temperature is therefore 3*273 K =m 819 K.
At state 3, the volume is 5.4 x 10-2 m^3, and P = 303900 Pa. The temperature at state 3 is therefore 2*819 K = 1638 K.
For this whole process, the work is -(303,900 Pa)(5.4 - 2.7)*10^-3 m^3 = -820.5 J.
The change of internal energy is 3/2*n*R(delta_T) = 17030 J.
From the 1st Law, this means that Q = 17030 J - (-820.5 J) = 17850 J.
Book Problems:
p. 440
32. H = k*A*delta_T/L = (0.84 J/s · m · C°)(3.0 m^2)[15°C - (- 5°C)]/(3.2 x 10-3 m) = 1.6 x 104 W.
p. 471 - 472:
Q3. It is enough information since the process is isothermal - delta_U = 0, so Q = -W.
Q6. In an adiabatic compression, Q = 0, and the work is positive. By the first law, this means delta_U is positive, which means delta_T is positive, an indication that temperature increases.
Problems:
6.
a) Volume doesn't change, so W = 0.
b)Q = -265 kJ, W = 0, so delta_U = -265,000 J.
7.
a) adiabatic means Q = 0.
b) Q = 0, W = 1350 J, so delta_U = 1350 J.
c) since delta_U is positive, the temperature must increase.
8.
a) Work is done only during the constant pressure process, and is equal to -P*delta_V = -1300 J.
b) Q = delta_U - W but since the final temperature is the same as the original, delta_U = 0.
Q = 0 - (-1300 J) = 1300 J.
9.
a) W = -3500 J
b)since the ttemperatures are equal, delta_U = 0.
c) Q = delta_U - W = 0 - (-3500 J) = 3500 J, which represents a heat flow INTO the gas.
1.
a) Tf = 400 K
b) The ratio of average kinetic energies becomes the ratio of the temperatures Tb/Ta = 3/4 since the Boltzmann constant divides out. (KE average = 3/2 kb*T)
c) W = -P*delta_V, but we have to find the pressure first. P = nRT/V = 8315 Pa.
W = -831.5 J
d) using 1st law, Q = delta_U - W = 1.5*(1 mole)(8.315 J/mol*K)(400 K - 300 K) - (-831.5 J) = 2,079 J
2.
a) 0.3 kg/(.004 kg/mol) = 75 moles
b) using 1st law, delta_U = Q + W. We can apply this to the entire process. During the first part, we can assume W = 0 and Q = +3100J. During the second, Q = 0, and W = +1000 J. Thus for the entire process, Q = 3100 J, and W = 1000 J. This means that delta_U = 4100 J.
c) Since delta_U = 4100 J = 1.5nRdelta_T, delta_T = 4.38 K. The final temperature is then 254.4 K.
3. It might help to draw a PV diagram on this problem. There are two segments, and three states to consider.
The initial temperature is 273 K, initial pressure = 101,300 Pa. We can use PV=nRT to find that the initial volume is 2.7 x 10-2 m^3.
At state 2, the volume is 2.7 x 10-2 m^3 (const. volume) and P = 303900 Pa. The temperature is therefore 3*273 K =m 819 K.
At state 3, the volume is 5.4 x 10-2 m^3, and P = 303900 Pa. The temperature at state 3 is therefore 2*819 K = 1638 K.
For this whole process, the work is -(303,900 Pa)(5.4 - 2.7)*10^-3 m^3 = -820.5 J.
The change of internal energy is 3/2*n*R(delta_T) = 17030 J.
From the 1st Law, this means that Q = 17030 J - (-820.5 J) = 17850 J.
Book Problems:
p. 440
32. H = k*A*delta_T/L = (0.84 J/s · m · C°)(3.0 m^2)[15°C - (- 5°C)]/(3.2 x 10-3 m) = 1.6 x 104 W.
p. 471 - 472:
Q3. It is enough information since the process is isothermal - delta_U = 0, so Q = -W.
Q6. In an adiabatic compression, Q = 0, and the work is positive. By the first law, this means delta_U is positive, which means delta_T is positive, an indication that temperature increases.
Problems:
6.
a) Volume doesn't change, so W = 0.
b)Q = -265 kJ, W = 0, so delta_U = -265,000 J.
7.
a) adiabatic means Q = 0.
b) Q = 0, W = 1350 J, so delta_U = 1350 J.
c) since delta_U is positive, the temperature must increase.
8.
a) Work is done only during the constant pressure process, and is equal to -P*delta_V = -1300 J.
b) Q = delta_U - W but since the final temperature is the same as the original, delta_U = 0.
Q = 0 - (-1300 J) = 1300 J.
9.
a) W = -3500 J
b)since the ttemperatures are equal, delta_U = 0.
c) Q = delta_U - W = 0 - (-3500 J) = 3500 J, which represents a heat flow INTO the gas.
Sunday, December 14, 2008
In the Thermodynamic Process of Decorating
Handout solutions:
5. We need to convert to Pascals and m^3: 4.0 L = 4 x 10-3m^3 and 2.0 L = 4 x 10-3m^3.
Since the initial gauge pressure is 3 atm, the absolute pressure is 4 atm = 405,200 Pa.
