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My bachelor's degree is in physics/mathematics. I studied these because I always wanted to understand how the things in the world work, and I thought physics was pretty much the key to everything. After graduation, reality set in and I quickly figured out it was grad school or a low-paying job...

Hello, I think a general demonstration could be given along these lines. Consider a general closed system operating between some heat source (possibly variable temperature ##T##) and the environment (temperature ##T_0##) as it evolves from state 1 to state 2. For this case we have: Energy...

Hello, I am looking to purchase laboratory samples for mechanical testing. For example, Charpy impact test samples and standard tensile test samples. Does anyone have a recommendation for a reliable source? Thanks

Yes. If you think neglecting the vapor is untenable, we could replace the liquid with a hot solid. It is and it isn't the same. In the process you suggest, one would perform two actions: place a weight and add heat. In the process I suggest, these are accomplished in one action. I believe that...

I am not sure why my previous attempt was unclear to some. I may have made a mistake (or many) somewhere, and that could be the reason. Please let me know where the error is if that is the case. This is meant to help with a "concrete" example. I believe all three conditions are possible to...

Here is some way to get at this that doesn't rely on having solved the system, but on some general considerations and a guess that the solution is sinusoidal. I am not sure this is what you are looking for, but hopefully it will contribute. :smile: Let's say you had discovered by your...

I thought I would chime into say that there is a perhaps easier way to see the reason for the outcomes questioned on stack exchange. Say we have an ideal gas as the working substance. So, we know that the internal energy is a function only of temperature, and thus in an isothermal process the...

Use this isentropic relation for each gas (a and b) , setting ##T_{2a} =T_{2b}## and similarly for the final pressures. See what relation must obtain between initial quantities in order for that to hold. ##\frac{T_2}{T_1}= \left( \frac{P_2}{P_1}\right) ^{\frac{\gamma-1}{\gamma}}##

http://www.aciscience.org/docs/physical_properties_of_glycerine_and_its_solutions.pdf Table 42

They don't have to mention a regenerator for there to be one... this is a Stirling engine we are discussing, right?? That's what a Stirling engine is... I suppose that one could make the same design with 4 thermal reservoirs instead of a regenerator, but then that really is a different kind of...

These expressions are equivalent. Perhaps it will help you to see how. Recall that: ##a=constant \implies v(t)=at+v_0 \implies x(t) = \frac{at^2}{2} + v_0t + x_0 \tag{1}## Now the first expression for the average velocity is: ##\frac{x_2-x_1}{t_2-t_1}\tag{2}## If we use the third...

Look carefully at your first equation. It applies to a rectangular wall with area A=wy. ##F = \rho g w \frac{y^2}{2} = \rho g w y \frac{y}{2} = \rho g A \frac{y}{2} = \rho g A y_c## They are the same equation, it is just that the depth of the centroid of a rectangular wall with one edge at...

We really can't help it if someone uses a formula naively, can we? I am not sure what you want from me on this... perhaps some method to prevent people from misusing an equation? There is no such method. The best we can do is explain what the symbols mean and what they don't mean. There are...

Compare a horizontal spring to a vertical spring with mass. If there is no difference here, then there will be no difference for intermediate angles. For a horizontal spring, the equation of motion is given by: ##m\frac{d^2x}{dt^2}=-kx \tag{1}## Where x is the displacement from the...

That wouldn't be naive, that would be wrong. Thermal efficiency for a power cycle is defined as the ratio of input heat to output work. Internal interactions are neither of these, and so can not be counted in the calculation. It's just a matter of definition. The first cycle you describe is...

With indefinite integration, you would get: ##U = C_v T - a/V## What this means is that the plot of U vs. T for some fixed volume VdW gass (think of a plot of U as a function of T for a fixed volume) is offset from the plot of U vs. T at that fixed volume for an IG, but the lines are parallel...

Substitute that expression, evaluated with VdW equation, into the one in post #4 and integrate.

Yes, you need to include the fact that regenerater heat interactions do not take place between the operating fluid and the thermal reservoirs, but are internal, and redo the calculation of the efficiency. You should arrive at the Carnot efficiency for the ideal Stirling engine.

"I find some versions of Carnot's theorem generically referring to 'a reversible thermodynamic cycle between two reservoirs'. Does that necessarily mean 'a Carnot cycle'? " No, there are other cycles that fit the bill. Among them: Ideal Stirling, Ideal Ericsson

Instead of guessing, why not work it out and compare? Start with what you know about each model and go from there. If you are stuck, post what you have so far and people will help.

FWIW, my text derives the basic relation (##c = \sqrt{\frac{\Delta p}{\Delta \rho}}##) using mass and momentum equations, then justifies the use of the isentropic relation (##\frac{p}{\rho ^k} = const##) by referring to experimental results which "indicate that the relationship between pressure...

In a Stirling cycle, the only heat added from the hot reservoir is during the isothermal process at ##T_H##, while all the heat rejected to the cold reservoir would take place during the other isothermal process at ##T_L##. The heat interactions during the isochoric processes are internal -...

Yes, the Bernoulli equation is very much related to the conservation of mechanical energy. Recall that pressure is Energy per unit Volume, and this is easy to see. It is important when interpreting/using Bernoulli to always keep in mind the assumptions used to derive it: 1. Along a...

O.k., it might be better to simply post the data you want to fit itself, because there are some missing functions that I don't have so your script won't run. I.e., Undefined function or variable 'rmmissing'. Since your question is really about the data... post the data in an csv or xlsx...

