§AERE 310 Homework 4

Sep 25, 2025

§1.

ψ(r,θ)=rnsin⁡(θ) \psi(r, \theta) = r^n \sin(\theta)

§1. (a)

vr=1r∂ψ∂θ=1rrncos⁡(θ)=rn−1cos⁡(θ) v_r = \frac{1}{r} \frac{\partial \psi}{\partial \theta} = \frac{1}{r} r^n \cos(\theta) = r^{n-1} \cos(\theta)
vθ=−∂ψ∂r=−nrn−1sin⁡(θ) v_\theta = -\frac{\partial \psi}{\partial r} = -nr^{n-1} \sin(\theta)
ω=∇×V=∂v∂x−∂u∂y=1r∂∂rrvθ−1r∂vr∂θ=0 \omega = \nabla \times V = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = \frac{1}{r} \frac{\partial}{\partial r} r v_\theta - \frac{1}{r} \frac{\partial v_r}{\partial \theta} = 0
0=−1r∂∂rrnrn−1sin⁡(θ)−1r∂(rn−1cos⁡(θ))∂θ 0 = -\frac{1}{r} \frac{\partial}{\partial r} r nr^{n-1} \sin(\theta) - \frac{1}{r} \frac{\partial (r^{n-1} \cos(\theta))}{\partial \theta}
0=−1r∂∂rnrnsin⁡(θ)+1rrn−1sin⁡(θ) 0 = -\frac{1}{r} \frac{\partial}{\partial r} nr^n \sin(\theta) + \frac{1}{r} r^{n-1} \sin(\theta)
0=−1rn2rn−1sin⁡(θ)+rn−2sin⁡(θ) 0 = -\frac{1}{r} n^2r^{n-1} \sin(\theta) + r^{n-2} \sin(\theta)
0=−n2rn−2sin⁡(θ)+rn−2sin⁡(θ)=−n2+1 0 = -n^2 \cancel{r^{n-2}} \cancel{\sin(\theta)} + \cancel{r^{n-2}} \cancel{\sin(\theta)} = -n^2 + 1
0=−n2+1 0 = -n^2 + 1
n2=1  ⟹  n=1 n^2 = 1 \implies \boxed{n = 1}

§1. (b)

ψuniform=U∞rsin⁡θ \psi_\text{uniform} = U_\infty r \sin \theta
ψdoublet=−κ2πrsin⁡θ \psi_\text{doublet} = -\frac{\kappa}{2 \pi r} \sin \theta
ψ=ψuniform+ψdoublet=U∞rsin⁡θ−κ2πrsin⁡θ \psi = \psi_\text{uniform} + \psi_\text{doublet} = U_\infty r \sin \theta - \frac{\kappa}{2 \pi r} \sin \theta
r=h  ⟹  ψ=0=U∞hsin⁡θ−κ2πhsin⁡θ r = h \implies \psi = 0 = U_\infty h \sin \theta - \frac{\kappa}{2 \pi h} \sin \theta
0=U∞hsin⁡θ−κ2πhsin⁡θ 0 = U_\infty h \cancel{\sin \theta} - \frac{\kappa}{2 \pi h} \cancel{\sin \theta}
0=U∞h−κ2πh 0 = U_\infty h - \frac{\kappa}{2 \pi h}
κ2πh=U∞h \frac{\kappa}{2 \pi h} = U_\infty h
κ=2πU∞h2 \kappa = 2 \pi U_\infty h^2
ψ=U∞rsin⁡θ−2πU∞h22πrsin⁡θ \psi = U_\infty r \sin \theta - \frac{\cancel{2 \pi} U_\infty h^2}{\cancel{2 \pi} r} \sin \theta
ψ=U∞rsin⁡θ−U∞h2rsin⁡θ \boxed{\psi = U_\infty r \sin \theta - \frac{U_\infty h^2}{r} \sin \theta}

§2. (a)

