Homework hints
Notes:
Assignments and due dates are in the Schedule section. Also note that answers are not guaranteed to be correct. I can make a numerical goof as easily as you can. Work towards understanding, not towards a number.
| Problem | Hint | Answer |
| 1-XXC | No format required for concept questions. | |
| 1-51 | (b) 39,312 kJ | |
| 1-71 | Remember that for convection you must use the surface area. What area is that? | 47 mW |
| 1-132 | 177 W in winter, 84.2 W in summer | |
| 2-27 to 2-31 | Format not required for these. Start with the most general form first, and then work your way down. (Do this for the other problems too.) | 2-28: a) steady b) 1-D c) yes d) variable |
| 2.43 | Boundary conditions (BCs) come from energy balances on the boundary surfaces, and boundaries can not store energy. Notation is very important here, too. T1 means something very different than T(r=r1). That matters! (And if that doesn't get your attention, points will go away when you mess it up.) | BC#1: -k·dT⁄dx)x=0 = 33,820 W/m2 |
| 2.49 | Again, boundary conditions (BCs) come from energy balances on the boundary surfaces, and boundaries can not store energy. | BC#1: -k·dT⁄dx)x=0 = 33,820 W/m2 |
| 3-25 | ||
| 3-51 | Remember the definition of contact conductance. | Qdot = 142 W, ΔT = 6.4 °C |
| 3-71 | Note that as a first approximation you can have resistors in parallel as well as in series. | Q̇=470 W |
| 3-81 | Careful with diameters vs. radii. | Q̇=83.4 W |
| 3-131 | 960°C at the tip | |
| 3-133E | Note the mistake in the units for k. Units should be B/hr·ft·°F, not B/hr·ft2·°F English units! Used exclusively by NASA. Weird, I know, but you have to be able to do it. Anyway, how are you going to model the spoon handle? It's long. It's skinny. Looks like a f... Also, "across the exposed surface of the spoon handle" means between where the spoon is in the soup and the tip of the spoon. What tip boundary condition do you want to use? | 125°C |
| 3-135 | You will need to do an energy balance on the whole darn motor, and also model the shaft as a fin to get an expression for Q̇. The fin this time is a fin with a specified tip temperature. | 87.7°C |
| HW, Extra fin problem [Download] | The calculations should be pretty short here. The problem really focuses on the concept of fin efficiency more than anything. Also note that most practical fin problems would look something like this. (I have made it a little easier by finding the efficiency for you rather than making you use that nasty equation with all the Bessel functions in it!) | |
| 3-142 | Neglect heat transfer from the tips to get these answers. I would also recommend that you find ε first, much like we did in the in-class example. I also highly recommend that you find the efficiciency for a single fin first! That intermediate answer is also given here. | ηf = 0.935 Q̇ = 17,400 ε = 7.10 |
| 4-33 | Model the thermocouple as a lump (as in lumped capacitance) to get velocity. But wait! Is the "V" in the time constant velocty, or is it volume? Hmm... There is a mistake in the problem. For the relation for convective heat transfer coefficient, D is the diameter of the thermocouple, not the duct. We'll learn more about this later. | V = 4.50 m/s |
| 4-26 | 422.3°C | |
| 4-54: Use lumped capacitance | Treat each plate as a "slab" subject to 1-D, transient conduction. | Tsurface, t=10 min = 448°C |
| 4-54: Do NOT used lumped capacitance | Treat each plate as a "slab" subject to 1-D, transient conduction. | Tsurface, t=10 min = 445°C |
| 4-68 | To find Q, you must find Qmax first. What is Qmax? (Q with no dot, that is.) | |
| 4-94 | This is a semi-infinite solid with a constant heat flux boundary condition. That BC gives a different solution than those in the notes, but it is still just a "I can assume this model, go look up the solution!" type problem. | T = 64.5°C |
| 4-98 | Where will the wood ignite first? You're not given any dimensions here, other than "thick." How are you going to model the thing? Hmmm.... There's one surface I'm interested in... | T = 360°C - won't ignite |
| 4-111 | The block is going to be the product of three other 1-D solutions. But it still should be easy, 'cuz how do those three solutions compare to each other?  Here's a little hint via Youtube! |
Block: 10 min - 323°C, 20 min - 445°C, 60 min - 500°C. Cylinder: 10 min - 331°C, 20 min - 449°C, 60 min - 500°C. |
| 6-XXC | Concept questions, concept questions, roly poly concept questions, concepts questions, concepts questions, eat them up - yum! | |
| 7-18 | Follow the steps on "How to do a generic convection calculation" and you'll be fine. Remember to use the film temperature for properties. Is the flow laminar or turbulent? At what Re does transition occur? Does that depend on it being a plate or a tube or a cylinder? Do you want the local or average Cf? Nus? | |
| 7-38 | You are told to assume turbulent flow over the whole "plate," but is it really? How would you determine it it were? Can you find the distance along the lfow direction where it goes from laminar to turbulent? (Mini-exam, anyone?) | Qdot,total = 1984 W (Inlcudes radiation) |
| 7-76 | I highly recommend you use Maple, MATLAB, or my favorite EES to automate the calculation of the Nusselt number here, as it will be one of the really nasty ones that makes it easy to screw up if doing it completely by hand. You can assume that the tube wall is very thin and has a very high k so that the average temperature of the outside surface is 75°C. | |
| 7-102 | Another "follow the recipe," just a different geometry. | 590 W |
| 8-52 | Water only! You will need to check for whether the flow is thermally fully developed or not. If it is, and the length of the pipe is long compared to the entry length, then assuming it is fully-developed throughout the entire pipe is a typical simplifying assumption. How does fluid temperature vary in internal flow for a constant temperature boundary condition? | |
| 8-62 | See hints for 8-52. | 86.3°C |