For a constant pressure process, W = -P*delta_V = -(405,200 Pa)(-2 x 10-3m^3) = 810.4 J
b) Since it's a constant pressure process, the final pressure is also 405,200 Pa.
c) Using ideal gas law, T2 = T1*(V2/V1) = 149 K
6.
a) Using ideal gas law, 5P0*3V0 = nRT0 so T0 = 15P0V0/nR
b) Since temperature is constant, the change of internal energy is zero.
c) Using ideal gas law again, P2V2 = nRT2
We know that T2 = T0 = 15P0V0/nR so P2V2 = 15P0V0.
V2 = 6V0
P2 = 15P0V0/6V0 = 5/2P0
7.
a) Ideal gas law - n = PV/nT = (7*101,300Pa)(10 x 10-3m^3)/(8.315.J/Mol*K * 500 K) = 1.706 moles
b) The container volume is constant, so the work done is zero.
c) T2 = T1*P2/P1 = 214.3 K
d) THe graph should be a vertical line at V = 10 x 10-3 m^3 from P = 7 atm to 3 atm (or the equivalent pressures in Pa.)
Book HW:
3,4,10abc, 53a,c
3.
4.
10
a) and c) PV diagram:
b) W = -2700 J, found since the process is isobaric, using W = -P*delta_V
53.
a) W = -2.2 x 105 J
c)
5. We need to convert to Pascals and m^3: 4.0 L = 4 x 10-3m^3 and 2.0 L = 4 x 10-3m^3.
Since the initial gauge pressure is 3 atm, the absolute pressure is 4 atm = 405,200 Pa.
For a constant pressure process, W = -P*delta_V = -(405,200 Pa)(-2 x 10-3m^3) = 810.4 J
b) Since it's a constant pressure process, the final pressure is also 405,200 Pa.
c) Using ideal gas law, T2 = T1*(V2/V1) = 149 K
6.
a) Using ideal gas law, 5P0*3V0 = nRT0 so T0 = 15P0V0/nR
b) Since temperature is constant, the change of internal energy is zero.
c) Using ideal gas law again, P2V2 = nRT2
We know that T2 = T0 = 15P0V0/nR so P2V2 = 15P0V0.
V2 = 6V0
P2 = 15P0V0/6V0 = 5/2P0
7.
a) Ideal gas law - n = PV/nT = (7*101,300Pa)(10 x 10-3m^3)/(8.315.J/Mol*K * 500 K) = 1.706 moles
b) The container volume is constant, so the work done is zero.
c) T2 = T1*P2/P1 = 214.3 K
d) THe graph should be a vertical line at V = 10 x 10-3 m^3 from P = 7 atm to 3 atm (or the equivalent pressures in Pa.)
Book HW:
3,4,10abc, 53a,c
3.
4.
10
a) and c) PV diagram:
b) W = -2700 J, found since the process is isobaric, using W = -P*delta_V
53.
a) W = -2.2 x 105 J
c)
Thursday, December 11, 2008
In Theory, We Trust - Kinetic Theory Solutions
A bonus assignment for tonight - if you check your answers, please log in and write a comment (go to the bottom of the post for the link) stating the name of your favorite ideal gas. If you don't have one, choose Helium.
50. The gas starts at 0 degrees C, or 273 K. Since v = (3RT/m)^1/2, in order to double v, the temperature must be multiplied by 4. Thus the temperature must be 1092 K.
52. Following the similar logic as with #50, If the pressure doubles, and volume stays the same, the ideal gas equation says that the temperature will also double. This will make the vrms increase by a factor of the square root of 2.
55. You could get to the solution of this equation by looking at the derivation we did today. We ended up with the fact that P = (1/3)m*N*v^3/(V).
The quantity m*N is the total mass of the gas contained in the container, as it's the mass of a single molecule times the number of molecules. The quantity M*N/V must then bethe density of the gas.
Also from the derivation, the v in our expression is the mean square speed of the molecules in the gas.
Thus P = (1/3)*density*v^2
There are other ways to derive it using the other formulas we came up with today, so if your method is different, that is fine.
80. Since you know the temperature, you can directly calculate the vRMS = 183.5 m/s.
Using the result of question 5, we have a relationship between the density, vRMS, and the pressure.
The density will be one atom/cm^3 = 1 amu * 1.66 x 10-27 kg*(106cm^3/m^3) = 1.66 x 10-21 kg/m^3.
Substituting, the pressure is 1.9 x -17 Pa = 1.84 x -22 atm
81. When 10 meters below the surface, the absolute pressure is Patm + rho*g*h = 199,300 Pa.
At the surface, the pressure is 101,300 Pa. You also know the initial volume underneath the water, as well as the fact that the moles of air in the diver's lungs are constant as he/she rises to the surface.
If we assume the rise happens, quickly, we can also say that the temperature is approximately constant.
Thus P1V1/P2V2 = nRT1/nRT2. The entire right side of the equation divides out to 1.
