§AERE 310 Homework 2

Sep 25, 2025

§1.

Fuel props: pfp_f, VfV_f, AfA_fOxidizer props: pop_o, VoV_o, AoA_oExit props: pep_e, VeV_e, AeA_eEnvironmental props: pap_a
∂∂t∭ρV⃗dV+∬ρV⃗(V⃗−V⃗b)⋅n^dS=∫−pn^dS+∭ρf⃗dV+F⃗vis+F \cancel{\frac{\partial}{\partial t} \iiint \rho \vec{V} dV} + \iint \rho \vec{V} (\vec{V} - \cancel{\vec{V}_b}) \cdot \hat{n} dS = \int -p \hat{n} dS + \cancel{\iiint \rho \vec{f} dV} + \cancel{\vec{F}_{vis}} + F
∬ρV⃗(V⃗⋅n^)dS=∫−pn^dS+F \iint \rho \vec{V} (\vec{V} \cdot \hat{n}) dS = \int -p \hat{n} dS + F
F=∬ρV⃗(V⃗⋅n^)dS+∫pn^dS F = \iint \rho \vec{V} (\vec{V} \cdot \hat{n}) dS + \int p \hat{n} dS
F=ρfVf2(j^⋅−j^)Af+ρoVo2(−j^⋅j^)Ao+ρeVe2(i^⋅i^)Ae+∫pn^dS F = \rho_f V_f^2 (\hat{j} \cdot - \hat{j}) A_f + \rho_o V_o^2 (-\hat{j} \cdot \hat{j}) A_o + \rho_e V_e^2 (\hat{i} \cdot \hat{i}) A_e + \int p \hat{n} dS
I am assuming here the cross section at the head of the rocket is the same as the exit nozzle as it does appear to be in the diagram. In that case, the force by atmospheric pressure on the x-axis should be paAep_a A_e and 00 on the y-axis:
F=−ρfVf2Af−ρoVo2Ao+ρeVe2Ae+peAe−paAe \boxed{F = -\rho_f V_f^2 A_f - \rho_o V_o^2 A_o + \rho_e V_e^2 A_e + p_e A_e - p_a A_e}

§2.

Nh=1200 N_h = 1200
Dh=5mm D_h = 5mm
Vs=8m/s V_s = 8m/s
T=20°C T = 20\degree C
V1=35m/s V_1 = 35m/s
D0=2.5m D_0 = 2.5m
Ds=0.8m D_s = 0.8m
Df=2.2m D_f = 2.2m
I am going to consider multiple bounding boxes for ease, staring from the right:
A0=π4D02=π4(2.5m)2=4.91m2 A_0 = \frac{\pi}{4} D_0^2 = \frac{\pi}{4} (2.5m)^2 = 4.91m^2
As=π4Ds2=π4(0.8m)2=0.5m2 A_s = \frac{\pi}{4} D_s^2 = \frac{\pi}{4} (0.8m)^2 = 0.5m^2
∂∂t∭ρdV+∬ρV⃗⋅n^dS=0 \cancel{\frac{\partial}{\partial t} \iiint \rho dV} + \iint \rho \vec{V} \cdot \hat{n} dS = 0
∬ρV⃗⋅n^dS=0 \iint \rho \vec{V} \cdot \hat{n} dS = 0
ρV0(i^⋅−i^)A0+ρV1(−i^⋅−i^)As=0 \cancel{\rho} V_0 (\hat{i} \cdot -\hat{i}) A_0 + \cancel{\rho} V_1 (-\hat{i} \cdot -\hat{i}) A_s = 0
−V0A0+V1As=0 -V_0 A_0 + V_1 A_s = 0
V0A0=V1As V_0 A_0 = V_1 A_s
V0=V1AsA0=35m/s∗0.5m24.91m2=3.56m/s V_0 = \frac{V_1 A_s}{A_0} = \frac{35m/s * 0.5m^2}{4.91m^2} = \boxed{3.56m/s}
Ah=Nhπ4Dh2=1200∗π4(5mm)2=0.02356m2 A_h = N_h \frac{\pi}{4} D_h^2 = 1200 * \frac{\pi}{4} (5mm)^2 = 0.02356m^2
∬ρV⃗⋅n^dS=0 \iint \rho \vec{V} \cdot \hat{n} dS = 0
ρV2(−i^⋅−i^)As+ρVs(n^⋅n^)Ah+ρV1(−i^⋅i^)As=0 \cancel{\rho} V_2 (-\hat{i} \cdot -\hat{i}) A_s + \cancel{\rho} V_s (\hat{n} \cdot \hat{n}) A_h + \cancel{\rho} V_1 (-\hat{i} \cdot \hat{i}) A_s = 0
V2As+VsAh−V1As=0 V_2 A_s + V_s A_h - V_1 A_s = 0
V2As=−VsAh+V1As V_2 A_s = -V_s A_h + V_1 A_s
V2=−VsAh+V1AsAs V_2 = \frac{-V_s A_h + V_1 A_s}{A_s}
V2=−8m/s∗0.02356m2+35m/s∗0.5m20.5m2=34.6m/s V_2 = \frac{-8m/s * 0.02356m^2 + 35m/s * 0.5m^2}{0.5m^2} = \boxed{34.6m/s}
Af=π4Df2=π4(2.2m)2=3.8m2 A_f = \frac{\pi}{4} D_f^2 = \frac{\pi}{4} (2.2m)^2 = 3.8m^2
∬ρV⃗⋅n^dS=0 \iint \rho \vec{V} \cdot \hat{n} dS = 0
ρVf(−i^⋅−i^)Af+ρV2(−i^⋅i^)As=0 \cancel{\rho} V_f (-\hat{i} \cdot -\hat{i}) A_f + \cancel{\rho} V_2 (-\hat{i} \cdot \hat{i}) A_s = 0
VfAf−V2As=0 V_f A_f - V_2 A_s = 0
Vf=V2AsAf=34.6m/s∗0.5m23.8m2=4.55m/s V_f = \frac{V_2 A_s}{A_f} = \frac{34.6m/s * 0.5m^2}{3.8m^2} = {4.55m/s}

