| 3.1 |
Don't ever, ever, ever start a problem by writing d∀sys/dt=Σ∀dot,in-Σ∀dot,out! Start with the general, conservation of mass first!
This is not a steady state problem. Learn what steady-state means, not just what term to cross out when you see it.
- The key here is the left hand side of the equation. Use the rate form of conservation of mass, and then focus on finding an expression for msys. You will then need to integate that equation to find msys as a function of time. This will be much easier if you do everything in symbolic form first!
- The msys term includes everything inside your system boundary. Would msys change, then? What about dmsys/dt?
|
At t = 4 min, dm/dt = -1.26 lbm/min and m = 991.77 lbm |
| 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 how you tackle any problem where you are using the accounting principle! |
mdot,3=-40 kg/s, mdot,8=100 kg/s, |
| 3.4 |
In all of these problems, visualize the flow area and how it relates to the geometries you are given.. |
|
| 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
|
| 3.7 |
For part b) be sure to use the normal velocity to find mdot. |
a) mdot,1 = 2.25 lbm/s |
| 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
|
| 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. |
|
| 3.11 |
Similar to in class examples. You should whip this baby. |
|
| 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 modelseveral times.
|
a) 0.49 lbm
|
| 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 cm3
|
| 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 |
| 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% |
| 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 = 2039 kg/h, mffruit = 0.0687, mfsugar = 0.599, mfPectin = 0.0023
|
| 3.17 |
|
mdot,5 = 29.99 kg/h, mdot,6 = 126.5 kg/h, mfsolid,3 = 93.0% |
| 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 |
| 3.20 |
Note that for stream 2 the weight percents should be oil 10% and solids 90%. In the end you should have 14 equations with 14 unknowns, which you can then solve swith your favorite piece od software. (I suggest taking a look at EES - Engineering Equation Solver, available on Tibia. It's easy to use, easy to learn, and a whole lot nicer than Maple for non-symbolic stuff, methinks.)
To guarantee my equations are independent, I always like to write species accounting for each species and then use mass fradtion realtions and/or problem constraints for the remaining equations, always leaving overall mass conservation as a check. Not always the easiest equation set to solve, but a scheme that should always work.
|
mfHx,3=75%, mfSol,3=22.5%
mdot,4=1200 kg/h, mfHx,4=24.19%
mdot,7=2710 kg/h
|
| 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
|
| 3.22 |
|
a) 17.4 kg
b) 29.3 m3
|
| 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 |
| 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. |
| 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 |
| 3.28 |
Will you assume steady-state?
When finding the flow rate at (2), remember that only the component of velocity normal to your system boundary contributes to the flow rate across. That is why it is so important (nay, imperative) to meticulously draw your system boundary!
|
∀dot,2⁄∀dot,1=0.70 |
| 3.29 |
Don't forget that you need the relative velocity normalto your system boundary for mass flow rate.
If you need density for water, use its standard value at STP.
|
|
| 3.30 |
Mass conservation, steady-state, one inlet, one outlet - the mass conservation part should be easy then, yes? The tricky part is, then, finding the exit ∀dot. Since V varies over the exit area, you must use ∀dot=∫V·dA. But you don't know V(A) (V as a function of A), you know V(y) (V as a function of y)! What to do, what to do? I suggest you use dA=W·dy. Why not dA=H·dx or dA=dx·dy?
|
b) 9.0 m/s |
| 3.31 |
If you are going to use mdot=∫ρVdA or mdot=ρVA, you must clearly define where you system boundary is. If you draw a system boundary in one place and use mdot=ρVA at a different place, NO POINTS! (Well, not very many...)
|
400 m |
| 3.32 |
In the first part of part a) you should pick a deforming system with one inlet and no outlets. Do not try to write an expression for d(msys)/dt all at once. Rather, write an expression for msys in terms of h, and then plug that msys into d(msys)/dt. For the second part of part a), you should pick a system with one inlet and one outlet. What is d(msys)/dt now?
Part b) is much like the first part of part a). This time you are draining the tank instead.
One common mistake students make is to integrate stuff too early. E.g., you might try
∫mdot,siphondt = mdot,siphon∫dt
= mdot,siphonΔt.
That is incorrect! mdot,siphon is not constant! Do no integration before it's time!
|
b) 8.56 min
|
| 3.34 |
Just becuase the volumetric flowrate at the exit is a function of time for part b) doesn't mean that your tools and/or method of attack is any different than in other problems. The tools are always pretty much the same; that is, conservation of mass (dmsys/dt=...) and the relations for mass flowrates (stuff like mdot=ρVnA).