Thus V2 = P1V1/P2 = 10.8 L. I personally would NOT want my lungs stretched to three times their original volume, nor would I expect it to be medically advisable for ANYONE to have this done to them.
This is one of the reasons why it's important for scuba divers to take their time rising to the surface after a dive. One of them is that it causes a condition called 'the bends', where dissolved gas comes out of solution in the blood stream. This is a bad thing. Another is that, without releasing any gas in the lungs during the resurfacing process, the lungs will expand rapidly during the rise, possibly causing tearing in the tissue.
50. The gas starts at 0 degrees C, or 273 K. Since v = (3RT/m)^1/2, in order to double v, the temperature must be multiplied by 4. Thus the temperature must be 1092 K.
52. Following the similar logic as with #50, If the pressure doubles, and volume stays the same, the ideal gas equation says that the temperature will also double. This will make the vrms increase by a factor of the square root of 2.
55. You could get to the solution of this equation by looking at the derivation we did today. We ended up with the fact that P = (1/3)m*N*v^3/(V).
The quantity m*N is the total mass of the gas contained in the container, as it's the mass of a single molecule times the number of molecules. The quantity M*N/V must then bethe density of the gas.
Also from the derivation, the v in our expression is the mean square speed of the molecules in the gas.
Thus P = (1/3)*density*v^2
There are other ways to derive it using the other formulas we came up with today, so if your method is different, that is fine.
80. Since you know the temperature, you can directly calculate the vRMS = 183.5 m/s.
Using the result of question 5, we have a relationship between the density, vRMS, and the pressure.
The density will be one atom/cm^3 = 1 amu * 1.66 x 10-27 kg*(106cm^3/m^3) = 1.66 x 10-21 kg/m^3.
Substituting, the pressure is 1.9 x -17 Pa = 1.84 x -22 atm
81. When 10 meters below the surface, the absolute pressure is Patm + rho*g*h = 199,300 Pa.
At the surface, the pressure is 101,300 Pa. You also know the initial volume underneath the water, as well as the fact that the moles of air in the diver's lungs are constant as he/she rises to the surface.
If we assume the rise happens, quickly, we can also say that the temperature is approximately constant.
Thus P1V1/P2V2 = nRT1/nRT2. The entire right side of the equation divides out to 1.
Thus V2 = P1V1/P2 = 10.8 L. I personally would NOT want my lungs stretched to three times their original volume, nor would I expect it to be medically advisable for ANYONE to have this done to them.
This is one of the reasons why it's important for scuba divers to take their time rising to the surface after a dive. One of them is that it causes a condition called 'the bends', where dissolved gas comes out of solution in the blood stream. This is a bad thing. Another is that, without releasing any gas in the lungs during the resurfacing process, the lungs will expand rapidly during the rise, possibly causing tearing in the tissue.
Wednesday, December 10, 2008
Expand your Mind, and Think of an Ideal Gas - Solutions
p.413, 12, 13, 34, 35, 38, 40 (What is V2/V1?)
12. Notice that you can do this problem without converting to meters, since you end up dividing delta_L by L0, which divides out the units anyway.
You should be able to find that delta_T = delta_L/(alpha*L0) = -89 degrees. (Remember change is final minus initial!) Adding this to the initial temperature of 20 degrees Celsius gives the final temperature of -69 degrees.
13. The key here is multiplying the final lengths of the plate after expansion.
(L + dL)(W + dW) = LW + W*dL + L*dW + dL*dW
As I mentioned in class, the last term is a higher order term, and can be neglected.
If we then substitute using the thermal expansion equation for dW = alpha*W*dT and dL = alpha*L*dT, we get that the final area is LW + 2*alpha*LW*dT.
The question asks for the change in area, so we just have to subtract the initial area LW to get what the text says should be the answer, 2*alpha*L*W*dT.
34. The first task is to determine the number of moles of the gas. Since there are 18.5 kg (18,500 g) of Nitrogen, and the molar weight of Nitrogen gas is 28 g/mol, we can divide to get that there are 661 moles of N2 gas in the tank.
We now have enough information to solve for V = nRT/P (remember that standard temperature and pressure means 273 K, 1.013 x 105 Pa)
V = 14.8 m^3.
For part b, the tank has not changed, so it has the same volume as you found for the first part. We do know there is a new number of moles in the tank (661 + additional 536 moles).
Since we are looking for pressure, we again must solve for P in the ideal gas equation.
P = nRT/V = 1.84 x 105 Pa
35. Remember we have to use absolute pressure, so the initial pressure in the container is (1 atm + .35 atm)*(1.013 x 105 Pa/atm) = 1.37 x 105 Pa
Solving, V = nRT/P = 0.439 m^3
b) Here we are changing states for the gas, so we need to use the division trick from class.
P1V1 = nRT1 and P2V2 = nRT2 , which can be combined to give
(P2/P1)(V2/V1) = T2/T1
Solving, T2 = 210 K = -63 degrees Celsius.
38. IN this case, since it is a change of state problem, we do NOT have to convert the pressure to Pascals before doing anything. We can use the same division method as before.