§3.

pa−pa=Δp=70kPa=70000Pa p_a - p_a = \Delta p = 70kPa = 70000Pa
Fx=6400N F_x = 6400N
A1=0.02m2 A_1 = 0.02m^2
A2=0.01m2 A_2 = 0.01m^2
ρ=1000kg/m3 \rho = 1000kg/m^3
θ=45° \theta = 45\degree

§3. (a)

I will definitely have to use simultaneous equations to solve this one. I will relate the flow rates of both ends and the forces on the xx axis.
∬ρV⃗⋅n^dS=0 \iint \rho \vec{V} \cdot \hat{n} dS = 0
ρV1(i^⋅−i^)A1+ρV2(n^⋅n^)A2=0 \cancel{\rho} V_1 (\hat{i} \cdot -\hat{i}) A_1 + \cancel{\rho} V_2 (\hat{n} \cdot \hat{n}) A_2 = 0
−V1A1+V2A2=0 -V_1 A_1 + V_2 A_2 = 0
V1A1=V2A2 V_1 A_1 = V_2 A_2
Now I am going to consider the statics of the xx axis where the viscosity is encapsulated by FxF_x. I will only consider the pressure forces on the exit since that's where the water spills out.
∂∂t∭ρV⃗dV+∬ρV⃗(V⃗−V⃗b)⋅n^dS=∫−pn^dS+∭ρf⃗dV+F⃗vis+F \cancel{\frac{\partial}{\partial t} \iiint \rho \vec{V} dV} + \iint \rho \vec{V} (\vec{V} - \cancel{\vec{V}_b}) \cdot \hat{n} dS = \int -p \hat{n} dS + \cancel{\iiint \rho \vec{f} dV} + \cancel{\vec{F}_{vis}} + F
∬ρV⃗(V⃗⋅n^)dS=∫−pn^dS+F \iint \rho \vec{V} (\vec{V} \cdot \hat{n}) dS = \int -p \hat{n} dS + F
I am condensing pressures of both ends into just Δp\Delta p because I am lazy haha. Note that I am using +Fx+ F_x which isn't negative since we are looking at the force from the perspective of the fluid.
ρV12(i^⋅−i^)A1+ρV22cos⁡(θ)A2=−Δpcos⁡(θ)A2+Fx \rho V_1^2 (\hat{i} \cdot -\hat{i}) A_1 + \rho V_2^2 \cos(\theta) A_2 = -\Delta p \cos(\theta) A_2 + F_x
−ρV12A1+ρV22cos⁡(θ)A2=−Δpcos⁡(θ)A2+Fx -\rho V_1^2 A_1 + \rho V_2^2 \cos(\theta) A_2 = -\Delta p \cos(\theta) A_2 + F_x
All of this gives me two equations with two unknowns:
V1A1=V2A2 V_1 A_1 = V_2 A_2
−ρV12A1+ρV22cos⁡(θ)A2=−Δpcos⁡(θ)A2+Fx -\rho V_1^2 A_1 + \rho V_2^2 \cos(\theta) A_2 = -\Delta p \cos(\theta) A_2 + F_x
Plugging the numbers in:
V1∗0.02m2=V2∗0.01m2 V_1 * 0.02m^2 = V_2 * 0.01m^2
−1000kg/m3∗V12∗0.02m2+1000kg/m3∗V22cos⁡(45°)∗0.01m2=−70000Pa∗cos⁡(45°)∗0.01m2+6400N -1000kg/m^3 * V_1^2 * 0.02m^2 + 1000kg/m^3 * V_2^2 \cos(45\degree) * 0.01m^2 = -70000Pa * \cos(45\degree) * 0.01m^2 + 6400N
Almighty WolframAlpha computes:
V1=26.6983m/s V_1 = 26.6983m/s
V2=53.3966m/s V_2 = 53.3966m/s
Finally, the flow rate:
Q=VA=V1A1=26.6983m/s∗0.02m2=0.534m3/s Q = VA = V_1 A_1 = 26.6983m/s * 0.02m^2 = \boxed{0.534m^3/s}

