My Attempt At Re-entry Temp. Calculation

  • Thread starter Thread starter Vodkacannon
  • Start date Start date
  • Tags Tags
    Calculation
AI Thread Summary
The discussion focuses on calculating the temperature (t2) of an object using its specific heat capacity and re-entry speed. The derived equation for t2 incorporates factors like dynamic pressure and energy-temperature relationships. The accuracy of this equation in estimating temperature gain during freefall is questioned, particularly regarding the omission of skin friction effects. The approach suggests using dynamic pressure to estimate adiabatic heating, but acknowledges limitations in the model. Overall, the conversation highlights the complexities of accurately predicting temperature changes during re-entry.
Vodkacannon
Messages
40
Reaction score
0
I wanted to figure out the temperature(t2) of any object based on things such as its specific heat capacity, re-entry speed, etc.

Work: W = Fy

Force by Air: FD = 1/2pv2CDA

Energy-temperature relationship: Q = mc(t2 - t1)

Setting things equal & solving for t2 gives:

t2 = (1/2pv2CDAy+mct1) / mc

(I'm sorry I could not get latex to format all of the formulas correctly so I formatted none of them...)

So, how accurate is this equation in describing temperature gain through freefall back down to earth?
 
Physics news on Phys.org
I'm going to take a guess and say you get an estimate by finding the dynamic pressure (q = 0.5 * density * velocity^2) and solving for the adiabatic heating due to the pressure rise. This doesn't take any skin friction into account however.
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'

Thread 'Gauss' law seems to imply instantaneous electric field propagation'

Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
View full post »

Thread 'Do electron density waves accompany EM waves in coaxial cables?'

Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
View full post »
Thread 'Recovering Hamilton's Equations from Poisson brackets'

Thread 'Recovering Hamilton's Equations from Poisson brackets'

The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But ##q_i## is a dynamical variable of ##t##, or ##q_i(t)##. In the derivation of Hamilton's equations from the Hamiltonian, viz. ##H = p_i \dot q_i-L##, nowhere did we assume that ##\partial q_i/\partial...
View full post »

Similar threads

Adiabatic Expansion of Pressurized Air in a Piston-Cylinder Setup
Replies
8
Views
2K
Open Gas Turbine - calculate T2, T3, T4, efficiency
Replies
1
Views
2K
Engineering Heat transfer from composite pipe -- Is my answer right?
Replies
1
Views
2K
I Warm-up time for a hollow cylinder
Replies
5
Views
2K
Calculating force of hydraulic cylinder
Replies
2
Views
3K
Calculating Insulation Thickness
Replies
2
Views
5K
Heat in a const. pressure/volume calculation
Replies
40
Views
4K
Mixing two ideal gases with different V, T at constant pressure
Replies
16
Views
3K
Using Constant-Volume Specific Heat for Calculating Change of Internal Energy?
Replies
2
Views
17K
Heat gain (or Loss) in steady state from Metal Rod
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top