(P2/P1)(V2/V1) = T2/T1
P2 = (T2/T1)(V1/V2)*P1
P2 = (323.2 K/291.2 K)*(55.0 L/48.8 L)* 1 atm = 3.03 atm
40. Again, using the same method:
(P2/P1)(V2/V1) = T2/T1
(V2/V1) = (T2/T1)*(P1/P2) = 1.4
12. Notice that you can do this problem without converting to meters, since you end up dividing delta_L by L0, which divides out the units anyway.
You should be able to find that delta_T = delta_L/(alpha*L0) = -89 degrees. (Remember change is final minus initial!) Adding this to the initial temperature of 20 degrees Celsius gives the final temperature of -69 degrees.
13. The key here is multiplying the final lengths of the plate after expansion.
(L + dL)(W + dW) = LW + W*dL + L*dW + dL*dW
As I mentioned in class, the last term is a higher order term, and can be neglected.
If we then substitute using the thermal expansion equation for dW = alpha*W*dT and dL = alpha*L*dT, we get that the final area is LW + 2*alpha*LW*dT.
The question asks for the change in area, so we just have to subtract the initial area LW to get what the text says should be the answer, 2*alpha*L*W*dT.
34. The first task is to determine the number of moles of the gas. Since there are 18.5 kg (18,500 g) of Nitrogen, and the molar weight of Nitrogen gas is 28 g/mol, we can divide to get that there are 661 moles of N2 gas in the tank.
We now have enough information to solve for V = nRT/P (remember that standard temperature and pressure means 273 K, 1.013 x 105 Pa)
V = 14.8 m^3.
For part b, the tank has not changed, so it has the same volume as you found for the first part. We do know there is a new number of moles in the tank (661 + additional 536 moles).
Since we are looking for pressure, we again must solve for P in the ideal gas equation.
P = nRT/V = 1.84 x 105 Pa
35. Remember we have to use absolute pressure, so the initial pressure in the container is (1 atm + .35 atm)*(1.013 x 105 Pa/atm) = 1.37 x 105 Pa
Solving, V = nRT/P = 0.439 m^3
b) Here we are changing states for the gas, so we need to use the division trick from class.
P1V1 = nRT1 and P2V2 = nRT2 , which can be combined to give
(P2/P1)(V2/V1) = T2/T1
Solving, T2 = 210 K = -63 degrees Celsius.
38. IN this case, since it is a change of state problem, we do NOT have to convert the pressure to Pascals before doing anything. We can use the same division method as before.
(P2/P1)(V2/V1) = T2/T1
P2 = (T2/T1)(V1/V2)*P1
P2 = (323.2 K/291.2 K)*(55.0 L/48.8 L)* 1 atm = 3.03 atm
40. Again, using the same method:
(P2/P1)(V2/V1) = T2/T1
(V2/V1) = (T2/T1)*(P1/P2) = 1.4
Monday, December 8, 2008
Sunday, December 7, 2008
Bernoulli and Friends - It's a Family Program!
Q 18,19 P37, 39, 41, 43, 72 (pressure head is the source of the water), 74
We will have a quiz Tuesday on fluid mechanics, so tomorrow (Monday) will be a day of practicing problems and deciding which method to use to solve parts of problems.
Questions:
18. When the streamlines of the wind curve over the roof, this creates a low pressure area above the roof, while the pressure below the roof remains high. This difference in pressure creates an upward force that pushes the roof off the house.
19. The increase in speed of the air passing between the sheets causes a lower static pressure between them. Since the air pressure on the outside of the sheets is the same, the outside pressure pushes into the space between where the pressure is lower.
Problems:
37. Using Bernoulli, with 1 at the ground and 2 at the top of the water flow, we can assume that at the MINIMUM pressure, the fluid is not moving.
P1 + 0 + 0 = Patm + 0 + rho*g*h
P1 - Patm = 1.176 x 105 Pa
39. The pressure is atmospheric underneath the roof. We can take 1 to be at a point far from the roof (where it is still), and 2 to be in the wind above the roof:
Patm + 0 + 0 = P2 + 1/2*rho*v^2 + 0
The net pressure difference will be the gauge pressure of P2, or P2 - Patm
P2 - Patm= - 1/2*rho*v^2 = 580.5 Pa
Since this is pressure, the force is this pressure times the area of the roof:
F = 580.5 Pa * 240 M^2 = 1.39 x 105 N
41.
Consider a streamline that goes from the outside of the hurricane (where P = Patm and the air is still) to the center. Remember to convert 300 km/h to 83.3 m/s.
Patm + 0 + 0 = P2 + .5*rho*v^2 + 0
P2 = 9.68 x 104 Pa
43. We have to use continuity to find the speed of the water in the pipe when the pipes have smaller diameter. Since it is a liquid, we can assume the flow is incompressible.
A1v1 = A2v2, so v2 = A1/A2*v1 = 2.22 m/s
The gauge pressure at the pump is 3.8 atm gauge, which means it is 4.8 atm = 4.86 x 105 Pa. The question asks for the static pressure at the top of the building.
Bernoulli:
Ppump + .5*rho*v1^2 + 0 = Ptop + .5*rho*v2^2 + rho*g*h
Ptop = 2.88 x 105 Pa absolute, or 1.87 x 105 Pa gauge.
