§AERE 310 Homework 1

Sep 25, 2025

§1.

Given:
f(x,y)=xy f(x,y) = xy

§1. (a)

f(x,y)=c=xy f(x,y) = c = xy
y=cx y = \frac{c}{x}
I chose all integers in [−3,3][-3, 3] to generate the following Desmos level surfaces :
Level SurfacesLevel Surfaces

§1. (b)

∇f(x,y)=⟨∂∂xxy,∂∂yxy⟩=⟨y,x⟩ \nabla f(x,y) = \langle \frac{\partial}{\partial x} xy, \frac{\partial}{\partial y} xy \rangle = \langle y, x \rangle
A GeoGebra applet conveys the gradient well:
Gradient VectorsGradient Vectors

§2.

Given:
V⃗=ui^=V0(1−e−ct)(1−xl)i^ \vec{V} = u \hat{i} = V_0 (1 - e^{-ct}) (1 - \frac{x}{l}) \hat{i}
DiffuserDiffuser

§2. (a)

We have already been given the velocity as a function of xx and tt:
V⃗(x,t)=V0(1−e−ct)(1−xl)i^ \vec{V}(x,t) = V_0 (1 - e^{-ct}) (1 - \frac{x}{l}) \hat{i}
Therefore, the acceleration can be derived directly:
A⃗(x,t)=∂V⃗∂t \vec{A}(x,t) = \frac{\partial \vec{V}}{\partial t}
A⃗(x,t)=V0i^[∂∂t((1−e−ct)(1−xl))] \vec{A}(x,t) = V_0 \hat{i} \left[ \frac{\partial}{\partial t} \left((1-e^{-ct}) (1-\frac{x}{l}) \right) \right]
A⃗(x,t)=V0i^[ce−ct(1−xl)+0] \vec{A}(x,t) = V_0 \hat{i} \left[ c e^{-ct} (1-\frac{x}{l}) + 0 \right]
A⃗(x,t)=cV0e−ct(1−xl)i^ \vec{A}(x,t) = \boxed{c V_0 e^{-ct} (1 - \frac{x}{l}) \hat{i}}

§2. (b)

Sep 25, 2025
V0=10ft/s V_0=10ft/s
l=5ft l=5ft
l=1s l=1s
c=0.49s−1 c=0.49s^{-1}
Here, x=lx=l since we are trying to find the "end" behavior by the end of the diffuser.
A⃗=0.49s−1×10ft/s×e−0.49s−1×1s(1−5ft5ft)i^=0 \vec{A} = 0.49s^{-1} \times 10ft/s \times e^{-0.49s^{-1} \times 1s} (1 - \frac{5ft}{5ft}) \hat{i} = 0
The 00 originates from the fact that x=lx=l. Thus, we end up dividing l/l=1l/l=1. The equation then subtracts 11 from 11 to get 00. Regardless of the values of other members of the product, the function returns 00.Intuitively, this behavior shouldn't be too otherworldly. An acceleration of 00 does not imply a velocity of 00. Sure, the liquid has stopped moving faster/slower, but it's still moving regardless. In other words, the diffuser has indeed diffused the liquid. Also the liquid is now starting to accelerate purely in the j^\hat{j} direction.

§3.

Given:
u=cxx2+y2 u = \frac{cx}{x^2+y^2}
v=cyx2+y2 v = \frac{cy}{x^2+y^2}

§3. (a)

V=⟨u,v⟩ V = \langle u, v \rangle
ΔδVΔt=∇⋅V=∂∂xcxx2+y2+∂∂ycyx2+y2 \frac{\Delta \delta V}{\Delta t} = \nabla \cdot V = \frac{\partial}{\partial x} \frac{cx}{x^2+y^2} + \frac{\partial}{\partial y} \frac{cy}{x^2+y^2}
ΔδVΔt=c(y2−x2)(x2+y2)2+c(x2−y2)(x2+y2)2=c(y2−x2)+c(x2−y2)(x2+y2)2=c(y2−x2+x2−y2)(x2+y2)2=0 \frac{\Delta \delta V}{\Delta t} = \frac{c (y^2 - x^2)}{(x^2 + y^2)^2} + \frac{c (x^2 - y^2)}{(x^2 + y^2)^2} = \frac{c (y^2 - x^2) + c (x^2 - y^2)}{(x^2 + y^2)^2} = \frac{c (y^2 - x^2 + x^2 - y^2)}{(x^2 + y^2)^2} = \boxed{0}
00 makes sense as this is incompressible flow after all.

§3. (b)

w=∇×V=∂∂xv−∂∂yu w = \nabla \times V = \frac{\partial}{\partial x} v - \frac{\partial}{\partial y} u
w=∂∂xcyx2+y2−∂∂ycxx2+y2 w = \frac{\partial}{\partial x} \frac{cy}{x^2+y^2} - \frac{\partial}{\partial y} \frac{cx}{x^2+y^2}
w=−2cxy(x2+y2)2−−2cxy(x2+y2)2=0 w = \frac{-2cxy}{(x^2 + y^2)^2} - \frac{-2cxy}{(x^2 + y^2)^2} = \boxed{0}
Once again, 00 makes intuitive sense as the diffusor doesn't seem to have any geometry that could cause the liquid to "turn on itself" and moving retrograde to all other fluid particles.

§4.

