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Mar
5
comment Questions about compressing a liquid in a piston
@FrenchKheldar There's a big difference between evaporation and boiling. Evaporation is a diffusion-driven process – it simply means that if there's a liquid and a gas phase, water molecules will constantly wander from one phase to the other such that the partial pressure of water in the gas phase tends toward vapor pressure. Boiling means that water molecules near a nucleation site will turn into a vapor bubble right there, all at once, and then try to find a gas phase. The "boiling curve" says when this may start, but metastable liquid states can last for weeks before boiling.
Feb
27
comment In what situations do I use the characteristic length of a fin to find the surface area?
@Greg Yes and mostly yes. You're both wrong, you having underestimated $q_f\ $ by ~5% and the solutions having overestimated it by ~5%. The point is that you were wrong in a consistent way, having made an assumption prescribed by the book; but the solutions were arbitrarily wrong, most likely having just confused variables.
Feb
26
comment What is the most efficient way to use hand dryer?
(Also @Manishearth) I'm not entirely convinced that putting your hands perpendicular to the flow is best for evaporation. For a cube at least, I've seen the correlation that $\text{Nu}=0.71\,\text{Re}^{0.52}\ $ for the front and $\text{Nu}=0.12\,\text{Re}^{0.70}\ $ for the sides. So for sufficiently turbulent flow, it'd make sense that the parallel arrangement offers better convection of heat and humidity.
Feb
26
comment In what situations do I use the characteristic length of a fin to find the surface area?
For the record, if I were a professor teaching from Incropera, I'd hand out permanent markers on the first day of class and tell all my students to cross out Eq. 3.92.
Feb
26
comment In what situations do I use the characteristic length of a fin to find the surface area?
@Greg You're working from Incropera, right? That book starts off with $R=\theta_b/q_f,\ $ (Eq. 3.83) and then introduces the equation you have up above (Eq. 3.92). Comparing the two you can see that $A_f\ $ comes only from the expression for $q_f,\ $ so you should use the corrected area, as if you were calculating $q_f\ $ first and then applying $R=\theta_b/q_f.\ $ But students are used to seeing actual areas in resistance equations, so they're tempted to use the actual area $A=PL+A_c\ $ even if they'd use the right area if they had worked from $R=\theta_b/q_f.\ $ Go for full credit.
Feb
26
comment In what situations do I use the characteristic length of a fin to find the surface area?
@Greg If you mean this equation:$$q_f=\sqrt{hPkA_c}\;\theta_b\,\frac{\sinh mL+\frac{h}{mk}\cosh mL}{\cosh mL+\frac{h}{mk}\sinh mL},$$then that doesn't use $A_f\ $ at all. And it'd give the most accurate answer, *if* you had the right value for $h.$
Feb
26
comment At what point can we assume the tip of a fin is adiabatic?
Using the corrected length does not assume an adiabatic tip. The corrected length is intended to make a non-adiabatic tip fit into equations used for an adiabatic tip. Not using the corrected length – i.e., taking $L_c=L\ $ – means assuming an adiabatic tip.
Feb
26
comment In what situations do I use the characteristic length of a fin to find the surface area?
If your professor wanted to be really rigorous, I'd suggest the correction $$L_c^\prime=L+\frac{t}{2}+\frac{h_l+h_t}{2hw}tL,\qquad A_f=2wL_c^\prime.$$ But nobody does that.
Feb
26
comment In what situations do I use the characteristic length of a fin to find the surface area?
@Greg Well, in the report I linked, that 16 mm² is ignored simply because "The fin is so long that the effect of the exposed ends... is negligible." The efficiency relation $\eta_f=\tanh(mL_c)/(mL_c)\ $ comes from that report, so if you use the relation, you automatically make the assumption. It's inconsistent to apply the assumption for one equation, but not for another. And if the fluid flows, then – like I said – the exposed ends "see" a quite different heat transfer coefficient than the rest of the fin, so you shouldn't lump all the sides together in any case. $A_f=2wL_c\ $ is correct.
Feb
23
comment What is the most efficient way to use hand dryer?
There are a couple of problematic things here. (1) Air flow can increase evaporation independent of temperature. Water will evaporate faster into dry air than into humid air. So if air flow keeps stripping away water from the boundary layer of air around your hands, the water will "see" less humidity and thus evaporate faster, and at a lower temperature. (2) Most of the latent heat is probably supplied by the skin – even if the blow dryer has to heat the skin up first. Evaporation has an absurdly high heat transfer coefficient and will sap heat from whatever is nearby.
Feb
3
comment Should I heat my room when I'm not here, energy-efficiently speaking?
Hence the "practical" bit I added. Basically, you'd want to lower the temperature a bit at first, then keep lowering it until you find a good trade-off between livability and savings. If you're comfortable shutting the heater off entirely, that's obviously the ideal – you don't need to measure your house's heat capacity in a series of silly experiments.
Dec
29
comment Newton's Second Law the real one. Is my theory correct?
$d\mathbf{p}/dt$ doesn't depend on the dimensions of $\mathbf{p}$; each coordinate is differentiated separately to get a result vector with the same dimensions as $\mathbf{p}$. $\nabla\mathbf{p}$ also gives a vector result, but it means something completely different – in $\mathbb{R}^3$, $\langle\;d\mathbf{p}/dx,\;d\mathbf{p}/dy,\;d\mathbf{p}/dz\;\rangle$.
Dec
18
comment This is going to sound really stupid. But how do I change my normal force?
(1) Seriously, you can pass out by doing this. And the paramedics will laugh at your squeaky voice. (4) The places on Earth with the lowest gravity are Zaire, Ceylon, and the Leeward Islands. Go to a doctor over there.
Dec
9
comment How much energy is contained in a 40 meter wave?
Also, are you assuming a perfectly square wave traveling at the arbitrary speed of 19.8 m/s?
Dec
8
comment How much energy is contained in a 40 meter wave?
See this site, which estimates that the southeast Asia tsunami had 0.5 terawatts total power – 1 gigawatt per kilometer of coastline. If the tsunami "lasted" a minute, that's about 7 kt. It also says that during the time the tsunami was only a meter high, it also traveled at 220 m/s and had a wavelength (i.e., in the direction of travel) of 300 km.
Dec
5
comment How does cooling scale with volume?
@Georg, vaporization and convection are both rolled up into the heat transfer coefficient. True, I assume that h is constant with temperature, but generally people aren't so anal about obvious back-of-the-envelope calculations. What's with the chip on your shoulder?
Dec
5
comment Figuring out the force using Newton's 2nd Law
Working from $F\cos(-30°)=\mu_sF_N=0.4\left[9.807\times 12-F\sin(-30°)\right]$ gives $F=70.7$. Check your math.
Nov
24
comment 2 streams going into an engine and 2 come out. Find the enthalpy?
Oh. Nevermind, then. But if you have two streams coming in at .01 kg/s, you can't have an exit stream at .109 kg/s.
Nov
24
comment 2 streams going into an engine and 2 come out. Find the enthalpy?
You seem to have made some typos in the question, but my first guess anyways is that you've made a mistake with the units on $\tfrac{v^2}{2}$: $1 \tfrac{\text{m}^2}{\text{s}^2}$=1 J/kg, so if you add it to enthalpies in kJ/kg you'll get a greater outlet enthalpy.
Nov
24
comment What type of substances allows the use of the Ideal Gas Law?
Wikipedia: Saturated fluid cites Çengel and Boles. Also, just about every steam table calculator I've run across lists boiling-point properties as saturation properties: example.