72. The pressure head is the source of the water coming out of the shower, a tank on the roof perhaps. We can assume that the water has no velocity initially in the tank, and that the initial pressure is atmospheric.
Patm + 0 + rho*g*H = Patm + .5*rho*v2^2 + 0
H = 2.64 m
74. If the raft holds the maximum number of people, the logs are completely submerged in the water, though not yet sinking. This means the logs and people are in equilibrium. Note the factor of 10 since there are ten logs.
Fb - Mtotalg = 0
10*rhowater*g*pi*R^2*L = (10*rhowood*pi*R^2*L + Mpeople)*g
Since we know the specific gravity of wood, we can solve for the mass of the people.
Mpeople = 2087 kg = N*70 kg
N = 29.8 people
Thus there can be no more than 29 people on the raft.
We will have a quiz Tuesday on fluid mechanics, so tomorrow (Monday) will be a day of practicing problems and deciding which method to use to solve parts of problems.
Questions:
18. When the streamlines of the wind curve over the roof, this creates a low pressure area above the roof, while the pressure below the roof remains high. This difference in pressure creates an upward force that pushes the roof off the house.
19. The increase in speed of the air passing between the sheets causes a lower static pressure between them. Since the air pressure on the outside of the sheets is the same, the outside pressure pushes into the space between where the pressure is lower.
Problems:
37. Using Bernoulli, with 1 at the ground and 2 at the top of the water flow, we can assume that at the MINIMUM pressure, the fluid is not moving.
P1 + 0 + 0 = Patm + 0 + rho*g*h
P1 - Patm = 1.176 x 105 Pa
39. The pressure is atmospheric underneath the roof. We can take 1 to be at a point far from the roof (where it is still), and 2 to be in the wind above the roof:
Patm + 0 + 0 = P2 + 1/2*rho*v^2 + 0
The net pressure difference will be the gauge pressure of P2, or P2 - Patm
P2 - Patm= - 1/2*rho*v^2 = 580.5 Pa
Since this is pressure, the force is this pressure times the area of the roof:
F = 580.5 Pa * 240 M^2 = 1.39 x 105 N
41.
Consider a streamline that goes from the outside of the hurricane (where P = Patm and the air is still) to the center. Remember to convert 300 km/h to 83.3 m/s.
Patm + 0 + 0 = P2 + .5*rho*v^2 + 0
P2 = 9.68 x 104 Pa
43. We have to use continuity to find the speed of the water in the pipe when the pipes have smaller diameter. Since it is a liquid, we can assume the flow is incompressible.
A1v1 = A2v2, so v2 = A1/A2*v1 = 2.22 m/s
The gauge pressure at the pump is 3.8 atm gauge, which means it is 4.8 atm = 4.86 x 105 Pa. The question asks for the static pressure at the top of the building.
Bernoulli:
Ppump + .5*rho*v1^2 + 0 = Ptop + .5*rho*v2^2 + rho*g*h
Ptop = 2.88 x 105 Pa absolute, or 1.87 x 105 Pa gauge.
72. The pressure head is the source of the water coming out of the shower, a tank on the roof perhaps. We can assume that the water has no velocity initially in the tank, and that the initial pressure is atmospheric.
Patm + 0 + rho*g*H = Patm + .5*rho*v2^2 + 0
H = 2.64 m
74. If the raft holds the maximum number of people, the logs are completely submerged in the water, though not yet sinking. This means the logs and people are in equilibrium. Note the factor of 10 since there are ten logs.
Fb - Mtotalg = 0
10*rhowater*g*pi*R^2*L = (10*rhowood*pi*R^2*L + Mpeople)*g
Since we know the specific gravity of wood, we can solve for the mass of the people.
Mpeople = 2087 kg = N*70 kg
N = 29.8 people
Thus there can be no more than 29 people on the raft.
Thursday, December 4, 2008
We MUST Maintain Continuity - The Continuity Equation
Q10,13
P35,36,71
Question 10. For the barge to float with its top lower, the bottom of the barge must be at a greater depth. This means the buoyancy force must be greater. For this to happen and the barge still be in equilibrium, the mass of the boat must be greater. Therefore sand must be added to the barge.
Q 13. The buoyancy force depends only on g, the fluid density, and the submerged volume. The buoyancy force must therefore be the same whether it is submerged completely just below the surface, or deep below, since the submerged volume is the same in both cases.
Problems:
35. We are given the volume of the room, and are told that the air in the room is replaced in 10 minutes. This gives us the volumetric flow rate into/out of the room: Q = 0.345 m^3/s.
This is equal to the volumetric flow rate in, with A1v1 = Q.
Solving, v1 = 3.80 m/s
36. The volume of the pool when it is filled is pi*(3.6 m)^2*(1.5 m) = 61.07 m^3
5/8 inch = .015875 m
Q = A1v1 = volume/delta_t
so delta_t = volume/(A1v1) = (61.07 m^3/(pi*(.015875 m)^2*.28 m/s) = 76.5 hours
71. Gauge pressure of -80 mm Hg means the atmospheric pressure is 680 mm Hg.
Using the conversion factor that 760 mm Hg = 1.013 x 105 Pa
The pressure in the lungs is 9.064 x 104 Pa
You should now imagine that the lungs are a closed container of gas on top of a column of water (like the problems we did on manometer day.)