In part c), think about what happens when you eait a really, really, really long time, like t→∞.
|
|
| 3.40 |
More species fun! See hints above for similar problems, like, say, 3.20.
|
(a) 2 composition & 6 species acctg
(b) mdot,3 =2510 lb/h
mf5,tomato = 0.741
|
| 5.1 |
Parts (a) and (b) are just kinematics. Take a good look at the review exercise on kinematics starting on page 5-5 and you'll be OK.
Since this is just a particle, it's easy to calculate it's linear momentum: Psys = mV. And since a particle is a closed system, the conservation of linear momentum equation is nice and simple too. |
|
| 5.2 |
Be careful how you draw your free-body (interaction) diagram. Do not put all the forces through a single point. Sketch a realistic system and place the transports of linear momentum (forces) where they actually occur. Remember that the drag force acts opposite to the direction of motion. The sled is moving in part a), but is its state of motion changing? Is anything about the system changing? It must be a ____-_____ system then, huh? In part b), however, we definitely do not have a ____-_____ system. In fact, the Psys term is where all the action is. |
(b) 68 s |
| 5.3 |
(a),(b),(c) Assume system is steady-state. Why is this a good assumption?
(b) Is the weight inside or outside of your system? How can you relate the weight to the tension in cables CA and CB if your system is just around the ring C? |
|
| 5.4 |
Follow the hints in the problem statement. Be careful with units. Note that in this problem you are using two different systems to study the same process. This is necessary because you are looking for two different pieces of information. This is a very common occurrence in problem solving. Be very careful to draw the forces on your free body diagram. |
(b) 4398 lbf in tension |
| 5.5 |
Like the in class example, Lesson 12 |
|
| 5.6 |
Your system should cut the flanges; thus you have both flow into and out of the reducing pipe bend. Carefully apply the conservation of linear momentum to this system paying particular attention to the net pressure force and the net transports of linear momentum with mass flow. (Ignore the words in parenthesis in the problem. You want to find the net force the bolts would apply to the elbow to hold it in place. Because you are neglecting the weight of the system, your answer will have only an x component.) |
Fbolts = 14.93 kN to the left |
| 5.10 |
Start with the general rate form of conservation of linear momentum and make the correct reductions. (It is a closed system, yes? You want the finite time, yes?) By looking for the average impulsive force, you can reduce ∫Fdt to FavΔt |
6.21Wathlete |
| 5.13 |
Be sure to define your system carefully. There is actually only one flow boundary where mass exits your system, not two. It's just an annular-shaped exit. |
0.108 kg |
| 5.14 |
You will need two very different systems here. For part (a) this should be a particle of sand (a cl___ system) as you travel with it. Is this system steady-state? For part (b) your system should be the region of space including the belt and the sand entering and leaving it (an op__ system). Is this a steady-state system?
To get the "advanced" answer for part (b), you have to eliminate the mass of the sand in favor of other quantities. How do you do this? Consider looking at conservation of mass for the time before the belt is completely covered with sand.
|
(a) 6.09° or 52.9°
(b) P = msandgsin(α) + mdotVo
Advanced answer: P = 0.041sin(α)gmdotL + 24mdot
|
| 5.15 |
(1) Recognize that the resistance force can be written as R= kx.
(2) Unfortunately this gives us two variables x and t. Change the variables of integration from t to x by using the following substitution:
If V = V(t) and x = x(t) then it is also true that V=f(x) so dV/dt = (dV/dx)(dx/dt) = V(dV/dx) since V = dx/dt.
Now you can integrate with respect to position since you know something about V vs x, e.g. V = 0 at x = 75 mm. |
V25 mm = 566 m/s |
| 5.16 |
Remember that f = μsN at impending motion only.
That lbm lbf thing is showing up again! You have to figure out how to handle this. (Remember that we are on the surface of the earth...)
In part (b) you find a min/max just like you learned in good ole Calc I>
|
(c) t = 17.88 s |
| 5.17 |
Remember, assume something about the state of motion and then check it. Here that might be assuming they stick together, or that they don't. |
aa = 0.997 ft/s2 |
| 5.18 |
As is the case with many problems involving friction, you do not know what the state of motion is a priori. You must therefore assume some state of motion first, apply CoLM, and then chack that assumption against something else. I assumed that block A did not slide relative to B, solved CoLM for the friciton force in between the blocks, and then compared that value to the critical value of friction when impending motion would occur (which is...)