Intuitively, 22 and 33 should be place perpendicular to 11, on the top and bottom.
p2=p3=p∞ p_2 = p_3 = p_\infty
For a circle (extrapolated to a cylinder) with radius RR:
ϕuniform=V∞rcos⁡θ \phi_\text{uniform} = V_\infty r \cos \theta
ϕdoublet=V∞R2rcos⁡θ \phi_\text{doublet} = V_\infty \frac{R^2}{r} \cos \theta
ϕ=ϕuniform+ϕdoublet=V∞rcos⁡θ+V∞R2rcos⁡θ=V∞(r+R2r)cos⁡θ \phi = \phi_\text{uniform} + \phi_\text{doublet} = V_\infty r \cos \theta + V_\infty \frac{R^2}{r} \cos \theta = V_\infty \left( r + \frac{R^2}{r} \right) \cos \theta
Vr=∂ϕ∂r=V∞(1−R2r2)cos⁡θ V_r = \frac{\partial \phi}{\partial r} = V_\infty \left( 1 - \frac{R^2}{r^2} \right) \cos \theta
Vθ=1r∂ϕ∂θ=−1rV∞(r+R2r)sin⁡θ=−V∞(1+R2r2)sin⁡θ V_\theta = \frac{1}{r} \frac{\partial \phi}{\partial \theta} = -\frac{1}{r} V_\infty \left( r + \frac{R^2}{r} \right) \sin \theta = -V_\infty \left( 1 + \frac{R^2}{r^2} \right) \sin \theta
r=R  ⟹  Vr=0 r = R \implies V_r = 0
r=R  ⟹  Vθ=−2V∞sin⁡θ r = R \implies V_\theta = - 2 V_\infty \sin \theta
V=Vr2+Vθ2=2V∞sin⁡θ V = \sqrt{V_r^2 + V_\theta^2} = 2 V_\infty \sin \theta
cp=1−V2V∞2=1−4V∞2sin⁡2θV∞2 c_p = 1 - \frac{V^2}{V_\infty^2} = 1 - \frac{4 V_\infty^2 \sin^2 \theta}{V_\infty^2}
p=p∞  ⟹  cp=0=1−4V∞2sin⁡2θV∞2 p = p_\infty \implies c_p = 0 = 1 - \frac{4 \cancel{V_\infty^2} \sin^2 \theta}{\cancel{V_\infty^2}}
1=4sin⁡2θ 1 = 4 \sin^2 \theta
±1=2sin⁡θ \pm 1 = 2 \sin \theta
±12=sin⁡θ  ⟹  θ=±30°, ±150° \pm \frac{1}{2} = \sin \theta \implies \theta = \pm 30 \degree, ~ \boxed{\pm 150 \degree}
±30°\pm30\degree with respect to +x+x isn't quite what I want judging by the fact the holes lie on the left side of the diagram so I am going with ±150°\pm150\degree.

§2. (b)

For incompressible, inviscid flow at the stagnation point:
cp=1 c_p = 1
cp=p−p∞12ρV∞2 c_p = \frac{p - p_\infty}{\frac{1}{2} \rho V_\infty^2}
At the stagnation point:
1=p1−p∞12ρV∞2 1 = \frac{p_1 - p_\infty}{\frac{1}{2} \rho V_\infty^2}
12ρV∞2=p1−p∞ \frac{1}{2} \rho V_\infty^2 = p_1 - p_\infty
ρV∞2=2p1−2p∞ \rho V_\infty^2 = 2p_1 - 2p_\infty
Since p2=p3=p∞p_2 = p_3 = p_\infty:
ρV∞2=2p1−p2−p3 \rho V_\infty^2 = 2p_1 - p_2 - p_3
V∞2=2p1−p2−p3ρ V_\infty^2 = \frac{2p_1 - p_2 - p_3}{\rho}
V∞=2p1−p2−p3ρ \boxed{V_\infty = \sqrt{\frac{2p_1 - p_2 - p_3}{\rho}}}

§3.