Comparing the pressure at the bottom of the water column with the atmosphere:
Plungs + rhowater*g*H = Patm
H = 1.088 m
P35,36,71
Question 10. For the barge to float with its top lower, the bottom of the barge must be at a greater depth. This means the buoyancy force must be greater. For this to happen and the barge still be in equilibrium, the mass of the boat must be greater. Therefore sand must be added to the barge.
Q 13. The buoyancy force depends only on g, the fluid density, and the submerged volume. The buoyancy force must therefore be the same whether it is submerged completely just below the surface, or deep below, since the submerged volume is the same in both cases.
Problems:
35. We are given the volume of the room, and are told that the air in the room is replaced in 10 minutes. This gives us the volumetric flow rate into/out of the room: Q = 0.345 m^3/s.
This is equal to the volumetric flow rate in, with A1v1 = Q.
Solving, v1 = 3.80 m/s
36. The volume of the pool when it is filled is pi*(3.6 m)^2*(1.5 m) = 61.07 m^3
5/8 inch = .015875 m
Q = A1v1 = volume/delta_t
so delta_t = volume/(A1v1) = (61.07 m^3/(pi*(.015875 m)^2*.28 m/s) = 76.5 hours
71. Gauge pressure of -80 mm Hg means the atmospheric pressure is 680 mm Hg.
Using the conversion factor that 760 mm Hg = 1.013 x 105 Pa
The pressure in the lungs is 9.064 x 104 Pa
You should now imagine that the lungs are a closed container of gas on top of a column of water (like the problems we did on manometer day.)
Comparing the pressure at the bottom of the water column with the atmosphere:
Plungs + rhowater*g*H = Patm
H = 1.088 m
Wednesday, December 3, 2008
The Problem is your Pressure Valve - I can get you a new one for $250
Handout problem solutions:
4.
a) The FBD has buoyancy force upwards, and weight (mg) downwards.
b) Fb = rho*g*Vsub where Vsub is the volume of the ball. Since the volume of a sphere is (4/3)*pi*R^3, and we know the density of water, we can calculate that the buoyancy force is
410.5 N 41.1 N.
c) Using the same method, but the density of the ball (800 kg/m^3) instead of water, we get that the mass is33.5kg 3.35 kg.
d) Fnet in y = Fb - mg = ma, so a = (Fb - mg)/m = 2.454 m/s^2
5.
a) tray is in equilibrium, mg = rho*g*Vsub
Vsub = area of tray A*depth d, where A = L*w
mg = rho*g*L*w*d, m = 0.3 kg
b) Fbmax = rho*L*W*h = 23.52 N
c) If the tray is just on the edge of sinking, then Fbmax - (mtotal)*g = 0
mtotal = mtray + mweights
Solving, mweights = 2.1 kg
Dividing by 0.005 kg per weight, this means there are 420 weights that could be added.
6.
a) Gauge pressure = 8 atm * (101.3 kPa/atm) = 8.103 x 105 Pa
b) Gauge pressure = P - Patm = rho*g*h = 8.103 x 105 Pa
c) F = P*A = 1.61 x 106 N directed upwards
d) Same formula, F = P*A = 1.621 x 106 N directed downwards
e) The difference between the two forces is 10,460 N directed downwards, so there needs to be an upwards force of 10,460 N to hold the window in place.
7.
a) The mercury will flow into the water since the pressure at the bottom is so much greater than the pressure of the water (P = P0 + rho*g*h, rho of mercury is much greater.)
b) When the mercury drops, we can say the distance it drops is H.
The height of the mercury column is then (1.0 m - H). Since the volume is conserved, the height of the water column is (1.0 m + H).
rhoWater*g*(1.0 m + H) = rhoMercury*g*(1.0 m - H)
Solving for H, H = 1.863 m.
4.
a) The FBD has buoyancy force upwards, and weight (mg) downwards.
b) Fb = rho*g*Vsub where Vsub is the volume of the ball. Since the volume of a sphere is (4/3)*pi*R^3, and we know the density of water, we can calculate that the buoyancy force is
c) Using the same method, but the density of the ball (800 kg/m^3) instead of water, we get that the mass is
d) Fnet in y = Fb - mg = ma, so a = (Fb - mg)/m = 2.454 m/s^2
5.
a) tray is in equilibrium, mg = rho*g*Vsub
Vsub = area of tray A*depth d, where A = L*w
mg = rho*g*L*w*d, m = 0.3 kg
b) Fbmax = rho*L*W*h = 23.52 N
c) If the tray is just on the edge of sinking, then Fbmax - (mtotal)*g = 0
mtotal = mtray + mweights
Solving, mweights = 2.1 kg
Dividing by 0.005 kg per weight, this means there are 420 weights that could be added.
