| Chapter 3 -- Conservation of Mass | |||
| Updated --- 11 Sept 06 0930 | |||
| # | Hint | Answer | Update Date |
| 3.1 | a) Remember that there is no conservation of volume principle. Apply the
conservation of mass principle to your system to find the instantaneous
rate of change. To find the amount of mass at any time you must
integrate the conservation of mass rate equation. To do this, separate
variables to get dm = [function(t)]dt and then
integrate both sides of the equation. Do not assume that t is a
constant when you do the integration. Suggest you use a definite integral,
but you can use an indefinite integral if you evaluate the constants
correctly. b) Think about what mass is actually in this larger system and how you would proceed. |
a) At t = 4 min |
05Sep05 |
| 3.2 | Clearly identify your system, develop a model using explicit assumptions, and show your work beginning with the rate-form of the appropriate governing equation. This is how you tackle any problem where you are using the accounting principle! | mdot,3=-40 kg/s, mdot,8=100 kg/s, | 05Sep05
|
| 3.3 | a) Use the definition of volumetric flow rate to set up the integral you
will need to evaluate. What should you use for dAc?
Note that u only depends on y, so which is better dAc
= d dy or H dx.
b) Apply conservation of mass. There is no conservation of volume. c) Consider the definition of mass flow rate when the velocity profile is uniform. |
a) 0.2531 m3/s |
05Sep05 |
| 3.4 | a) How can you perform an integral over this area if you only know the
velocity at discrete points? What if the velocity was assumed to be uniform
over each sub region? b) Apply conservation of mass. What's your system? |
a) Vdot=8.58 ft3/s ; |
05Sep05
|
| 3.5 | For a) you should use the definition of volumetric flow rate
( For b) keep in mind the definition of specific gravity - ρstuff/ρwat. |
a) (2/3)Uhb b) 0.174 kg/s |
05Sep05
|
| 3.6 | |||
| 3.7 | For part b) be sure to use the normal velocity to find mdot. | a) mdot,1 = 2.25 lbm/s | 05Sep05 |
| 3.8 | Remember that only the component of velocity normal to your system boundary contributes to mdot. (And you, if you haven't been drawing system boundaries, you betta' start right away!) | a) 471 kg/s c) With θ=60°, V2=2.5 m/s |
05Sep05 |
| 3.9 | It's all about choosing your system and making the conservation of mass reflect what's going on in the picture of that system. Since we've given you hints on choosing systems, be sure you actaully draw the system boundaries. Pay particular attention to where your conservation of mass terms are on those pictures. | 05Sep05 | |
| 3.10 | |||
| 3.11 | Similar to in class examples. You should whip this baby. | 05Sep05 | |
| 3.12 | When using the ideal gas model, be careful to handle units
correctly. In part b) see if you can't solve this using ratios of pressures and temperatures instead of writing the full blown ideal gas model several times. |
a) 0.49 lbm | 05Sep05 |
| 3.13 | a) You can base your mass fraction calculations on the actual number of moles or on a hypothetical amount. Also keep in mind that there is the universal gas constant (mole basis) and the gas constant specific to each ideal gas (mass basis). | mfO2 = 0.169, |
05Sep05 |
| 3.14 | It's probably easiest to assume 1 lbmol of mix for analysis. You'll need density for part c. Why not use the form of the ideal gas equation with density in it? |
a) mfCH4 = 0.484, mfN2 =
0.065 c) 771 lbm/min |
05Sep05 |
| 3.15 | On any accounting of species type problem remember that you will
always have an extra equation left over. Use it as a check. One method that should always work is to use the accounting of species equation(s) for all species along with composition equation(s) (∑mfi = 1). Then use overall mass as a check. |
mdot,2 = 870 kg/h, mfbenzene,3 = 38.7% | 05Sep05 |
| 3.16 | Again, if you want a method that should always work, you can use species accounting for each species and then any necessary composition equations (∑mfi = 1). (Of course, you also need any given information as well.) In this problem you should end up with 10 equations and 10 unknowns. | Stream 6: mdot = 2036.5 kg/h, mffruit = 6.87%, mfsugar = 60.00% mfPectin = 0.13 % |
13Sep05 |
| 3.17 | mdot,5 = 29.99 kg/h, mdot,6 = 126.5 kg/h, mfsolid,3 = 93.0% | 05Sep05 | |
| 3.18 | Is there anything fundamentally different with accounting for species on a mol basis rather than a mass basis? Practically speaking, doesn't this just mean writing n and ndot instead of m and mdot? | nfCS2 = 0.0626 | 05Sep05 |
| 3.19 | |||
| 3.20 | Note that for stream 2 the weight percents should be oil 10% and solids
90%. You have 14 unknowns and require 14 independent equations to solve.