cp, front=1−4sin⁡2θ c_\text{p, front} = 1 - 4 \sin^2 \theta
pback=p∞  ⟹  cp, back=p∞−p∞12ρV∞2=0 p_\text{back} = p_\infty \implies c_\text{p, back} = \frac{\cancel{p_\infty - p_\infty}}{\frac{1}{2} \rho V_\infty^2} = 0
cD=12∫02πcpcos⁡θ dθ=12∫−π2π2cp, backcos⁡θ dθ+12∫π23π2cp, frontcos⁡θ dθ c_D = \frac{1}{2} \int_0^{2\pi} c_p \cos \theta ~ d\theta = \cancel{\frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} c_\text{p, back} \cos \theta ~ d\theta} + \frac{1}{2} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} c_\text{p, front} \cos \theta ~ d\theta
cD=12∫π23π2(1−4sin⁡2θ)cos⁡θ dθ c_D = \frac{1}{2} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (1 - 4 \sin^2 \theta) \cos \theta ~ d\theta
cD=12∫π23π2cos⁡θ−4sin⁡2θcos⁡θ dθ=12∫π23π2cos⁡θ dθ−∫π23π22sin⁡2θcos⁡θ dθ c_D = \frac{1}{2} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \cos \theta - 4 \sin^2 \theta \cos \theta ~ d\theta = \frac{1}{2} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \cos \theta ~ d\theta - \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} 2 \sin^2 \theta \cos \theta ~ d\theta
cD=12[sin⁡θ]π23π2−∫π23π22sin⁡2θcos⁡θ dθ c_D = \frac{1}{2} [\sin \theta]_{\frac{\pi}{2}}^{\frac{3\pi}{2}} - \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} 2 \sin^2 \theta \cos \theta ~ d\theta
cD=12[−2]−∫π23π22sin⁡2θcos⁡θ dθ c_D = \frac{1}{2} [-2] - \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} 2 \sin^2 \theta \cos \theta ~ d\theta
cD=−1−∫π23π22sin⁡2θcos⁡θ dθ c_D = -1 - \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} 2 \sin^2 \theta \cos \theta ~ d\theta
cD=−1−23[sin⁡3θ]π23π2 c_D = -1 - \frac{2}{3} [\sin^3 \theta]_{\frac{\pi}{2}}^{\frac{3\pi}{2}}
cD=−1−23[−1−1] c_D = -1 - \frac{2}{3} [-1 - 1]
cD=−1+43 c_D = -1 + \frac{4}{3}
cD=13 \boxed{c_D = \frac{1}{3}}

§4.