6.
a) Gauge pressure = 8 atm * (101.3 kPa/atm) = 8.103 x 105 Pa
b) Gauge pressure = P - Patm = rho*g*h = 8.103 x 105 Pa
c) F = P*A = 1.61 x 106 N directed upwards
d) Same formula, F = P*A = 1.621 x 106 N directed downwards
e) The difference between the two forces is 10,460 N directed downwards, so there needs to be an upwards force of 10,460 N to hold the window in place.
7.
a) The mercury will flow into the water since the pressure at the bottom is so much greater than the pressure of the water (P = P0 + rho*g*h, rho of mercury is much greater.)
b) When the mercury drops, we can say the distance it drops is H.
The height of the mercury column is then (1.0 m - H). Since the volume is conserved, the height of the water column is (1.0 m + H).
rhoWater*g*(1.0 m + H) = rhoMercury*g*(1.0 m - H)
Solving for H, H = 1.863 m.
Tuesday, December 2, 2008
The Pressure is Getting to Me... - Manometers and Buoyancy
Solutions:
Sphere problem #3 on the handout from today:
a) When the sphere is not submerged, T - mg = 0 so m = T/g = 2 kg.
Once the sphere is submerged, T + Fb - mg = 0 so Fb = mg - T. Since Fb = rho*g*Vsub, V = (mg - T)/(rho*g) = .00153 m^3
b) density is m/V = 2 kg/(.00153 m^3) = 1307 kg/m^3
Book Problems:
22. Apparent mass comes from apparent weight. If you know the apparent weight (Fn) you can divide by g to get the apparent mass.
In this case, Fn = mapparent * g = 60.6 N
Drawing a FBD of the rock at the bottom of the container, Fn + Fb - mg = 0. Fb = mg - Fn = rho*g*V. This enables us to get that the volume is .002016 m^3.
The density is therefore 8.2 kg/.002016 m^3 = 4067 kg/m^3
24. Answer: 3.0 x 104 kg
Here's the FBD:
Since the balloon is in equilibrium, Fb - mcargo*g - mballoon*g - mHelium*g = 0.
We can find the buoyancy force because we can find the volume of a sphere from the radius, and the density of air is 1.29 kg/m^3 (from p. 276). The same thing is true for knowing the mass of the Helium.
(I think in tutoring, I forgot about the mass of the Helium...)
The only unknown is the mass of the cargo, so you can solve!
26. Using the same method as in #9, we can find that the density of the metal is 8.94 x 103 kg/m^3. From page 276, we can see that this is copper.
27. Same method, again. Density = 1.03 x 103 kg/m^3
29. Finding the definition of specific gravity (SG) is a bit tricky in the chapter, which is why coming to tutoring made this a bit easier.
SG is the ratio of a density and the density of water. For example, if the SG = 0.79 for the alcohol, it means that the density is 790 kg/m^3.
Using this info, this is another problem that is along the lines of 9, 26, and 27. You should find that the density of the wood is 850 kg/m^3, which means that its specific gravity is 0.85.
30. This problem is a bit tricky because it's hard to picture what's happening. If we say that the iceberg is a cylinder, with cross section A, we can say that the part below the water is of depth D, and the part above the water is height H. The fraction the problem is asking for is H/(H+D).
Knowing the SG of both the iceberg and the ocean water, we know the densities of both. From a FBD, we can see that Fb - mg = 0.
Fb = rhoocean*g*Vsubmerged = rhoocean*g*A*(D)
mg = rhoice*g*V = rhoice*g*A*(H+D)
Setting these equal to each other, we get that D/(H+D) = rhoocean/rhoice = .895. The fraction above the water is then 1 - D/(H+D) = .105
Sphere problem #3 on the handout from today:
a) When the sphere is not submerged, T - mg = 0 so m = T/g = 2 kg.
Once the sphere is submerged, T + Fb - mg = 0 so Fb = mg - T. Since Fb = rho*g*Vsub, V = (mg - T)/(rho*g) = .00153 m^3
b) density is m/V = 2 kg/(.00153 m^3) = 1307 kg/m^3
Book Problems:
22. Apparent mass comes from apparent weight. If you know the apparent weight (Fn) you can divide by g to get the apparent mass.
In this case, Fn = mapparent * g = 60.6 N
Drawing a FBD of the rock at the bottom of the container, Fn + Fb - mg = 0. Fb = mg - Fn = rho*g*V. This enables us to get that the volume is .002016 m^3.
The density is therefore 8.2 kg/.002016 m^3 = 4067 kg/m^3
24. Answer: 3.0 x 104 kg
Here's the FBD:
Since the balloon is in equilibrium, Fb - mcargo*g - mballoon*g - mHelium*g = 0.
We can find the buoyancy force because we can find the volume of a sphere from the radius, and the density of air is 1.29 kg/m^3 (from p. 276). The same thing is true for knowing the mass of the Helium.
(I think in tutoring, I forgot about the mass of the Helium...)
The only unknown is the mass of the cargo, so you can solve!
26. Using the same method as in #9, we can find that the density of the metal is 8.94 x 103 kg/m^3. From page 276, we can see that this is copper.