You must use at least two systems here. Just one is not enough. And if you screw up your free body diagram, there is no hope. (Assume the pulley is weightless and fricitonless so that the force P has the same magnitude wherever your system boundary cuts through it.)
|
(a) No sliding. |aA| = |aB| = 0.667 m/s2
(c) Sliding. |aA| = 5.1 m/s2, |aB| = 0.980 m/s2 |
| 5.19 |
Can you pick the same system for parts (a) and (b)? Why? |
(a) 0.742 m/s
(b) Rx,avgΔt = -18.56 N-s
Ry,avgΔt = (15.00 N-s) + (98.1 N)Δt
|
| 5.20 |
The best you can do is find the ratio of the speeds of the two cars. To get actual numbers for the speeds, you have to assume that one driver is not lying and was actually traveling the speed limit. |
VA=192 km/hr (And the driver of A is dirty rotten liar.) |
| 5.21 |
To find the tension in the rope, where should your system boundary be? What information have you been given by looking at the speed reduction over time?. |
(a) 8.51 km/h
(b) 6.67 N |
| 5.22 |
If you keep in mind the difference between gage pressure (units written as psig) and absolute pressure (units written as psia) life will go more smoothly. When you have an object that is almost completely surrounded by the atmosphere, it's almost always easier to use gage pressures, since the atmospheric contribution will cancel out everywhere else. (See your example from class.) |
Fx = 161 lbf →
Fy = 122 lbf ↓ |
| 5.23 |
You will need several systems here. One should be a closed system over a finite time consisting of... Since we know there is sliding between the blocks after impact, would it make sense to make both blocks and the bullet together as you next system? You can do it, but it's not what i would recommend.
Anyway, if you find that after you look at a single system you don't have enough equations to solve for all your unknowns, that's OK! Just keep going. Hopefully by now you are getting more comfortable with working on problems whose solution you can't see right away, but are nonetheless confident about your approach. |
|
| 5.24 |
Parts (a) through (c) are just kinematics. Take a good look at the review exercise on kinematics starting on page 5-5 and you'll be OK.
Since this is just a particle, it's easy to calculate it's linear momentum: Psys = mV.
Note in part (d) that dPsysdt is proportional to Fexternal. Since this is a closed system, this should make sense. That is, Fexternal is transporting linear momentum per unit time into our system, and it shows up as a rate of change of system linear momentum!
|
(a) t=0 s and 4 s
(b) 169 m
|
| 5.25 |
It is no longer OK just to draw a dotted line around something and say, "I identified my system!" No siree Bobby Boo! You must now identify your system boundary and draw all the external forces that cross that system boundary. Don't let me ever ever ever catch you even thinking about writing Conseration of Linear Momentum before you clearly draw a picture of your system in this way.
OK, since you've done the above, now write CoLM. It's a vector, isn't it? And so, you will next need to identify two cooridinate directions, and write CoLM for each direction. (Is this a steady-state system? I think it is...)
|
TAC=1376 N |
| 5.26 |
You can make your coordinate system anything you want. Why not pick one in line with the block?
The block could move by sliding up or down the plane. Also remember that f = μsN at impending motion only, not just no motion. |
(a) 57.0 N to 319 N |
| 5.27 |
Remember the force in the link connecting the sphere and the block can only act along the direction of the link compression or tension. A little trick: If a single system fails, change it ---- expand it, move one boundary, etc. |
(a) 204 N |
| 5.30 |
Ah, the power of ConAps! You can not solve this problem using just F=ma, can you? You see that F=ma is really just a special case of our more general conservation of linear momentum for a closed system, and the system here is open. And so be sure to draw momentum transport rates at non-flow boundaries (forces) on you diagram, as well as the ones at flow boundaries (mdotV terms).