M=N+1 M = N + 1
a=1 a = 1
Γ0=10π \Gamma_0 = 10\pi
P+=(1,10) P_+ = (1, 10)
P−=(1,−10) P_- = (1, -10)
M=11  ⟹  N=10 M = 11 \implies N = 10
M=101  ⟹  N=100 M = 101 \implies N = 100
M=1001  ⟹  N=1000 M = 1001 \implies N = 1000
ψ=−Γ2πln⁡r=−Γ2πln⁡x2+y2 \psi = - \frac{\Gamma}{2\pi} \ln r = - \frac{\Gamma}{2\pi} \ln \sqrt{x^2 + y^2}
The vortices will be offset from the origin:
ψn=−Γn2πln⁡(x−xn)2+(y−yn)2 \psi_n = - \frac{\Gamma_n}{2\pi} \ln \sqrt{(x - x_n)^2 + (y - y_n)^2}
After much experimentation with the equations, I have settled on an iterator nn in a summation that goes from −N2-\frac{N}{2} to N2\frac{N}{2} which gives me N+1N + 1 summands instead of going from 00 to NN which introduces a summation inside the ln⁡\ln.xx is a function of nn which just paces right with a step size of aa as nn:
xn=an=n x_n = \cancel{a}n = n
All the vortices are on the yy axis:
yn=0 y_n = 0
Γ\Gamma is the same for all:
Γn=Γ0=10π \Gamma_n = \Gamma_0 = 10 \pi
And, finally the whole summation:
ψ=∑n=−N2N2ψn=∑n=−N2N2ln⁡−Γn2π(x−xn)2+(y−yn)2 \psi = \sum_{n = -\frac{N}{2}}^{\frac{N}{2}} \psi_n = \sum_{n = -\frac{N}{2}}^{\frac{N}{2}} \ln -\frac{\Gamma_n}{2\pi} \sqrt{(x - x_n)^2 + (y - y_n)^2}
ψ=−10π2π∑n=−N2N2ln⁡(x−n)2+(y−yn)2 \psi = -\frac{10 \cancel{\pi}}{2 \cancel{\pi}} \sum_{n = -\frac{N}{2}}^{\frac{N}{2}} \ln \sqrt{(x - n)^2 + (y - \cancel{y_n})^2}
ψ=−5∑n=−N2N2ln⁡(x−n)2+y2 \psi = -5 \sum_{n = -\frac{N}{2}}^{\frac{N}{2}} \ln \sqrt{(x - n)^2 + y^2}
ψ=−52∑n=−N2N2ln⁡((x−n)2+y2) \psi = -\frac{5}{2} \sum_{n = -\frac{N}{2}}^{\frac{N}{2}} \ln ((x - n)^2 + y^2)
u=∂ψ∂y=−52∑n=−N2N22y(x−n)2+y2 u = \frac{\partial \psi}{\partial y} = -\frac{5}{2} \sum_{n = -\frac{N}{2}}^{\frac{N}{2}} \frac{2y}{(x - n)^2 + y^2}
v=−∂ψ∂x=−52∑n=−N2N22(x−n)(x−n)2+y2 v = -\frac{\partial \psi}{\partial x} = -\frac{5}{2} \sum_{n = -\frac{N}{2}}^{\frac{N}{2}} \frac{2(x - n)}{(x - n)^2 + y^2}
u+=u(P+)=−52∑n=−N2N220(1−n)2+100 u_+ = u(P_+) = -\frac{5}{2} \sum_{n = -\frac{N}{2}}^{\frac{N}{2}} \frac{20}{(1 - n)^2 + 100}
v+=v(P+)=−52∑n=−N2N22(1−n)(1−n)2+100 v_+ = v(P_+) = -\frac{5}{2} \sum_{n = -\frac{N}{2}}^{\frac{N}{2}} \frac{2(1 - n)}{(1 - n)^2 + 100}
u−=u(P−)=−52∑n=−N2N2−20(1−n)2+100 u_- = u(P_-) = -\frac{5}{2} \sum_{n = -\frac{N}{2}}^{\frac{N}{2}} \frac{-20}{(1 - n)^2 + 100}
v−=v(P−)=−52∑n=−N2N22(1−n)(1−n)2+100 v_- = v(P_-) = -\frac{5}{2} \sum_{n = -\frac{N}{2}}^{\frac{N}{2}} \frac{2(1 - n)}{(1 - n)^2 + 100}
V(P)=⟨u(P),v(P)⟩ V(P) = \langle u(P), v(P) \rangle
N+1=11  ⟹  V+=⟨−4.9988,−0.4206⟩ N + 1 = 11 \implies V_+ = \langle -4.9988, -0.4206 \rangle
N+1=11  ⟹  V−=⟨4.9988,−0.4206⟩ N + 1 = 11 \implies V_- = \langle 4.9988, -0.4206 \rangle
N+1=101  ⟹  V+=⟨−13.7524,−0.1906⟩ N + 1 = 101 \implies V_+ = \langle -13.7524, -0.1906 \rangle
N+1=101  ⟹  V−=⟨13.7524,−0.1906⟩ N + 1 = 101 \implies V_- = \langle 13.7524, -0.1906 \rangle
N+1=1001  ⟹  V+=⟨−15.5082,−0.0200⟩ N + 1 = 1001 \implies V_+ = \langle -15.5082, -0.0200 \rangle
N+1=1001  ⟹  V−=⟨15.5082,−0.0200⟩ N + 1 = 1001 \implies V_- = \langle 15.5082, -0.0200 \rangle
Wolfram Alpha had a stroke calculating the above so here's the Python code I used where I changed summand(n) for every PP and between uu and vv (this is snippet calculates vv for P−P_- which I did last):
def summand(n): return (2*(1 - n))/((1 - n)**2 + 100) def compute_summation(N): total = 0 for n in range(-N // 2, N // 2 + 1): total += summand(n) return -5 / 2 * total N = 1000 result = compute_summation(N) print(result)
Python • 247B

§5.