§3. (b)

The process for finding FyF_y should be identical to the equation for the statics of the xx axis. Note that the properties of upstream are not included since that exclusively on the xx axis.
∂∂t∭ρV⃗dV+∬ρV⃗(V⃗−V⃗b)⋅n^dS=∫−pn^dS+∭ρf⃗dV+F⃗vis+F \cancel{\frac{\partial}{\partial t} \iiint \rho \vec{V} dV} + \iint \rho \vec{V} (\vec{V} - \cancel{\vec{V}_b}) \cdot \hat{n} dS = \int -p \hat{n} dS + \cancel{\iiint \rho \vec{f} dV} + \cancel{\vec{F}_{vis}} + F
∬ρV⃗(V⃗⋅n^)dS=∫−pn^dS+F \iint \rho \vec{V} (\vec{V} \cdot \hat{n}) dS = \int -p \hat{n} dS + F
ρV22sin⁡(θ)A2=−Δpsin⁡(θ)A2+Fy \rho V_2^2 \sin(\theta) A_2 = -\Delta p \sin(\theta) A_2 + F_y
1000kg/m3∗(53.3966m/s)2sin⁡(45°)∗0.01m2=−70000Pa∗sin⁡(45°)∗0.01m2+Fy 1000kg/m^3 * (53.3966m/s)^2 \sin(45\degree) * 0.01m^2 = -70000Pa * \sin(45\degree) * 0.01m^2 + F_y
Fy=20656N \boxed{F_y = 20656N}

§4.

