- #1
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- Homework Statement
- Write the First Law for an Eulerian-non inertial system
- Relevant Equations
- q+w_t=(h_2-h_1)+(c'_2^2-c'_1^2)/2+g(z_2-z_1)-(w^2*r_2^2-w^2*r_1^2)/2 (a)
Hello to everyone, I'm trying to demonstrate (a) starting from a Lagrangian system. However I have found some difficoulties so I hope you can help me.
First of all, I'm going to illustrate a demonstration I found of the First Law, for a Eulerian -inertial system:
q+w_t=(h_2-h_1)+(c_2^2-c_1^2)/2+g(z_2-z_1) (b)
applying the First Law for a Lagrangian system:
Q+W_e=ΔE_e+ΔE_g+ΔU=ΔE_t (c)
Now I want to apply this method for a non inertial system, which rotates with the rotor. However in this case I have to consider all the forces I see from this system:
Q+W'_e=ΔE'_e+ΔE'_g+ΔU'=ΔE'_t
I can write this equation:
A'=A+W x (W x R')-2*W x V'
https://it.wikipedia.org/wiki/Inter...un'interazione,dei momenti reali, o effettivi. and that sign was one of my dubts. Anyway I'll continue so that you can give me your opinions]
From (2):
A'=A-W x (W x R')-2*W x V'
m*A'=m*A-m*W x (W x R')-m*2*W x V'
So I can see that:
dW'_e=m*A'*dR'=m*A*dR'-m*[W x (W x R')]*dR'-m*2*[W x V']*dR'
For the term (m*A*dR') I have done these considerations for this specific case: this term includes the work real forces which act on the system so that they consist in the forces of pressure (probably they include also the work of rotor but I'll consider it then). This forces, in this case, are parallel to V and perpendicular to (W x R') as we can see in this pic and so:
m*A*dR'=R*V'dt=m*A*(V-W x R')*dt=m*a*v*dt (equal to the inertial system)
W_apparent=∫m*w^2*r'*dr'=m*w^2*(r_2^-r_1^2)/2
In term of work/mass:
w_apparent=w^2*(r_2^-r_1^2)/2
dW_pressure=m*a*v*dt
in term of work/mass:
w_pressure= dW_pressure/(dt*m)=a*v=p_1*v_1-p_2*v_2
So, including the tecnique work (which is zero in this case) we obtain:
q+w'_t+p_1*v_1-p_2*v_2+w^2*(r_2^-r_1^2)/2=(u_2-u_1)+(c'_2^2-c'_1^2)/2+g(z_2-z_1)
q+w'_t=(h_2-h_1)+(c'_2^2-c'_1^2)/2+g(z_2-z_1) -w^2*(r_2^-r_1^2)/2
where z'Ξz and h'Ξh.
I don't think many of these steps are formal but I have tried to obtain it by myself. Also I don't speak english very well so I can't explain so clearly but I hope you can understand the steps and help me.
Thank you so much
First of all, I'm going to illustrate a demonstration I found of the First Law, for a Eulerian -inertial system:
q+w_t=(h_2-h_1)+(c_2^2-c_1^2)/2+g(z_2-z_1) (b)
applying the First Law for a Lagrangian system:
Q+W_e=ΔE_e+ΔE_g+ΔU=ΔE_t (c)
- E_c=kinetic energy
- E_g=gravitational potential energy
- U=internal energy
- E_t=total energy
- W_e=work of external forces (except weight which is still included in E_g)
Now I want to apply this method for a non inertial system, which rotates with the rotor. However in this case I have to consider all the forces I see from this system:
Q+W'_e=ΔE'_e+ΔE'_g+ΔU'=ΔE'_t
I can write this equation:
- V'=V-V(OO')-W x R'
- A'=A-A(OO')-W x (W x R')+dW/dt x R'+2*W x V'
- O≡O'→V(OO')=A(OO')=0
- Stationary regime→dW/dt=0
- V'=V-W x R'
- A'=A-W x (W x R')-2*W x V'
A'=A+W x (W x R')-2*W x V'
https://it.wikipedia.org/wiki/Inter...un'interazione,dei momenti reali, o effettivi. and that sign was one of my dubts. Anyway I'll continue so that you can give me your opinions]
From (2):
A'=A-W x (W x R')-2*W x V'
m*A'=m*A-m*W x (W x R')-m*2*W x V'
So I can see that:
- R'_e=m*A'=Risultant of real and apparent forces
- m*A=Resultant of real forces (This forces are the same I considered in the inertial system)
- -m*W x (W x R')-m*2*W x V'=resultant of apparent forces
dW'_e=m*A'*dR'=m*A*dR'-m*[W x (W x R')]*dR'-m*2*[W x V']*dR'
- dR'=V'dt→[W x V']*dR'=0
- m*A*dR' is not equal to the work of forces in the inertial system because dR'=/=dR
- [W x (W x R')]*dR'=-w^2*r'*dr'
For the term (m*A*dR') I have done these considerations for this specific case: this term includes the work real forces which act on the system so that they consist in the forces of pressure (probably they include also the work of rotor but I'll consider it then). This forces, in this case, are parallel to V and perpendicular to (W x R') as we can see in this pic and so:
m*A*dR'=R*V'dt=m*A*(V-W x R')*dt=m*a*v*dt (equal to the inertial system)
W_apparent=∫m*w^2*r'*dr'=m*w^2*(r_2^-r_1^2)/2
In term of work/mass:
w_apparent=w^2*(r_2^-r_1^2)/2
dW_pressure=m*a*v*dt
in term of work/mass:
w_pressure= dW_pressure/(dt*m)=a*v=p_1*v_1-p_2*v_2
So, including the tecnique work (which is zero in this case) we obtain:
q+w'_t+p_1*v_1-p_2*v_2+w^2*(r_2^-r_1^2)/2=(u_2-u_1)+(c'_2^2-c'_1^2)/2+g(z_2-z_1)
q+w'_t=(h_2-h_1)+(c'_2^2-c'_1^2)/2+g(z_2-z_1) -w^2*(r_2^-r_1^2)/2
where z'Ξz and h'Ξh.
I don't think many of these steps are formal but I have tried to obtain it by myself. Also I don't speak english very well so I can't explain so clearly but I hope you can understand the steps and help me.
Thank you so much