Hey! The weight of the thing is a force and should be on there too. So should the normal force the ground exerts on the shredder. You don't have numbers for these. Hmmmm...
|
90.6 N → |
| 5.42 |
I would choose my system to be just the ball. I would (as always) work only in symbols until the bitter end. If you do, you should find that the resulting equations look familiar, say, like a grain of sand in 5.22? or perhaps a droplet of water from a fireboat? The only really new thing here is that there are several velocities floating around. For every term in conservation of linear momentum, all velocties need to be referenced to the same inertial reference frame. The easiest one is almost always the surface of the earth. One key for keeping these straight is to use Vb = Va + Vb/a. Here we might write that Vball = Vcar + Vball/car, where what we really need in linear momentum is Vball, not Vball/car. (Now this is vector relation, and so when you go to look at components of it [like in the x or y direction] be careful with negative signs showing up!) |
(a) 0.792 s
(b) 44.9 ft/s |
| 5.43 |
When you use mdot=ρAV the V you need is always relative to the system boundary. And so, see the hint for 5.42 about that relative velocity jazz. Another juicy tid bit of uniqueness in this problem is that the mass of your system is not constant! And so when you go to look at dPsys/dt, things are different than usual. In particular
d(msysVsys)/dt ≠ msys·dVsys/dt
No, no, no, no, no, nonnie no! Rather,
d(msysVsys)/dt = msys·dVsys/dt + Vsys·dmsys/dt
What equation will help you find dmsys/dt?
|
|
| 5.44 |
The system I picked was both the swimmers and the boat together, just before and after the dive, making it a clo___ system over a fin__ time, yes? You should be able to reduce linear momentum to a pretty simple form using this system. All the fun comes in paying very close attention to absolute velocities and relative velocities. Remember, you are given the relative velocities of the diving swimmers to the boat, not the absolute. You want the absolute velocities in the linear momentum. And so, once again remember
Vb = Va + Vb/a
also keeping in mind that this is a vector relation. When you go to look at the horizontal components negative signs may show up, depending on your assumed direction of the boat's velocity.
In part b) you get to so all this good stuff again, but you will need to pick two time frames, which will be....
|
(a) 1.154 m/s ←
(b) 1.292 m/s ←
(c) 1.290 m/s ← |
| 6.1 |
If you are going to find the tension in the cable, then your system boundary must cut through the cable. Otherwise the force in the cable in internal to your system. Remember, everything onthe right hand side of our accounting equations has to with what's crossing the boundary of our system. |
Tcable = 2000 lbf |
| 6.2 |
The drawing is misleading as the chute BC is NOT straight but curved. All you know is that the mass leaving the chute at B makes angle of 10o with the horizontal as shown in the figure. (A straight line passing through BC actually makes an angle of 27o with the horizontal.) |
FB = 423 lbf ↑ |
| 6.3 |
After picking your system the next most important thing is to consider the point about which you should sum moments. Note that there is no friction between the horizontal bar and the inclined surface; however, there is a reaction force acting on the bar and normal to the inclined surface. Also assume that the two chains are parallel to the inclined surface. (Otherwise the system wouldn't be stationary!) |
F2 = 116.7 lbf
R = 17.32 lbf acting on the bar and normal to the inclined surface |
| 6.4 |
Ok, so you're thinking "This thing is moving. Why am I given the static coefficient of friction?" Remember that friction is a contact force between surfaces. What makes it kinetic or static friction is the relative motion between the surfaces. Here there is no relative motion between the wheel and the rail, otherwise the wheel would be slipping, like skidding you tires on the pavement. That might sound cool, but it doesn't make your car go.
Also don't make the fatal mistake of assuming that since the monorail is translating it has no angular momentum. Oh contraire mon ami! Since LP is "moment of linear momentum", you can always find some P for which LP is not zero. |
0.340g ← |
| 6.5 |
(a) Try CoLM first. Not enough equations? Go on to CoAM. Note that the reaction of the ground on the wheel is directly below G.
(b) With the trailer translating to the right with an acceleration of 4.5 m/s2, does dP/dt = 0 anymore? (No...) How about dLsome point/dt? (Maybe, maybe not... Does it depend on what some point is?) Anyway, you should end up with 3 unknowns and 3 equations. |
(b) Vertical force of ground on wheel is 10,179 N up. Vertical force of hitch on the trailer is 1350 N down. The hitch has to pull down to keep the trailer translating. Horizontal force on trailer is 4050 N to the right. |
| 6.9 |
Assume that the outline of the system corresponds with the outline of the fire hydrant. This is an open, steady-state system with water flowing in at the bottom and flowing out at the horizontal discharge.
(a) Application of conservation of LM in the x direction. (Recall that the pressure at the discharge from the hydrant is equal to the atmospheric pressure.)
(b) Same as (a) only this time in the vertical direction and you now must cope with the pressure at the inlet.
