Recent content by BeeKay

I am looking at the derivation for the Entropy equation for a Newtonian Fluid with Fourier Conduction law. At some point in the derivation I see \frac{1}{T} \nabla \cdot (-\kappa \nabla T) = - \nabla \cdot (\frac{\kappa \nabla T}{T}) - \frac{\kappa}{T^2}(\nabla T)^2 K is a constant and T...

Homework Statement In an attempt to find a transfer function of the system, I need to come up with equations that I can use to solve for unknowns. See the attached image to see the diagram of the pulley system. J is the moment of inertia, r is the radius. The smaller radius on pulley 2 is r1...

Homework Statement Hello, I have a question about using Eulers Method to approximate a solution to a differential equation. The problem lists forces that would be applied on an object and influences its velocity and therefore its position. I believe I am doing the Euler method correct to...

I'll clarify in case you ask exactly to see that work. h+(T1/a)-h1-------->h+((T-a(h-h1))/a)-h1 h+(T/a)-h+h1-h1 This leaves me with 1/a in the denominator, which when brought out of the integrand cancels out the a that was there from when I divided the outside by a. When I divided the...

I think I'm nearly there, but I seem to have an extra a somewhere. I took it from the ln(h+(T1/a) -h1) and then substituted T1=T-a(h-h1). This cancels out the h and h1, but I still end up with T/a. Is this right? Or do I not have to multiply the inside by a and divide the outside?

Equations given. 1) (dP/P) = (-g/(RT))dh 2) T=T1+a(h-h1) First substitute T from equation 2 into equation 1. (dP/P)=(-g/(R(T1+a(h-h1)))) dh Multiply by a on inside of integrand, while dividing on the outside. I will also take constants out of integrand here. (dP/P) = (-g/(aR)) * Integral...

Okay. Back to post 11 with better parentheses. I am trying to integrate 1/(h-(T1/a)-h1) wrt h This, as far as I know, should be solved as if it were just like constant/ (x+constant) because t1, a, and h1 are all constants because we would be given them in the problem. This integrates to...

Just curious, you said you agreed with their result in your first post and that you wanted to lead me to it rather than just give me the answer, but did you make a mistake in the calculations or are you just sick of helping me haha. I have just been staring at it trying to figure it out but I...

I can't see any way to get 1/T by itself in the integrand and still end up getting the right final formula. You at some point said multiply by a on the inside of the integral and divide on the outside, but doesn't this just cancel out and do nothing because you end up bringing out the inside...

I have been under the impression that you could set both bounds on the integral, but it makes more sense that h1 would always be a constant, I'd assume 0? But even then, I don't see what you do with the constant portion of the integral because none of that is involved in the derived equation.

Integral of 1/x+b = ln(x+b). Terms collected in the denominator is 1/(h-h1)+((t1/a))? That means the integral is ln(h+(t1/a)), right?

You can divide it by a, making it 1/(t1/a+(h-h1)). I just don't see how that h-h1 in the denominator gets canceled with something. If I look at it as 1/x+1 and try to integrate, it is ln(x+1). So what makes the h-h1 cancel?

So I'll have -g/aR*Integral of (a*dh/(T1+a(h-h1)). I don't see how that is comparable to 1/(h+const). Where did the T go? Is "h" the T in this case? Doesn't the integrand have to end up as something that will integrate to ln(t)-ln(t1) so I can use log rules to make that ln(t/t1) and then...

Okay, that makes a little bit more sense. But as for a being a constant, I know it is. But I cannot just bring it outside the integral because it is not multiplied by the T1 term, right? Once you pull out g and R, how do you integrate 1/(T1+a(h-h1)) and somehow get a to end up as a constant on...

The part where I get stuck is pretty much once I substitute T in terms of T1 and H. Once I get to -g/(R(T1+a(h-h1))*dh I have no idea where to go from there. I don't even see how I could separate the variables. From what I remember when I did limited stuff with differential equations, it was...