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Each partial differential equation shows how ρ changes as one of P and T varies but the other stays constant.Tales Ferraz said:Homework Statement: Can someone please help me go from equation 2.2 and 2.3 to 2.4a (see image)? I mean, how Beta and Alpha show together in the exponencial? I can solve for only one coefficient (alpha or beta), but not for both. What steps am I missing? Thanks in advance.
Homework Equations: α = − (1/ρ)(∂ρ/∂T) (2.2)
β = (1/ρ)(∂ρ/∂P) (2.3)
ρ = ρr.e^β(P−Pr)−α(T −Tr) (2.4a)
See image
haruspex said:Each partial differential equation shows how ρ changes as one of P and T varies but the other stays constant.
You can treat one as an ODE and solve it in isolation, e.g. ##\rho=Ae^{-\alpha T}##, but still bearing in mind that this is with the other constant. So A here is a function of the other independent variable, β.
Likewise, ##\rho=Be^{-\beta P}##, where B is a function of T.
Can you work it out from there?
mitochan said:Hi. Assuming ##\alpha## and ##\beta## are constant,
[tex]d\rho=(\frac{\partial \rho}{\partial T})_p dp +(\frac{\partial \rho}{\partial p})_T dp=-\rho \alpha dT + \rho\beta dp[/tex]
[tex]\frac{d\rho}{\rho}=- \alpha dT + \beta dp[/tex]
[tex]ln\ \rho - ln\ \rho_r = - \alpha (T-T_r) + \beta( p-p_r) [/tex]
That should bemitochan said:Hi. Assuming ##\alpha## and ##\beta## are constant,
[tex]d\rho=(\frac{\partial \rho}{\partial T})_p dp +(\frac{\partial \rho}{\partial p})_T dp[/tex]
Hi again, actually, I'm still stuck. Can you show more steps after the ρ=Ae^−αT, please? I mean, what would you do next. Thank you, in advance!haruspex said:Each partial differential equation shows how ρ changes as one of P and T varies but the other stays constant.
You can treat one as an ODE and solve it in isolation, e.g. ##\rho=Ae^{-\alpha T}##, but still bearing in mind that this is with the other constant. So A here is a function of the other independent variable, β.
Likewise, ##\rho=Be^{-\beta P}##, where B is a function of T.
Can you work it out from there?
First, see my correction to post #4.Tales Ferraz said:Hi again, actually, I'm still stuck. Can you show more steps after the ρ=Ae^−αT, please? I mean, what would you do next. Thank you, in advance!
The relationship between density and pressure is an inverse one. As pressure increases, the volume of a substance decreases, leading to an increase in its density. This is known as Boyle's Law.
Temperature and density have a direct relationship. As temperature increases, the particles within a substance have more energy and move faster, causing the substance to expand and become less dense. Conversely, as temperature decreases, the substance contracts and becomes more dense.
Air density decreases with altitude because as altitude increases, there is less air above pushing down on the air below. This decreases the pressure, causing the air molecules to spread out and the air to become less dense.
Density plays a crucial role in the behavior of fluids. Heavier fluids, or those with a higher density, will sink below lighter fluids. This is why oil, which has a lower density than water, will float on top of water. Additionally, the density of a fluid can affect its ability to flow and its resistance to pressure and temperature changes.
Density is a measure of how much mass is contained in a given volume of a substance. Specific gravity, on the other hand, is a ratio of the density of a substance to the density of a reference substance, typically water. It is a unitless number that is useful for comparing the densities of different substances.