Λ=25πm2s \Lambda = 25 \pi \frac{m^2}{s}
h1=3m h_1 = 3m
h2=4m h_2 = 4m
d1=4m d_1 = 4m
d2=3m d_2 = 3m
Λ1=Λ1′=Λ \Lambda_1 = \Lambda_1' = \Lambda
Λ2=Λ2′=−Λ \Lambda_2 = \Lambda_2' = -\Lambda
For the method of images, this is what I came up with where the reflections have the same strengths (which I am indicating with a prime above):
I am also treating A=(0,0)A = (0, 0) as the origin.
P1=(−d1,h1) P_1 = (-d_1, h_1)
P1′=(−d1,−h1) P_1' = (-d_1, -h_1)
P2=(d2,h2) P_2 = (d_2, h_2)
P2′=(d2,−h2) P_2' = (d_2, -h_2)
For simplicity (note the 44 in the denominator instead of 22, it'll be useful soon):
C≡Λ4π C \equiv \frac{\Lambda}{4 \pi}
Superpositioning ψ\psi will introduce arctan⁡\arctan when going to rectangular so I rather deal with the ln⁡\ln in ϕ\phi:
ϕ=Λ2πln⁡r=Λ2πln⁡r=Λ2πln⁡x2+y2=Λ4πln⁡(x2+y2)=Cln⁡(x2+y2) \phi = \frac{\Lambda}{2 \pi} \ln r = \frac{\Lambda}{2 \pi} \ln r = \frac{\Lambda}{2 \pi} \ln \sqrt{x^2 + y^2} = \frac{\Lambda}{4 \pi} \ln (x^2 + y^2) = C \ln (x^2 + y^2)
ϕ1=Cln⁡((x+d1)2+(y−h1)2) \phi_1 = C \ln ((x + d_1)^2 + (y - h_1)^2)
ϕ1′=Cln⁡((x+d1)2+(y+h1)2) \phi_1' = C \ln ((x + d_1)^2 + (y + h_1)^2)
ϕ2=−Cln⁡((x−d2)2+(y−h2)2) \phi_2 = -C \ln ((x - d_2)^2 + (y - h_2)^2)
ϕ2′=−Cln⁡((x−d2)2+(y+h2)2) \phi_2' = -C \ln ((x - d_2)^2 + (y + h_2)^2)
ϕ=ϕ1+ϕ1′+ϕ2+ϕ2′ \phi = \phi_1 + \phi_1' + \phi_2 + \phi_2'
∂ϕ∂x=∂ϕ1∂x+∂ϕ1′∂x+∂ϕ2∂x+∂ϕ2′∂x \frac{\partial \phi}{\partial x} = \frac{\partial \phi_1}{\partial x} + \frac{\partial \phi_1'}{\partial x} + \frac{\partial \phi_2}{\partial x} + \frac{\partial \phi_2'}{\partial x}
u=u1+u1′+u2+u2′ u = u_1 + u_1' + u_2 + u_2'
Since it's a wall at A=(0,0)A = (0, 0), vv has to be 00, saving me some work:
v=0 v = 0
All hell breaks loose for uu:
u1=∂ϕ1∂x=2C(x+d1)(x+d1)2+(y−h1)2 u_1 = \frac{\partial \phi_1}{\partial x} = \frac{2 C (x + d_1)}{(x + d_1)^2 + (y - h_1)^2}
u1′=∂ϕ1′∂x=2C(x+d1)(x+d1)2+(y+h1)2 u_1' = \frac{\partial \phi_1'}{\partial x} = \frac{2 C (x + d_1)}{(x + d_1)^2 + (y + h_1)^2}
u2=∂ϕ2∂x=−2C(x−d2)(x−d2)2+(y−h2)2 u_2 = \frac{\partial \phi_2}{\partial x} = \frac{-2 C (x - d_2)}{(x - d_2)^2 + (y - h_2)^2}
u2′=∂ϕ2′∂x=−2C(x−d2)(x−d2)2+(y+h2)2 u_2' = \frac{\partial \phi_2'}{\partial x} = \frac{-2 C (x - d_2)}{(x - d_2)^2 + (y + h_2)^2}
u1(A)=2Cd1d12+h12 u_1(A) = \frac{2 C d_1}{d_1^2 + h_1^2}
u1′(A)=2Cd1d12+h12 u_1'(A) = \frac{2 C d_1}{d_1^2 + h_1^2}
u2(A)=2Cd2d22+h22 u_2(A) = \frac{2 C d_2}{d_2^2 + h_2^2}
u2′(A)=2Cd2d22+h22 u_2'(A) = \frac{2 C d_2}{d_2^2 + h_2^2}
u(A)=2u1(A)+2u2(A) u(A) = 2u_1(A) + 2u_2(A)
u(A)=4Cd1d12+h12+4Cd2d22+h22 u(A) = \frac{4 C d_1}{d_1^2 + h_1^2} + \frac{4 C d_2}{d_2^2 + h_2^2}
u(A)=4C[d1d12+h12+d2d22+h22]=Λπ[d1d12+h12+d2d22+h22] u(A) = 4C \left[ \frac{d_1}{d_1^2 + h_1^2} + \frac{d_2}{d_2^2 + h_2^2} \right] = \frac{\Lambda}{\pi} \left[ \frac{d_1}{d_1^2 + h_1^2} + \frac{d_2}{d_2^2 + h_2^2} \right]
u(A)=25πm2sπ[4m(4m)2+(3m)2+3m(3m)2+(4m)2]=7m/s u(A) = \frac{25 \pi \frac{m^2}{s}}{\pi} \left[ \frac{4m}{(4m)^2 + (3m)^2} + \frac{3m}{(3m)^2 + (4m)^2} \right] = \boxed{7m/s}