(c) Application of conservation of AM around the center of the base of the hydrant. |
Mbase = 324.0 lbf-ft (CW) |
| 6.10 |
Application of cons. of AM and LM to a closed SS, system, yes? Since you want to know information about the reactions at E, your system must cut the frame where the vertical support goes into the ground at E What kind of reactions are possible at E? You can either memorize the reaction types given in your text, or think about how the support restrins the motion of the system (my preference). |
Ex=90 kN ←
Ey=200 kN ↑
ME=180 kN-m CCW |
| 6.12 |
Consider a system that includes the massless rod and the ball. What forces and moments act on this closed system during the motion of the ball from position A to position B? |
|
| 6.13 |
You need all your toys for this one, including CoM, CoLM, and CoAM for an open, steady-state system. Your choice for a system should consist of the conveyor belt device and only the coal on the belt. Why? Because you do not know the weight of the coal dropping onto the belt, or its G either. As such, you'll have to find the velocity components of the coal as it is hitting the belt. (Vx = 3 m/s →; Vy = 3.27 m/s ↓ If this doesn't make since to you, think about this. Is the angular momentum of the coal at 1 the same as at A? No siree!) As always with CoAM, look for a point about which to take moments so that your life is easier. |
Dx = 0; Dy = 2900 N ↑> |
| 6.14 |
Treat the rod as the system and apply AM and LM to the rod. Also recognize that the rod is translating to the left at a known velocity. (Hint: Fc should be less than the value for when there is no acceleration. Another hint: Do you know the direction of the reaction at C? I think you do...) |
At B: Bx = 6.971 N ←; and
By = 23.36 N ↑ |
| 6.15 |
In part (a) it is best to take only the crate as your system. As always, start with the general principle first (here, CoLM) and then reduce it for your particular system. In which term does the acceleration show up?
In part (b) you gotsta' take both forklift and crate as your system. Is there any point about which you can look at CoAM such that dLpoint/dt = 0? Careful! You know G for the crate and G for the forklift, but do you know G for the two combined? |
(a) 9.66 ft/s2
(b) 2.88 ft/s2 |
| 6.18 |
This is simply "a moment is a force multipied by a perpendicular distance" problem. (Don't forget that you should always write M with a subscript to refer to the point about which you are taking the moment.)
Part (b) is greatly simplified if you break the forces into components and then find net moment as the sum of each component, e.g. break force A into an x and y component. |
(a) About Point O: 5600 ft-lbf CW ; About Point P: 2600 ft-lbf CW
(b) 135.7 N-m CW; 200.0 N-m CW |
| 6.19 |
For part (a), use your knowledge of rotational velocity: V = rω.
Part (b) is just the idea that the angular momentum of a particle is the "moment of linear momentum," i.e., LP = r×(mV). To get the direction of the angular momentum, try the thought experiment in which you tie a thread to the force and tack the other end to the point about which you are calculating L. Which way does the thread wind up?
|
(b) LP,A = 0.313 (kg-m2/s) CW; LP,B = 0.625 (kg-m2/s) CW |
| 6.24 |
There are several different systems you can use here. Probably the easiest one is just the bus by itself. Be sure to draw the friction force between the pad and the bus in the correct direction, which is opposed to relative motion (or potential relative motion) between the surfaces in contact. If you just draw friction opposite plain-ole motion and forget the relative part of it, you're bound to get the direction wrong here. You can write conservation of angular momentum about any point you want. Why not pick one that makes some moments go away, say, perhaps, about the tipping point? You can run, you can hide, but eventually you will have to deal with how to calculate Lsys. If you follow the hints above, you'll have to do that in this problem. It's not so hard! One more time everybody! Angular momentum is the "moment of linear momentum," i.e., LP = r×(mV).
|
(a) 24.2 ft/s2
(c) 0 .75
|
| 7.1 |
Consider the lunar module as the system for both Parts (a) and (b). Both approaches to the solution should give the same answer. Be very careful with signs. The first answer may seem strange since it says that the answer does not depend upon the direction of motion as long as the vertical speed of the lunar module is 3 m/s. |
If V1 = 3 m/s ↑ or ↓, then z1-z2 = 4.89 m |
| 7.6 |
a) Apply the definition of mechanical work.
