At the thermodynamic critical point the pressure first and second derivatives with specific volume vanish.
Part a
To prove that a gas behaves according to the perfect gas equation of state does not have a thermodynamic critical point is done by using the definition of the location of a material’s critical point: where both the first $\frac{\partial P}{\partial v}$ and second derivatives of pressure $P$ with respect to specific volume $v = \frac{V}{n}$ are $0, (\frac{\partial P}{\partial v})$
Clearly, as neither of the above expressions have roots, the ideal gas law cannot support a critical point on the $PvT$ diagram.
Part b
We are asked to derive formulas for the parameters $a$ and $b$ in the Van der Waals equation of state as a function of the critical pressure and temperature of the gas, and the specific gas constant and determine the value of the compressibility factor of the Van Der Waals gas. To do so, the Van der Waals equation of state is written, first as a function of density and then of specific volume with some algebraic manipulation:
P=1−bρρRT−aρ2=v(1−vb)RT−a(v1)2=v−bRT−v2d
With this expression, we can more easily find the first and second derivatives which are both set to 0:
The Redlich-Kwong equation of state is given by the expression:
P=v−bRT−[v(v+b)]aT−1/2
Again, we are asked to determine the values of $a$ and $b$ as a function of the critical pressure and temperature of the gas, and the specific gas constant; then determine the value of the compressibility factor of the Redlich-Kwong gas at the critical point. The critical point equation is evaluated using binomial expansion to compare with expression $[\ref{eq:compare}]$:
(v−vc)3=0=v3+3v2⋅vc+v⋅3vc3+vc3.
Multiplying the both sides of the equation by the LHS denominator, all fractions become polynomials in terms of $v$:
By comparison of first and third terms, which express $v_c$, we can isolate one of the parameters, $a$, in terms of the other parameter $b$:
Substitution of $a$ into the middle term $b$, allows isolation first of $b$. Then , $a$ is then derived using using $[\ref{eq:a}$]:
Critical point compressibility $Z_c$, is found by substitution back into the original Redlich-Kwong equation of state. Unlike the Van der Waals equation, this does not require the coefficients $a$ or $b$ to be known in terms of the critical state properties.
Zc=RTcPcvc=RTcPc(3PcRTc)=31
Problem 2
Which of the following expressions are allowed in index notation?
$x = a_i b_i$ … $\textcolor{green}{\texttt{Right}}$ as it is a contraction on a single index
$uj = T{ij}v_j$ … $\textcolor{red}{\texttt{Wrong}}$ as the free index on the LHS is $j$, whereas the RHS has free index $i$
$ui = T{ij}v_j$ … $\textcolor{green}{\texttt{Right}}$ as this is the proper method for the previous subproblem
$ei = A{ij}v_j + t_i$ … $\textcolor{green}{\texttt{Right}}$ addition of two first order tensors
$a{lk} = t{lk}+v_{kwl}$ … $\textcolor{red}{\texttt{Wrong}}$ as the equation is trying to add different rank tensors
$bi = a{ij}c_j + t_i$ … $\textcolor{green}{\texttt{Right}}$ same expression as subproblem 4
$ci= \epsilon{ijk}b_{jck}$ … $\textcolor{red}{\texttt{Wrong}}$ as the expressions on either side do not match indices.
$t = a{ii} + g{ij} + e_{kk}$ … $\textcolor{red}{\texttt{Wrong}}$ as the first and third terms on the RHS is of rank zero, where as the middle term is of rank 2.
Problem 3
Like a matrix, a tensor ${\bf A}=a{ij}$ of second order is symmetric when $a{ij} = a{ji}$ and skew- symmetric when $a{ij} = -a{ji}$. Also, the transpose tensor, ${\bf A}^T = a{ji}$
Write its symmetric, ${\bf S } = \frac{1}{2}\left({\bf T + T}^T\right) $, and skew-symmetric, ${\bf R } = \frac{1}{2}\left({\bf T + T}^T\right)$ tensors.
Let $\bf S$ be a symmetric tensor and $\bf R$ be a skew-symmetric tensor of arbitrary tensor $\bf T$. Show that ${\bf S} : {\bf R}=0$.
Problem 5
Use tensor analysis developed in class to prove the identity:
Investigation begins with substitution of the rotor of $\bf v$ into the Eqn. $[\ref{eq:rotor}]$, so that the identity is expressed solely in terms of the velocity vector. Then, the vector notation is substituted for indicial notation
On the LHS, we see a contraction of the second order tensor resulting from the gradient of the velocity vector, begin dotted with itself from the left. On the right hand side, the gradient of the divergence subtracted by the cross product of the velocity vector with its curl. Investigate each term individually. First the gradient of the inner product can be conducted using chain rule. As order of differentiation does not matter:
(vjvj),i=vj,ivj+vjvj,i=2vjvj,i.