27. Same method, again. Density = 1.03 x 103 kg/m^3
29. Finding the definition of specific gravity (SG) is a bit tricky in the chapter, which is why coming to tutoring made this a bit easier.
SG is the ratio of a density and the density of water. For example, if the SG = 0.79 for the alcohol, it means that the density is 790 kg/m^3.
Using this info, this is another problem that is along the lines of 9, 26, and 27. You should find that the density of the wood is 850 kg/m^3, which means that its specific gravity is 0.85.
30. This problem is a bit tricky because it's hard to picture what's happening. If we say that the iceberg is a cylinder, with cross section A, we can say that the part below the water is of depth D, and the part above the water is height H. The fraction the problem is asking for is H/(H+D).
Knowing the SG of both the iceberg and the ocean water, we know the densities of both. From a FBD, we can see that Fb - mg = 0.
Fb = rhoocean*g*Vsubmerged = rhoocean*g*A*(D)
mg = rhoice*g*V = rhoice*g*A*(H+D)
Setting these equal to each other, we get that D/(H+D) = rhoocean/rhoice = .895. The fraction above the water is then 1 - D/(H+D) = .105
Monday, December 1, 2008
Under Pressure - Fluid Properties
Remember, knowing the definitions and equations is CRUCIAL in fluid mechanics since the tough parts come from applying them
Question 3. The pressure is the same in each case, as well as the area, and this makes the force exerted on the bottom face of the container the same for all three. There is some downward pressure exerted on the diagonal sides in the second and third container. This makes up for the additional weight of water in the second and third container.
Problems:
9. This is straight plug-and-chug - the force is just P*A = 3.08 x 105 N. The force on the bottom of the table is the same, as the atmospheric pressure is the same and the area is as well.
11. Make sure you convert the area into square meters! 200 cm^2 *(1 m/100 cm)^2 = .02 m^2
The total force is 4*P*A = 19,200 N = mg. Thus m = 1959 kg.
16. The key to this problem is in the hint - the pressure is the same at both a and b.
rhooil*g*H1 + P0 = rhowater*g*H2 + P0
rhooil = 654 kg/m^3
17. The total height of the water over the bottom of the hill is 5.0 m + 100 m sin 60 = 91.6 meters.
The total pressure at the bottom is given by the equation we derived today:
P = Patm + rho*g*h which is the absolute pressure at the bottom.
If we now subtract Patm from both sides, we get that the gauge pressure is rho*g*h = 8.98 x 105 Pa.
18. This is similar to 17. Pabsolute = Patm + rho*g*h so the gauge pressure is Pabsolute - Patm = rho*g*h = 4.02 x 105 Pa
19.
a) The volume of the water in the tube is pi*R^2*H = 3.39 x 10-4 m^3. (Don't forget to convert to meters!) Since rho = M/V, M = rho*V = 0.339 kg.
b) The pressure on the inside surface of the barrel lid is given by Patm + rho*g*h. The pressure on the outside surface of the lid is Patm. Thus the net pressure will be the gauge pressure, rho*g*h. The force is therefore equal to the pressure times the area: rho*g*h*A = 1000 kg/m^3 * 9.8 m/s^2 * 12 m * pi*(.2 m)^2 = 1.48 x 104 N
Question 3. The pressure is the same in each case, as well as the area, and this makes the force exerted on the bottom face of the container the same for all three. There is some downward pressure exerted on the diagonal sides in the second and third container. This makes up for the additional weight of water in the second and third container.
Problems:
9. This is straight plug-and-chug - the force is just P*A = 3.08 x 105 N. The force on the bottom of the table is the same, as the atmospheric pressure is the same and the area is as well.
11. Make sure you convert the area into square meters! 200 cm^2 *(1 m/100 cm)^2 = .02 m^2
The total force is 4*P*A = 19,200 N = mg. Thus m = 1959 kg.
16. The key to this problem is in the hint - the pressure is the same at both a and b.
rhooil*g*H1 + P0 = rhowater*g*H2 + P0
rhooil = 654 kg/m^3
17. The total height of the water over the bottom of the hill is 5.0 m + 100 m sin 60 = 91.6 meters.
The total pressure at the bottom is given by the equation we derived today:
P = Patm + rho*g*h which is the absolute pressure at the bottom.
If we now subtract Patm from both sides, we get that the gauge pressure is rho*g*h = 8.98 x 105 Pa.
18. This is similar to 17. Pabsolute = Patm + rho*g*h so the gauge pressure is Pabsolute - Patm = rho*g*h = 4.02 x 105 Pa
19.
a) The volume of the water in the tube is pi*R^2*H = 3.39 x 10-4 m^3. (Don't forget to convert to meters!) Since rho = M/V, M = rho*V = 0.339 kg.
b) The pressure on the inside surface of the barrel lid is given by Patm + rho*g*h. The pressure on the outside surface of the lid is Patm. Thus the net pressure will be the gauge pressure, rho*g*h. The force is therefore equal to the pressure times the area: rho*g*h*A = 1000 kg/m^3 * 9.8 m/s^2 * 12 m * pi*(.2 m)^2 = 1.48 x 104 N
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