b) To make life easier, apply the mechanical energy balance to a system that includes the collar and the spring. (If the spring is outside the system, then you'll have to calculate work done by the spring too.) Remember that spring energy is related to the spring extension from the unextended length; i.e., in State 1 the spring already has energy stored in it since it is extended 0.5 m. |
W1→2 = 65.0 N-m |
| 7.9 |
Be sure to pay attention to the sign of the work for each process. When we write Wa→b = -∫(pdV), the resulting work is postive if it is into (on) the system. |
W1→2 = 129 kJ, Wnet = -191 kJ |
| 7.10 |
The energy change of the system used in the energy balance is the total energy change of the system, including KE, PE and now, U. And pay attention to units (ft-lbf ≠ Btu) |
U2 - U1 = 0.462 Btu |
| 7.11 |
Another closed system energy conservation, except this time you need the rate form. (If you have been sloppy with your dots - writing min when you mean mdot,in, for example - you'll need to cut that out real quick. Conservation of energy is exceedingly unforgiving [i.e., wrong!] if you mix the finite time form and the rate form.)
In part (c) you'll need to integrate the Qdot term.
| (b) dE/dt = 2.0 kW at 1.2 h
(c) ΔE = 5.00 kW-h = 18,000 kJ |
| 7.12 |
When looking at CoE, remember that Win is Win,net. That is, the work term is the sum of all works into the system.
A commom mistake is to confuse u with U. The Esys term on the left hand side of CoE is an extensive property - energy, not specific energy. Sticking with the lower-case-letter/specific-property, upper-case-letter/extensive-property notation helps you avoid this problem.
|
a) WPW = 2790 J |
| 7.13 |
We do have compression expansion work going on here, don't we. But we don't have P = f(V), and so we have to calculate the work the old-fashioned way; i.e., W1→2 = ∫Fds. Don't forget the contribution of Fatm on the work. |
a) Wout = 0.116 kJ |
| 7.15 |
The only tricky thing here is the application of the ideal gas equation for part b). Other than that, this is similar to the problem(s) we worked in class. |
c) 3.67 HP |
| 7.17 |
This problem involves shaft power and electrical power. |
c) 112 lbf-ft |
| 7.18 |
System, system, system! I suggest you try both the heat exchanger and the turbine as your system in part a). Is this the best system for part b)? Maybe, maybe not... |
b) 3163 MW |
| 7.19 |
Just an exercise in the different forms of work and power. (By the way, don't forget that time frame is still an important thing to think about when you apply the accounting principle. Here you will want the rate form, yes? |
(a) Wdot,shaft,out = 800 Btu/s
Wdot,elec,out = 780.8 kW |
| 7.20 |
You'll need both CoM and CoE. There are several systems that you can pick. In fact, you must pick at least two different sytems to solve for everything asked. |
a) 704 kg/min
c)2990 N-m |
| 7.21 |
You can get the work from good ole W1→2 = ∫pdV. Where do you get your temperatures? What does PV = C mean for an ideal gas? |
|
| 7.24 |
More energy fun. We only jazz it up bit with a little Wdot,elec = VI action. |
(b) 35.1 kW |
| 7.25 |
More of the same... (Repetition is important for learning. Repetition is important for learning.)
This problem emphasizes the important of thinking about your system. The ouputs of some components are inputs to other components, but you won't be able to put numbers on any of them unless the cross of the boundary of your system. And so, do you think you will have to look at several different systems? Is this a rhetorical question? |
(a) 120 rpm and 100 rpm
(c) 9.520 hp
(e) 10.58 hp
(f) 30.87 ft-lbf |
| 7.26 |
Pay close atention to units here. h usually ends up in kJ/kg. V2/2 and gz, however, are usually in m2/s2? Are these the same units?
The density of air is not constant here. Vdot,3 = Vdot,2 + Vdot,1 won't work, will it? That's why you should always start with the general conservation principle first. |
(b) 0.249 m3/s |
| 7.27 |
A common mistake is always to blindly write h2 - h1 = cp(T2 - T1) without thinking. That equation is strictly true for and ideal gas with constant specific heats. You've got an incompressible substance instead. That means h2 - h1 = c(T2 - T1) + something else. Don't forget about "something else." |
(b) 218 N-m
(c) 95.5 amps |
| 7.29 |
Nice application of pdV work and conservatin of energy for closed systems. Can you ignore the KE and PE of the system? Is the piston cyclider flying through space? Are you running it up or down the stairs in New Hall? |
W1→2,in = -113 kJ |
| 7.31 |
The clever thing to do here is to make just the air your system so that you now have a closed system. Now the water is pushing on the air, and therefore does _____ on it,yes? What kind of ____ is this? How do you find the expression to integrate, then? How do you find u2 - u1 if you need it? Ah, my young ConApsters... 'tis an ideal gas, for which PV = mRT, and a couple of new equations from section 7.4 apply as well. |
Win = 16.6 kJ |
| 7.32 |
Conservation of mass and energy are both helpful in this problem.