As for the cross product of the curl, application of indicial notation allows for the use of the $\epsilon-\delta$ identity, as the index $k$ is shared with both. Beginning with grouping the terms, an even permutation on the second Levi-Civita symbol follows so the first index matches on each:
Substitution of these results back into the identity being proven shows that the condition which makes the identity true: that ${\bf v}\times{\bf \omega}$
Graded
MANE 6960-01 Fluid Mechanics:
Conservation Equations in Radial Cylindrical Coordinate System, assuming incompressible flow so that $\rho$ is constant. The resulting generic form of continuity is now independent density.
General Conservation of Mass
r1∂r∂(rur)+r1∂θ∂(uθ)+∂z∂uz=0
Operations
There are an non-unit coefficients in the plane of rotation, due to the cylindrical coordinate system.
In an annulus formed between a rod of radius $R_0$ and a concentric tube with radius $R_1$, the rod is rotated with an angular velocity $Ω$ while the tube is stationary. The upstream and downstream pressures are the same. Find the expression for the velocity components of the fluid contained within the annulus.
Conservation of Mass
Form uses application of incompressibility to reach this generic form. The RHS is $0$ because there is no mass generation term.
r1∂r∂(rur)+r1∂θ∂(uθ)+∂z∂uz=0
Apply differentiation of the first term, then boundary conditions of the spinning problem:
r∈[R0,R1],uθ(r=R0)=Ω,uθ(r=R1)=0
additionally, the problem is both radially and axially symmetric
From this we conclude that there is no variation in the radial direction.
Conservation of Momentum
Each time this is applied for this and for subsequent problems, 2 of the 3 vector components are useful and one is trivial. In this case, it is the axial direction which is not useful.
Axial Direction
Since their symmetry in this direction, that this direction’s momentum equation yields no useful results.
ρDtDuz=−∂z∂p+fz+μ∇2uz→0=0
Radial Direction
The boundary conditions for this direction are: $r\in [R0,R_1], u\theta|{r=R_0} = \Omega, u\theta|_{r=R_1} = 0$.
incompressibility allows for density to drop out. This simplifies into an ODE
−ruθ2=−drdp→ruθ2=drdp.
Tangential Direction
There is no variation as the problem is symmetric about with respect to the tangential direction. However, one integration shows the axial velocity is varying with respect to the radial direction.
The boundaries conditions (BC’s) are: $r=R0, V{\theta}|_{r=R_0} = \Omega_0R_0$ and apply the problem assumptions of no body forces and no pressure gradient. Density does not disappear due to incompressibility as before, but because it must be nonzero, because the fluid is not inviscid:
Boundary Conditions
Apply the reverse chain rule and then factor a single derivative in the $\theta$ direction. Again, an ODE results which is solved by direct integration.
Considering the same geometry as in Problem 1, However, now $Ω = 0$ so there is no vorticity imparted. The rod is pulled in the axial direction with a speed $V_x = V_0$. Find the velocity of the fluid while neglecting gravity and assuming that there is no pressure gradient.
Conservation of Mass
Form uses application of incompressibility to reach this generic form.
Conservation of Momentum
Previously the axial direction yielded no useful results. Now, radial direction produces the trivial equation.
Again, boundary conditions at $r = R0$ then $u\theta|{r=R_0} = u\theta|{r=R_1} = 0 $ is good indication that there is no variation in the $u\theta$
Axial Direction
Assumptions of no pressure gradient and no body forces shows that the Laplacian of the axial velocity is 0. This is a stricter than the similar first order differential operations in the continuity equation and yields new terms due to the coordinate system:
Again, the viscosity must be nonzero or the fluid would be inviscid. Apply the harmonic operation. This is simplified by as there is no loading in the tangential or axial directions.
Use the reverse integration and derivative factoring from the previous problem to find an ODE
0=(∂r2∂2+r1∂r∂)(uz)=r1drd(rdrduz)
With two integrations we reach
uz=C1ln(r)+C2
Boundary Conditions
Use that both velocities are no slip at each face.
at $r = R0$ then $u_z|{r=R_0} = V_0 = C_1 \ln{(R_0)}+C_2$
at the far end
at $r = R1$ then $u_z|{r = R_1} = 0 = C_1 \ln{(R_1)}+C_2$
Solve the system of equations so that
C2=−C1ln(R1)C1=ln(R1R0)V0=V0ln(R0R1)
Substitute back into $C_2$
C2=−V0ln(R0R1)ln(R1)
So that
uz=V0ln(R1R0)ln(R1r)=V0ln(R1r)
Problem 3: Pressure Gradient
Consider the same geometry as in Problem 1. However, now $Ω = 0$, and a constant pressure gradient $ΔP/L$ is applied along the axis of the rod. Find the velocity of the fluid while neglecting gravity.
Conservation of Momentum
Radial Direction
For similar reasons to the previous problem, the radial direction produces trivial equation:
Apply reverse product rule and factor a derivative. The result: that the pressure gradient is a constant and can be replace with a total difference as one would see in a headloss term.
Consider the same geometry as in Problem 1, but now $Ω ≠ 0, Vx ≠ 0, $ and $ ΔP/L ≠ 0.$ Is the solution to this problem simply the combination of solutions of Problems 1-3? If yes, then why?