Are you saying that the data in TideHeight1s needs the curve fit? It looks very smooth for data. Rather than being data, which I would expect to have some spread, it looks like it was generated as a trig function.

You replied before I caught my omission of a subscript. The expression I used does indeed give the correct answer as supplied by CheesyPeeps. I thought it was vague enough that CheesyPeeps would have to do some thinking and figuring, and yet it might help push the solution process forward.

These are what I said above in post #2. Centroid is independent of mass - it's a geometric average. Center of mass is an average taking into account mass distribution. Just like you said and showed...

I think you need to integrate instead of using the results of the integration from a simple constant-density case. For example: ##F = \int_0^{height} \int_0^{width} \gamma(y)_{avg} \cdot y \;dx dy##

This looks divergent. ##\int_{-\infty}^\infty \frac{1}{e^{ix}-1}dx = -x-i\cdot log \left( 1- e^{ix}\right)\bigg|_{-\infty} ^\infty##

These terms are somewhat mixed in common parlance. One useful distinction is that the centroid is a geometric center, or average point, assuming equal mass distribution or density. The center of mass is the average point taking into account mass distribution. If the mass is distributed...

The scale is likely reporting your mass with the assumption that you live where the gravitational acceleration is equal to the standard value. In other words, the scale detects a force of 931.95 N and tells you that your mass is ##\frac{931.95 N}{9.81 \frac{m}{s^2}} = 95 kg##. If you took that...

When you have a combination of units that is not one of the standard combinations, it is hard to tell what is going on without a context because there are many ways to get the same resultant units from combining base and named combinations. I have seen all kinds of weird units working in...

Right. They leave that as an unknown. So you are to find the force per unit width. That is what you did! That is why the units look like they do, and why we can find the total force if we ever find out the width (as my examples showed). They are just saying length instead of width, but it is...

That is right. Since BvU answered this, I will provide an example of how this is used in an effort to clarify for you. The force you calculated is Newtons per meter. This means Newtons per unit width. Since we are not given the width, your solution is as much as can be given. It is used like...

That procedure looks good, except I wonder if the total mass is 4x what you have used. Is it 4 slabs of 700 kg each or 4 slabs with a combined mass of 700 kg? Seems like the problem wouldn't mention the number of slabs if it gave the total mass.

Yes, it seems the teacher made an error in writing: ##\Delta U = m c_p (T_2 - T_1)## This should be written (for an ideal gas) as: ##\Delta U = m c_v (T_2 - T_1)## Thus the work should be written, in terms only of the given parameters, as: ##W = \frac{P_1 V_1 c_p}{R_{air} k}...

I think an easy approach is to simply differentiate the fundamental relationship between volume, mass and density. ##m = \rho V\tag{1} \rightarrow V = m \rho^{-1}## If we differentiate this for a unit of mass: ##dV = -m\rho^{-2} d \rho \tag{2}## Now simply plug (1) and (2) into the relation...

Obviously this comes down to an opinion about how to interpret the question. I would interpret it as your professor did. If the question had said, "The air is originally at 1 atm," then I would have interpreted it as you did. The reason is that by saying "all the air," I think the description...

Yes, that's what I imagine happens in the limit. It also explains why the temperatures are the same even though the tanks aren't thermally connected. When we consider a reversible and adiabatic process it is in the limit as a quasi-static, or infinitely slow, process. Thus, the gas temperature...

That's great. But now I noticed something about my solution in post 23. I left out the mass in the final step of my isentropic solution. When I include the mass (0.1165 kg), calculated from the ideal gas law on the initial contents of tank 1, our solutions closely match. Unfortunately, I can't...

Ah, that explains the pressure discrepancy. The second problem remains, however.

No, for your analysis is very detailed! One thing I notice is that the pressures aren't equal at the end, unless I have misapplied your results? (Perhaps this is round-off error.) ##P_1=P_{10}\left(\frac{n_1}{n_{10}}\right)^{\gamma}=997 \left(\frac{1.258 }{4.024 }\right)^{1.4}=195.7kPa##...

I assumed no other equipment than the tanks and turbine, so I am confused about why you would point this out. "As close as possible to reversible" is very near reversible. I assumed exactly reversible, which puts an upper limit on the "very near" part. What am I missing?

Right. But if we are looking for the maximum theoretical work, we should consider all processes reversible. Of course this is never achievable, but this would establish an upper limit. So it seems to me.

I am not sure why one couldn't simply use the change in entropy formula considering the gas as the system undergoing a process from state 1 to 2 - with the turbine volume negligible. If so, then we have ##\Delta S = \int_{1}^{2} \frac{\delta Q}{T} + \sigma## Since there is no heat transfer...

I agree with Chestermiller. I think a better definition is the following, from Structural and Stress Analysis by Megson: "Axial loads are applied along the longitudinal or centroidal axis of a structural member."

I guess it will be up to Kaushik to understand his problem as stated. The first time I saw this was in high school physics class a long time ago, and we used linear density with a simple linear Hooke's law. As he is in high school, that is how I interpret the problem. Either way, he has both...

Not if the units of ##\rho## are, as I said, kg/m.

Your problem statement did not mention M anywhere, so why introduce it? I see a density, which I assume is a linear density (kg/m) since we are told the ring is thin. So from the FBD I posted, ##dm = 2 \rho rd\theta## Now plug this into the expression you got previously (##2Td\theta =...

Yes, a FBD is essential to one approach for solving this. Perhaps the below image will help. It shows a small length ##2rd\theta## of the ring. Now equate the tension forces with the centrifugal force, use the small angle approximation for the sin function, and then solve for T. Once you have T...