Recent content by Bavid

\frac{P^{\gamma-1}}{T^\gamma} is a conserved quantity for the adiabatic process. No matter what instance of time you take during the process, the instantaneous pressure and temperature will yield a constant value of \frac{P(t)^{\gamma-1}}{T(t)^\gamma} = \frac{{P_1}^{\gamma-1}}{{T_1}^\gamma} ...

Ok, so it IS adiabatic expansion. The formula I have written in my last post relates P and V at different times. However, you should also remember that the equation of state PV=NRT is always valid for an ideal gas. So you can easily transform equation P_1{V_1}^\gamma=P_2{V_2}^\gamma in terms of...

First things first. If you have decided your construction will be based on wood, you can find the approximate material properties in Michael Ashby's book on Materials Selection (https://www.amazon.com/dp/1856176630/?tag=pfamazon01-20). The properties you are looking for are the following...

Your question needs more detail. It is not apparent which quantities are being held fixed here. If it is adiabatic expansion then the volume V is obviously changing. There is no thermodynamic process (except diffusion) where P,V, and T all remain constant. I'll assume you have an insulated...

If it is plain carbon steel, Austenite is way too unstable to exist at the tempering temperature. I obviously do not know the tempering time you are using, but if you steel is plain carbon, you can very quickly convert ALL of the retained austenite to ferrite+pearlite, so when you quench for...

First, I have seen flat asbestos sheets at places, so the corrugated shape must be based on utility, rather than a fabrication limitation. I'd imagine corrugated sheets are much more efficient at diverting dirt and rainwater from inclined roofing. If you had flat sheets, they might suffer...

During heating, as in tempering, the retained austenite may form pearlite but will NOT form martensite. Martensite formation is diffusionless and shear-dominated, and therefore some combination of quenching and/or deformation is needed. In short, it is not possible to 'heat' a phase and convert...

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Thanks everyone.

Measure of the "Sharpness of a curve" I have a set of curves that belong to the family of curves y=\frac{c}{x^m}, where m and c are parameters. The attached picture (save.png) shows three such curves for different values of m and c. Now these curves have different 'sharpenss' of curvature (to...

It appears to me that rank 1 is necessary but not sufficient for a matrix to be represented as a_ib_j. For example, rank one matrices may have rows differing by a constant, as in (row_i)=m*(row_j)+c which could be rank 1 but not decomposable to a_ib_j. Vector outer product ALWAYS produces a...

Consider the matrix a_ib_j. This matrix is rank 1 since only one row and one column of this matrix is independently determined. The others rows(columns) bear constant ratios to this one row(column). Now if you take the ratio of successive elements in each row, it is easy to calculate...

Hi, I have been trying to understand the fluctuation-dissipation theorem of statistical physics. I understand the objective: to predict the dissipated and stored power in a system driven to oscillate about thermodynamic equilibrium. Most formulations I have come across assume a sinusoidal...

With its current mass, can Jupiter become a stable star? Suppose Jupiter was artificially compressed to only a fraction of its present size (a hypothetical situation occurring in at least one work of science fiction), would the gravity-radiation pressure equilibrium be reached at all? It appears...

over the interval -pi to pi, the integral of cos(mx)cos(nx)dx is zero, as long as m and n are integers. Therefore, if you select ANY pair of elements from the set, the 'integral of their product' will be zero, thereby satisfying the condition of orthogonality.