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 dm/dt = -1.26 lbm/min and m = 991.77 lbm	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 m^3/s c)  20.2 m/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 ft^3/s ; Vavg= 3.9 ft/s	05Sep05
3.5	For a) you should use the definition of volumetric flow rate (Vdot, not mdot) for an incompressible substance. You will have to evaluate an integral. Keep your wits about you. Think tiny first, like we did in class. For b) keep in mind the definition of specific gravity - ρstuff/ρwat.	a) (2/3)Uhb b) 0.174 kg/s	05Sep05
3.6	Part (a) is a math problem; just use the given velocity equation and plug in the values given to solve for delta. If you define your system as shown and then follow the CoM method for that system, Parts (b) and (c) are not too hard as long as you keep track of your units. Remember to write down all of your assumptions when computing your mass flow rates!	a) delta = 4.99 mm b) Vtop_dot = 0.25 m^3/s c) vave = 1.25 cm/s
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 degs, 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.	a) V_dot = 0.00471 m^3/s or 74.7 gpm, v = 15 m/s b) dh/dt = -0.00471 m/s c) d) h = 0.353 m	05Sep05
3.10	For an open system you will have a mass flowing across a boundary, so you probably want your system boundary to cut through the pipe. For a closed system there can be no matter entering or leaving the system, so you probably want to choose a system that accounts for all of the gasonline.	a) 3 m^3/min, 6.4 m/s b) 0.06 m/min c) 0.06 m/min
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, V= 4690 cm^3	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
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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 1200 kg/h 24.19% Hx State 6 -- Oil 90.32 kg/h 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 m^3	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	Your plot should be neat and legible with labeled axes. The definition of density will help here. Be sure to consider all of the mass inside this bigger system.	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	Standard open system with a moving boundary. Nothing tricky here other than keeping your units consistent.	a) dh/dt = 0.0143 m/s (up) b) V2_dot = 0.046 m^3/s (out) c) V2_dot = 0.034 m^3/s (out)
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	Clearly identify your open system and remember to use the system boundaries defined in the problem (even if you do not think they are a great choice!). Remember, this is a Conservation of Mass problem, so your diagram should only contain mass or mass flow rate terms. Be sure to list all the assumptions you make about the flows.	a) V3_dot = -0.5m^3/s b) v1 = 0.71 m/s
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 m^3/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	It is very important to understand the chain rule in this problem! For part (b), try assuming that steady-state will exist and then see if you can find a value of h which satisfies the equation you derive. For part (c), you need to use some more of those calculus skills: separate and integrate!	a) V_dot = 3h^2 b) h = 5 m c) h = sqrt(H^2 - 100t/3)
3.34	What does it mean in the language of CoM if the fluid stays at the same height? That is the key to Part (a). Remember that your CoM diagram must only show mass or mass flow rates! List all your assumptions. For Part (b), look at the units to figure out if the time varying rate is a mass flow rate, a volumetric flow rate, or a velocity. Part (c) is a calculus problem; due to the equation the child can drink for an infinite amount of time but the glass will never empty. If you spot that, the rest is just math.	a) 1.59 m/s b) dh/dt = -0.667exp(-t/5) cm/s c) 2.67 cm
3.35	Part (a): choose your system so the liquid melt pouring through the top, and the solid PVC on the right, both cross through your system boundaries. Remember to start with Conservation of Mass, and keep track of the densities because they change between the liquid and solid state. Part (b): choose a new system to make the math a little easier.	a) Vdot = (pi/4)(rhos/rhom)v(Do^2-Di^2) b) Ve = (rhos/rhom)v(Do^2-Di^2)/(De^2-Di^2)
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3.37	Parts (a) and (b) are pretty standard. For Part (c), notice that it is only the top of speed lake that is draining; the bottom of the lake maintains the same volume throughout.	a) h1 = 2m b) h1 = 2.21 m c) dh2/dt = -0.99 m/hr d) t = 1.01 hr
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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 mf5,tomato = 0.741	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
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