Source of equations: 3 equations from composition on streams 3, 4, 5;
2 cons. of mass on mixing tee and condenser (only one species); 2 species
acctg on evaporator (no solids); 6 species acctg for extractor and filter; 1
constraint eqn about the ratio of hexane mass to bean oil mass in the filter
cake.
Note: You can only write as many species equations for any given system as their are species within or crossing the boundary of the system. |
State 3 -- Effluent 75% Hx, 22.5% Solid State 4 -- Filter Cake State 6 -- Oil State 7 -- 2710 kg/h |
14Sept05 |
| 3.21 | a) Start with cons. of mass first. Remember there is no general law
"conservation of volume" b) SS means nothing in the system is changing with time. All d()sys/dt terms are therefore 0. c) You must integrate dh/dt with respect to time to find time for liquid to drop from 80 to 70 ft. |
b) 80 ft c) 48.1 min |
05Sep05 |
| 3.22 | a) 17.4 kg b) 29.3 m3 |
05Sep05 | |
| 3.23 | Clearly identify your system, develop a model using explicit assumptions, and show your work beginning with the rate-form of the appropriate governing equation. This how you tackle any problem where you are using the accounting principle! | -5 lbm/s; 3 lbm/s; -17 lbm/s | 05Sep05 |
| 3.24 |
|
At t = 2 min, m = 146.7 kg and dm/dt = -50.0 kg/min. | 05Sep05 |
| 3.26 | a) Try an open system with one moving boundary and one
outlet. b) Remember this is a closed system. |
b) 2.81 cm/s | 05Sep05 |
| 3.27 | |||
| 3.28 | Clearly identify your open system and apply conservation of mass. Where necessary use the definition of mass flow rate to relate density, normal velocity, and flow area. Remember that volume is not usually conserved. | (a) 0.7 (b) 2.05 | 07Sept05 |
| 3.29 | |||
| 3.30 | Clearly identify your open system and apply conservation of mass. Where necessary use the definition of mass flow rate to relate density, normal velocity, and flow area. Because the outlet velocity profile is non-uniform, you must integrate the product of local velocity V over the flow area. [Hint: You can always pick dA = dxdy, but if velocity only depends on y use dA = Wdy. Why would dA = Hdx be a BAD choice? So you want the largest dA for which the velocity is essentially constant! If V was a function of both x and y, then you must use dA = dxdy.]] | (a) 9.00 m3/s (b) 9 m/s |
07Sept05 |
| 3.31 | Clearly identify your open system and apply conservation of mass. Where necessary use the definition of mass flow rate to relate density, normal velocity, and flow area. | L = 400 m | 07Sept05 |
| 3.32 | Select an open, deforming system that consists of the gasoline in the tank, so the volume can be written in terms of h and tank diameter D. Apply conservation of mass to this system. Simplify assuming that the density is uniform and constant. Be careful with the math. It is suggested that when required you separate variables and then use a definite integral. It's possible to use and indefinite integral but you must evaluation the constants. | (a) 6.41 min; 5.44 m (b) 8.56 min |
07SEpt05 |
| 3.33 | |||
| 3.34 | |||
| 3.35 | |||
| 3.36 | |||
| 3.37 | |||
| 3.38 | |||
| 3.39 | |||
| 3.40 | |||
| 3.41 | Eight unknowns. Apply species accounting to get 6 equations and then use composition at 3 and 5 to obtain remaining equations. | (a) 2 composition & 6 species acctg (b) m3
=2510 lbm/h |
14Sept05 |
| 3.42 | Try four systems: the mixing tee, the humidifier, the room, and the splitter tee. Other combinations will also work. Remember that for a splitter tee the mass fractions do not change between the inlet and the outlet. This should help you generate some additional equations. | mdot8 = 503.5 kg/min mfw4 = 0.0124 |
11Sept06 |
| 3.43 | |||
| 3.44 | |||
| 3.45 | |||
| 3.46 | |||
| 3.47 |