| 8-83 | Your answers may vary depending on which correlation you use. | hair=21.7 W/m2-°C, hwat=11,960 W/m2-°C |
| 8-90 | When you see how much uncertainty there is in the value of h you can see why using an adiabatic tip boundary condition with a corrected length for a fin isn't such a big deal. (That is, any loss of accuracy by not assuming a convective tip boundary condition for a fin pales in comparison to the uncertainty in h.) | At Re=104: NuCol=69.7, NuPet=86.4, NuGniel=79.5 |
| 8-88 | Now you get to use all you toys at once! Those include conservation of energy (like in ConApps), Nusselt number correlations, and even frciton factor correlations (or the Moody Chart) too! How does the local and average value of h compare for a constant surface heat flux BC? How does the temperature difference between the surface and the fluid vary with the flow direction (at least in fully developed flow)? The answer to those questions will be helpful in finding Tsur,exit. | Tsur,exit= 94.6°C |
| 9-31 | Follow the recipe once again! This time however, there are no correlations on your recipe sheet! You have no choice but to use your textbook this time! Remember that β must be in K-1 and not °C-1. | h=3.92 W/m2·K (Answers may vary based on Nu correlation.) |
| 9-54 | See hints for 9-31. | Ra=2.27×106 for the cylinder |
| 9-28 (a) | As the book says, this requires an iterative solution. One way to approach this is to guess Tsur,average, find all your properties, fins Nu, find h, and then find qdot. Does the calculated qdot equal the real qdot? If not, then guess better! I don't recommend that method, however, because to just "guess better" is hard. Instead, I recommend doing everything in the above procedure save the last step before guessing. After you have h, pretend like you haven't guessed Tsur,average. Use the real qdot to calculate a new value of Tsur,average. That is a much better guess than you started with. Repeat until it stops changing. That is usually two or fewer times. Lastly, you will need to inlcude radiation heat transfer as part of the solution! You can use the "baby form" of the equation for radiation we learned back in the first few classes with Tsurr=T∞. Note, however, that you are estimating the value of Tsurr to be the same as T∞, not that Tsurr is the air temperature! | |
| 9-104 | Your text has cookbook formulas for both keff and Qdot. You don't need either one of them! You should recognize the Nusselt number in the keff formula, and you should also recognize Rth,sphere for steady state conduciton in the r direction for a sphere in the Qdot formula. Remember, you are really doing nothing more that pretending like this is steady state, 1-D conduction in a sphere where you have replaced k with keff = k·Nu | 4.87 W |
| 9-115 | Liquid water is not an ideal gas! And so you cannot use β=1/Tfilm. You will need to get β somewhere else. It is not an assigned problem, but 9-114 is the same problem with the same numbers, same dimesions, same fluid, etc. The only difference between the two is the orientation of the rod. That will change the result! | 9-119 | You should first check to see if 0.1 < Gr/Re2 < 10. Also note that the whole "between 3 and 4" thing is pretty wishy washy. And so I, being a "This porridge is just right!" kinda' guy, picked n=3.5. | Nucomb=23.5 |
| 12-25 | I would find the wavelength where the max occurs first, and then find the spectral BB emissive power. Hey! How is is that the moonlight can be modeled as from a BB at 4000K? Is the surface of the moon 4000K? | Match flame: 1.81×105 W/m2-μm |
| 12-27 | (b) 123,563 W/m2 | |
| 12-33 | You are going to have to perform conservation of energy on something here. What should your system be? A common mistake we want to nip in the bud right here is that Q̇rad for a blackbody is σAT4. That is only what is emits. There is also radiation incident upon it (irradiation) which figures into the mix. Most of the time what we want for a surface is the Q̇rad,net. That "net" is really important. | (b) 70°C < Ts < 100°C |
| 12-81 | ε=0.513 | |
| 12-90 | Gtrans=566 W/m2 | |
| 12-108 | q˙net=141 W/m2 | |
| 13-13 | What must F2→1 be for the semicircular groove? (Also note that if you tried the cross string method here [which you don't need to do, by the way] you would see that the figure in your text is wrong, namely that the string on surface 1 would need to be lying flat on the surface, not pulled tight.) | 0.64 for semicircle to surroudnings |
| 13-16 | Save your work, as you will need these view factors in a later problem. | F1→2= |
| 13-26 | Cross those strings! | F1→2=0.3424 |
| 13-34 | Draw some resistors first and find Qdot using our method. Then you can find the radiation heat trasfer coefficient, the one that looks like a convection coefficient. Also note, that since it is a function of the temperature of both surfaces, it's kinda' stupid to write Qdot=hrad(T1-T2), isn't it? | |
| 13-50 | What, are you going to draw a puppy dog? Maybe a nice landscape? NO! Draw some resistors! | Qdot=205 W |
| 13-56 | (We all know you are going to draw a circuit, yes?) Note that there is a mistake in the problem statement in the book. Surfaces (1) and (2) ar not insulated! That would be impossible! |
Q˙1→3=1342 W |
| 13-91 | Q˙one shield/Q˙no shield~1/6 | |
| 11-23 | Your answer may vary depending on the convection correlation you use. Regardless, you should find the flow to be fully developed turbulent flow. | 3838 W/m2°C |
| 11-50 | Now you will be happy that I taught you how to calculate the LMTD differently way back during internal convection. | (c) 3.99 m2 |
| 11-75E | Another example of what is called a HXR design problem: you are calculating how large the HXR must be in order to accomplish a certain Qdot. | F≈0.94 |
| 11-101 | This in an example of what is called a HXR analysis problem: you have a fixed, already built HXR and you are calculating the outlet temperatures. You could use either method (LMTD-F or ε-NTU) but one will involves iteration and the other doesn't. | ε=0.108 |
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