Unlike ideal gases, the change in specific enthalpy of incompressible substances is a function of temperature and pressure. |
(b) 4.08 HP |
| 7.33 |
Again, air can be treated as an ideal gas, which will help you through the missing equations once you've exhausted mass and energy for all they're worth.
Now we have often gotten rid of V2/2 terms in CoE, claiming that they are small compared to the other stuff in the equation. Are they small here? If the whole purpose of a nozzle is to do something with these terms, then my guess is no...
| A2/A1 = 0.236 |
| 7.34 |
This looks a lot like 7.33, but now we have liquid water instead of air. Is liquid water and ideal gas? What substance model will you use, then?
| (b) V2 = 17.8 m/s |
| 7.35 |
In CoE problems (like all accounting equation problems) start with your system and basic principle. Then make assumptions and whittle it down. If you find you still don't have enough, use some extra constituitive relations. In (b), the constituitive relation is Ohm's "Law". In (d), it's Qdot,conv = hA(Tsurf-Tamb). |
(b) 0.0144 watts; 1.2 mA
(d)43°C |
| 7.36 |
The power factor here for both input and output is one. The temperature is a function of the number of sides you assume convect thermal energy away. |
If number of sides for Qdot is 5: 95.0°C
If number of sides for Qdot is 6: 83.3°C |
| 7.37 |
It's OK to treat He as an ideal gas.
Similar to 7.39 except this time you must find the change in internal energy, work and heat transfer for each process yourself. There are four processes at which to look. Be careful with your time frame here. Qin=Wout only works for net values for the entire cycle time frame, yes? |
Win,net=128 kJ |
| 7.38 |
Note that we are doing everything on a per unit mass basis in this problem. In analyzing this cycle, you are not doing anything new. Simply apply CoE and/or the definition of compression/expansion work to the process going from (1) to (2), and then repeat for (2) to (3), (3) to (4) and (4) back to (1).
When you total Qin for the cycle, how should it compare to Win for the cycle? What should ΔU for the entire cycle be?
|
(a) Win,1→2/m = 459 kJ/kg
(b) Power cycle
(c) MOP = η = 56.6%
|
| 7.39 |
There is really nothing new here in terms of analysis. Apply pdV work and conservation of energy to each process in the cycle and you're good to go. |
|
| 7.40 |
Note that the COP for the resistance heating is one. Also note that this is the smallest the COP can be for a heat pump! |
(c) $167.52
(d) $506.37 |
| 7.42 |
Apply the work-energy principle to either the OH crane plus the bucket and rope or just the bucket. Be careful to explain how the WEP simplifies for your model.
There are at least 3 possible systems for this problem: System I is just the bucket (neglecting any rotation of the bucket); System II is the bucket and the cable up to the pivot point (neglecting the mass of the cable); and System III is the bucket, the cable, and the traveling crane. If you understand why each of these systems gives you the same answer you will have a deeper understanding of the work-energy principle (and also the conservation of energy - coming soon!) |
|
| 7.43 |
As with many examples at which we looked in class, you probably want to break this up into several steps, applying the work energy principle to each. |
20.3 ft |
| 7.44 |
This is yet aother application of the work-energy principle. Now the cord acts like a spring, and you can put it in your system, or make just the bungee jumper you system. Though you'll get the same answer for either systems, the former system is easier to use. Why? Because it's the least amount of ... work... (hint, hint)
Hey! If you try to set the weight of the jumper equal to the force in the bungee cord, you're gonna' get oh so many wrong answers, and the jumper might go splat! (Not a good outcome...) Why? |
(a) 533 lbf/ft |
| 7.45 |
Can you use work energy for the entire process? No! There is an impact going on here. And so, at least for the impact part, you will have to use linear momentum. Also be careful with the spring energy terms. There is an initial deflection in the spring. Don't you make those algebraic boo-boos either... |
0.271 ft |
| 7.46 |
This is an application of the mechanical energy balance. At least three systems seem plausible: Block A, Block A and Spring, Block A and Spring and Cable past the pulley. The first two systems have the disadvantage that the forces doing work vary with position. The last system when applied correctly has a constant force that moves through an easily computed distance giving the mechanical work done on the system. The answer for all three systems should be the same; but the latter system requires the least amount of calculation. |
|
| 7.48 |
At first look you might assume that this is only a MEB problem with a system including both blocks and the spring; however, it has the telltale impact (friction inside the system) that makes the MEB invalid. (Mechanical energy is not conserved in this process; however, the all inclusive energy is conserved.) Thus, this requires analysis as a two-step process: 1 → 2 Using a closed system of the two blocks, apply the finite time form of conservation of linear momentum. In addition, you must assume that the time interval is so small that all external forces are negligible during this process. (The terms perfectly plastic or perfectly inelastic means that the two blocks stick together after they contact.)