m˙1=9slugs \dot{m}_1 = 9\frac{slug}{s}
V1=300ft/s V_1 = 300ft/s
V2=V3=900ft/s V_2 = V_3 = 900ft/s
D1=4ft D_1 = 4ft
θ=30° \theta = 30\degree
The equations later on are bound to involve inlet and outlet areas so I will compute it:
A1=π4D12=π4∗(4ft)2=12.57ft2 A_1 = \frac{\pi}{4} D_1^2 = \frac{\pi}{4} * (4ft)^2 = 12.57ft^2
The other two areas will require flow rates:
∬ρV⃗⋅n^dS=0 \iint \rho \vec{V} \cdot \hat{n} dS = 0
−ρV1A1+ρV2A2+ρV3A3=0 -\cancel{\rho} V_1 A_1 + \cancel{\rho} V_2 A_2 + \cancel{\rho} V_3 A_3 = 0
−V1A1+V2A2+V3A3=0 -V_1 A_1 + V_2 A_2 + V_3 A_3 = 0
If V2=V3V_2 = V_3, in incompressable flow, their areas would intuitively be equal too:
A2=A3 A_2 = A_3
−V1A1+2V2A2=0 -V_1 A_1 + 2V_2 A_2 = 0
Solving for A2A_2:
A2=A3=V1A12V2=300ft/s∗12.57ft22∗900ft/s=2.095ft2 A_2 = A_3 = \frac{V_1 A_1}{2V_2} = \frac{300ft/s * 12.57ft^2}{2 * 900ft/s} = 2.095ft^2
Another thing I will need is the density of the fluid. Ω\Omega here represents volume since VV is already used for velocity:
ρ=mΩ=m˙Ω˙=m˙AV=m˙1A1V1=9slugs12.57ft2∗300ft/s=0.002387slugft3 \rho = \frac{m}{\Omega} = \frac{\dot{m}}{\dot{\Omega}} = \frac{\dot{m}}{A V} = \frac{\dot{m}_1}{A_1 V_1} = \frac{9\frac{slug}{s}}{12.57ft^2 * 300ft/s} = 0.002387\frac{slug}{ft^3}
Onto the statics. I will only consider the xx axis since the yy cancels out.
∂∂t∭ρV⃗dV+∬ρV⃗(V⃗−V⃗b)⋅n^dS=∫−pn^dS+∭ρf⃗dV+F⃗vis+F \cancel{\frac{\partial}{\partial t} \iiint \rho \vec{V} dV} + \iint \rho \vec{V} (\vec{V} - \cancel{\vec{V}_b}) \cdot \hat{n} dS = \cancel{\int -p \hat{n} dS} + \cancel{\iiint \rho \vec{f} dV} + \cancel{\vec{F}_{vis}} + F
∬ρV⃗(V⃗⋅n^)dS=F \iint \rho \vec{V} (\vec{V} \cdot \hat{n}) dS = F
−ρV12A1+ρV22cos⁡(θ)A2+ρV32cos⁡(θ)A3=−Fx -\rho V_1^2 A_1 + \rho V_2^2 \cos(\theta) A_2 + \rho V_3^2 \cos(\theta) A_3 = -F_x
Since both reverse thrusters are identical:
−ρV12A1+2ρV22cos⁡(θ)A2=−Fx -\rho V_1^2 A_1 + 2 \rho V_2^2 \cos(\theta) A_2 = -F_x
Fx=ρV12A1−2ρV22cos⁡(θ)A2 F_x = \rho V_1^2 A_1 - 2 \rho V_2^2 \cos(\theta) A_2
Fx=0.002387slugft3∗(300ft/s)2∗12.57ft2−2∗0.002387slugft3∗(900ft/s)2cos⁡(30°)∗0.69832 F_x = 0.002387\frac{slug}{ft^3} * (300ft/s)^2 * 12.57ft^2 - 2 * 0.002387\frac{slug}{ft^3} * (900ft/s)^2 \cos(30 \degree) * 0.6983^2
Fx=1067.43lb \boxed{F_x = 1067.43lb}

§5.

u1(y)=U∞yHD u_1(y) = \frac{U_\infty y}{H_D}
u2(y)=−u1(y)=−U∞yHD u_2(y) = -u_1(y) = \frac{-U_\infty y}{H_D}
HD=0.025c=140c H_D = 0.025c = \frac{1}{40} c

§5. (a)