Given:
dp=55ft=16.76m d_p=55ft=16.76m
Dp=950lbf=4226N D_p=950lbf=4226N
up=12mi/h=5.364m/s u_p=12mi/h=5.364m/s
Tp=20°C=293.2K T_p=20\degree C=293.2K
pp=1atm=101325Pa p_p=1atm=101325Pa
dm=1.7m d_m=1.7m
ρ=1kg/m3 \rho=1kg/m^3
μ=2∗10−5Pa∗s \mu=2*10^{-5}Pa*s

§4. (a)

Rem=Rep Re_m=Re_p
ρumdmμ=ρupdpμ \frac{\rho u_m d_m}{\mu} = \frac{\rho u_p d_p}{\mu}
um=ρupdpμμρdm=updpdm=5.364m/s∗16.76m1.7m=52.9m/s u_m = \frac{\cancel{\rho} u_p d_p \cancel{\mu}}{\cancel{\mu} \cancel{\rho} d_m} = \frac{u_p d_p}{d_m} = \frac{5.364m/s * 16.76m}{1.7m} = \boxed{52.9m/s}

§4. (b)

D=12ρu2CDA D = \frac{1}{2} \rho u^2 C_D A
CD=2Dρu2A=2Dpρup2π4dp2=2∗4226N1kg/m3∗(5.364m/s)2π4∗(16.76m)2=1.3315 C_D = \frac{2D}{\rho u^2 A} = \frac{2D_p}{\rho u_p^2 \frac{\pi}{4} d_p^2} = \frac{2 * 4226N}{1kg/m^3 * (5.364m/s)^2 \frac{\pi}{4} * (16.76m)^2} = 1.3315
Dm=12ρum2CDπ4dm2=12∗1kg/m3∗(52.9m/s)2∗1.3315∗π4(1.7m)2=4230N D_m = \frac{1}{2} \rho u_m^2 C_D \frac{\pi}{4} d_m^2 = \frac{1}{2} * 1kg/m^3 * (52.9m/s)^2 * 1.3315 * \frac{\pi}{4} (1.7m)^2 = \boxed{4230N}

§4. (c)

DmD_m is suspiciously close to DpD_p and I suspect this is due to some secret cancellations within the many fractions.

§5.

F[N]  ⟹  MLT−2 F[N] \implies MLT^{-2}
D[m]  ⟹  L D[m] \implies L
Ω[rad/s]  ⟹  T−1 \Omega[rad/s] \implies T^{-1}
V[m/s]  ⟹  LT−1 V[m/s] \implies LT^{-1}
ρ[kg/m3]  ⟹  ML−3 \rho[kg/m^3] \implies ML^{-3}
μ[Pa∗s]  ⟹  ML−1T−1 \mu[Pa*s] \implies ML^{-1}T^{-1}
FF is dependant (11 variables). DD, Ω\Omega, VV, ρ\rho, and μ\mu are independent (55 variables). This makes a total of N=6N=6 variables. MM, LL, and TT are the fundamental dimensions (k=3k=3 dimensions). Hence, there are N−k=6−3=3N-k=6-3=3 dimensionless groups.I am making the function gg use FF as an input; it should be solvable for it implemented. Also, I am using gg and GG for function names because ff and FF would be confusing as we're already using FF for force.
g(F,D,Ω,V,ρ,μ)  ⟹  G(Π1,Π2,Π3) g(F, D, \Omega, V, \rho, \mu) \implies G(\Pi_1, \Pi_2, \Pi_3)
After much research, I settled on DD, VV, and ρ\rho for my repeating variables:
Π1=g1(D,V,ρ,F) \Pi_1 = g_1(D, V, \rho, F)
Π2=g1(D,V,ρ,Ω) \Pi_2 = g_1(D, V, \rho, \Omega)
Π3=g1(D,V,ρ,μ) \Pi_3 = g_1(D, V, \rho, \mu)
The question inquires about FF so I am going to only mess around with Π1\Pi_1:
Π1=D Va ρb Fc \Pi_1 = D ~ V^a ~ \rho^b ~ F^c
[Π1]=[D] [V]a [ρ]b [F]c [\Pi_1] = [D] ~ [V]^a ~ [\rho]^b ~ [F]^c
1=L (LT−1)a (ML−3)b (MLT−2)c 1 = L ~ (LT^{-1})^a ~ (ML^{-3})^b ~ (MLT^{-2})^c
1=LLaT−aMbL−3bMcLcT−2c 1 = L L^a T^{-a} M^b L^{-3b} M^c L^c T^{-2c}
1=L1+a−3b+cT−a−2cMb+c 1 = L^{1 + a - 3b + c} T^{-a - 2c} M^{b + c}
1+a−3b+c=0 1 + a - 3b + c = 0
−a−2c=0 -a - 2c = 0
b+c=0 b + c = 0
1+a−3b+c−a−2c=0 1 + a - 3b + c - a - 2c = 0
1−3b−c=0 1 - 3b - c = 0
1−3b−c+b+c=0 1 - 3b - c + b + c = 0
1−2b=0 1 - 2b = 0
1=2b 1 = 2b
b=12 b = \frac{1}{2}
12+c=0 \frac{1}{2} + c = 0
c=−12 c = -\frac{1}{2}
−a−2∗−12=0 -a - 2 * -\frac{1}{2} = 0
−a+1=0 -a + 1 = 0
a=1 a = 1
Π1=DVρ1/2F−1/2=DVρF \Pi_1 = D V \rho^{1/2} F^{-1/2} = \boxed{D V \sqrt{\frac{\rho}{F}}}
Plugging in all base SI units does indeed give you 11:
1m∗1m/s∗1kg/m31N=1 1m * 1m/s * \sqrt{\frac{1kg/m^3}{1N}} = 1