2 → 3 Now using a closed system consisting of the two blocks and the spring, you can apply the MEB for a finite time interval and solve for the final displacement. [Again this last portion could have also been done using CofLM for just the two blocks and then treat the spring as an external force that varies with position.] |
(b) 0.817 m |
| 7.69 |
Can you use the Work-Energy Principle across a collision? No! Can you use Conservation of Energy? Yes! But is it helpful? No! That's becuase you have no information about internal energy during the collision. And so, what will you use across the collision? Linear Momentum, perhaps? |
|
| 8.1 |
Similar to the in-class example with the motor. Don't forget that you need absolute temperature. (°R = °F + 460) What is the source of extra irreversibility, and thus Sdot,gen, in part d? |
b) 0.451 B/h-°R
d) 0.480 B/h-°R |
| 8.2 |
Steady-state means that no system quantity changes with time, including dSsys/dt. This is true even if Sdot,gen ≠0. |
Sdot,gen=-2.50 kW/K, which means this Mission is... |
| 8.3 |
With entropy accounting you must be especially careful with to signs on energy transfer, as well as using the temperature at the boundary where Q or Qdot crosses. This T may or may not be the same as the rest of the system.
How do you think the electrical power input and output would compare to each other if the device were reversible, and thus Sdot,gen? |
|
| 8.6 |
More entropy fun! Assume system is steady-state. |
|
| 8.7 |
OK you potential patent examiners, whatdya' think? |
I think... ...no. |
| 8.8 |
Drawing a picture and properly identifying the energy transfers with correct directions is key here. You need both energy conservation and entropy accounting. For S-accounting, be sure you specify a boundary for which you know the temperatures where your Qdot terms cross. |
|
| 8.9 |
Drawing a picture and properly identifying the energy transfers with correct directions is key here. You need both energy conservation and entropy accounting. For S-accounting, be sure you specify a boundary for which you know the temperatures where your Qdot terms cross. |
b) COPmax = 12.7, Wdot,ideal = 7895 B/h
c) It's valid alright. |
| 8.10 |
This is similar to the previous problems, although it's a finite time frame, where your time frame is one cycle. The common thread amoung these last few problems is that Sgen or Sdot,gen > 0 always hurts performance. |
|
| 8.15 |
This is similar to our in-class pump example, and so you should have some physical insight here. |
(a) s2–s1=0; 30 kW
(b) 50.9 kW; 0.0697 kW/K |
| 8.16 |
Similar to our in-class nozzle example. Be sure to think about what's happening in part c. A nozzle is an energy conversion device. What kind of energy to what other kind of energy? |
b) Vout = 604 m/s, Sdot,gen =0.204 W/K |
| 8.17 |
Careful! You need the finite time form of S-accounting, not the rate form. In thinking about part d, remember that Sgen measures irreversibility, and Qdot through a finite ΔT is a big one. |
a)17,560 KJ
c) 56.4 KJ/K |
| 8.19 |
Cool! We're blowing stuff up! What does reversible and adiabatic imply from S-accounting? (Ans:, s(little s) = constant) From knowledge of this fact, can you somehow get a relationship for P=P(V) or at least P=P(v) so that you can evaluate W1→2=-∫PdV? Sure you can. Air is an ideal ga... |
|
| 8.21 |
S-accounting is most useful when used in conjunction with energy conservation. Since we know that entropy generation always hurts performance, what value of Sgen is going to maximize power out? |
987.8 kW |
| 8.22 |
|
|
Copyright © 2017 | Thomas M. Adams, PhD | www.rose-hulman.edu/~adams1