Surfaces 11 and 22 will have flux as they clearly have velocities. Flux through surfaces 33 and 44 will be 00 since they lie on streamlines. The flux through a new surface I created labeled as surface 55 will also have a flux of 00 since the airfoil is a solid surface and the other two infinitesimally close parallel horizontal surfaces have equal and opposite fluxes.I am assuming HUH_U is exactly half of the upstream. HDH_D is definitely half of the downstream for the illustration to make geometric sense. All my math will be in units of "per length," allowing extension into the 3rd3^\text{rd} dimension by simply multiplying by the length of the airfoil.Considering only the flow rates:
∬ρV⃗⋅n^dS=0 \iint \rho \vec{V} \cdot \hat{n} dS = 0
2ρU∞(i^⋅−i^)HU+ρ(i^⋅i^)∫0HDu1(y)dy+ρ(i^⋅i^)∫HD0u2(y)dy=0 2 \rho U_\infty (\hat{i} \cdot -\hat{i}) H_U + \rho (\hat{i} \cdot \hat{i}) \int_0^{H_D} u_1(y) dy + \rho (\hat{i} \cdot \hat{i}) \int_{H_D}^0 u_2(y) dy = 0
−2ρU∞HU+2ρ∫0HDu1(y)dy=0 -\cancel{2 \rho} U_\infty H_U + \cancel{2 \rho} \int_0^{H_D} u_1(y) dy = 0
−U∞HU+∫0HDU∞yHDdy=0 -\cancel{U_\infty} H_U + \int_0^{H_D} \frac{\cancel{U_\infty} y}{H_D} dy = 0
−HU+1HD∫0HDydy=0 -H_U + \frac{1}{H_D} \int_0^{H_D} y dy = 0
−HU+1HD[12y2]0HD=0 -H_U + \frac{1}{H_D} \left[ \frac{1}{2}y^2 \right]_0^{H_D} = 0
−HU+1HD12HD2=0 -H_U + \frac{1}{\cancel{H_D}} \frac{1}{2}H_D^{\cancel{2}} = 0
−HU+12HD=0 -H_U + \frac{1}{2}H_D = 0
HU=HD2=0.025c2 H_U = \frac{H_D}{2} = \frac{0.025c}{2}
HU=0.0125c \boxed{H_U = 0.0125c}

§5. (b)

Using the same bounds as above:
∂∂t∭ρV⃗dV+∬ρV⃗(V⃗−V⃗b)⋅n^dS=∫−pn^dS+∭ρf⃗dV+F⃗vis+F \cancel{\frac{\partial}{\partial t} \iiint \rho \vec{V} dV} + \iint \rho \vec{V} (\vec{V} - \cancel{\vec{V}_b}) \cdot \hat{n} dS = \cancel{\int -p \hat{n} dS} + \cancel{\iiint \rho \vec{f} dV} + \cancel{\vec{F}_{vis}} + F
∬ρV⃗(V⃗⋅n^)dS=F \iint \rho \vec{V} (\vec{V} \cdot \hat{n}) dS = F
−2ρU∞2HU+2ρ∫0HDu12(y)dy=F -2 \rho U_\infty^2 H_U + 2 \rho \int_0^{H_D} u_1^2(y) dy = F
−2ρU∞2HU+2ρ∫0HDU∞2y2HD2dy=F -2 \rho U_\infty^2 H_U + 2 \rho \int_0^{H_D} \frac{U_\infty^2 y^2}{H_D^2} dy = F
−2ρU∞2HU+2ρU∞2HD2∫0HDy2dy=F -2 \rho U_\infty^2 H_U + 2 \rho \frac{U_\infty^2}{H_D^2} \int_0^{H_D} y^2 dy = F
−2ρU∞2HU+2ρU∞2HD213HD3=F -2 \rho U_\infty^2 H_U + 2 \rho \frac{U_\infty^2}{H_D^2} \frac{1}{3} H_D^3 = F
−2ρU∞2HU+23ρU∞2HD=F -2 \rho U_\infty^2 H_U + \frac{2}{3} \rho U_\infty^2 H_D = F
F=23ρU∞2HD−2ρU∞2HU F = \frac{2}{3} \rho U_\infty^2 H_D - 2 \rho U_\infty^2 H_U
F=23ρU∞2(0.025c)−2ρU∞2(0.0125c) \boxed{F = \frac{2}{3} \rho U_\infty^2 (0.025c) - 2 \rho U_\infty^2 (0.0125c)}

§5. (c)

U∞=100m/s U_\infty = 100m/s
c=1m c=1m
ρ=1kg/m3 \rho = 1kg/m^3
F=−D=23∗1kg/m3∗(100m/s)2(0.025∗1m)−2∗1kg/m3∗(100m/s)2(0.0125∗1m) F = -D = \frac{2}{3} * 1kg/m^3 * (100m/s)^2 (0.025 * 1m) - 2 * 1kg/m^3 * (100m/s)^2 (0.0125 * 1m)
D=83.33N/m \boxed{D = 83.33N/m}
The units are per meter, hooray!