# Notes
# [[ME 308 HW 05.pdf#page=1|Problem 1]]
$
\sum_{CS}v_{y}'(\rho v_{y}'\cdot dA)=F_{sy}+F_{by}-\int_{CV}a_{y}'\rho dV
$
$
-(v_{2}-u)^2\rho A=-mg-ma\to a=\frac{(v_{2}-u)^2\rho A-mg}{m}
$
$
\text{1D flow: } \frac{1}{2}v_{1}^2=\frac{1}{2}v_{2}^2+gz\to v_{2}=\sqrt{ v_{1}^2-2gz }=12.89m/s
$
$
\boxed{ \text{So }a=.565 \hat{j}~ m/s^2 }
$
# [[ME 308 HW 05.pdf#page=2|Problem 2]]
$
\frac{dm}{dt}=\rho A(v-u)=\dot{m}
$
$
\cancelto{0}{\frac{d}{dt}\int_{CV}u\rho dV}+\sum_{CS}v_{x}'(\rho\vec{v}'\cdot d\vec{A})=-\int_{CV}a_{x}'\rho dV
$
$
-(v-u)^2\rho A=-m\frac{du}{dt}\to m\frac{du}{dt}=\rho A(v-u)^2=\frac{dm}{dt}(v-u)
$
$
m\,du=(v-u)\,dm\to \frac{dm}{m}=\frac{du}{v-u}
$
$
\int_{m_{0}}^{m}\frac{dm}{m}=\int_{0}^{u}\frac{du}{v-u}
$
$
\ln\left(\frac{m}{m_{0}}\right)=-\ln\left(\frac{v-u}{v}\right)=\ln\left(\frac{v}{v-u}\right)
$
$
\boxed{m=\frac{m_{0}v}{v-u}}
$
$
\frac{m_{0}v}{v-u}\frac{du}{dt}=\rho A(v-u)^2\to \frac{du}{(v-u)^3}=\frac{\rho A}{m_{0}v}dt
$
$
\int_{0}^{u}\frac{du}{(v-u)^3}=\frac{\rho A}{m_{0}v}\int_{0}^{t}dt
$
$\dots$
$
\boxed{\frac{u}{v}=1-\frac{1}{\sqrt{1+\frac{2\rho Av}{m_{0}}t}}}
$
# [[ME 308 HW 05.pdf#page=3|Problem 3]]
$
\frac{dM}{dt}=-\rho Av
$
$
\cancelto{0}{\frac{d}{dt}\int_{CV}u\rho dV}+\sum_{CS}v_{x}'(\rho\vec{v}'\cdot d\vec{A})=-\int_{CV}a_{x}'\rho dV
$
$
(-v)(\rho vA)=-M\frac{du}{dt}\to -\rho Av^2=-M\frac{du}{dt}
$
$
M\frac{du}{dt}=\rho Av^2
$
$
\frac{dM}{dt}=-\rho Av\to \rho Av=-\frac{dM}{dt}
$
$
M\frac{du}{dt}=-v\frac{dM}{dt}\to M\,du=-v\,dM
$
$
du=-v\frac{dM}{M}
$
$
\int_{0}^{u}du=-v\int_{M_0}^{M}\frac{dM}{M}
$
$
u=-v\ln\left(\frac{M}{M_0}\right)=v\ln\left(\frac{M_0}{M}\right)
$
$
\frac{dM}{dt}=-\rho Av
$
$
\int_{M_0}^{M}dM=-\rho Av\int_{0}^{t}dt
$
$
M=M_0-\rho Avt
$
$
\boxed{
u(t)=v\ln\left(\frac{M_0}{M_0-\rho Avt}\right)
}
$
# [[ME 308 HW 05.pdf#page=4|Problem 4]]
$
\frac{p_{1}}{\rho g}+\frac{V_{1}^2}{2g}+z_{1}
=
\frac{p_{A}}{\rho g}+\frac{V_{A}^2}{2g}+z_{A}
$
$
p_{1}=p_{atm},\qquad V_{1}\approx 0,\qquad z_{1}=0,\qquad z_{A}=h
$
$
V_{A}=\frac{Q}{A}=\frac{Q}{\pi D^2/4}=\frac{4Q}{\pi D^2}
$
$
\frac{p_{atm}}{\rho g}
=
\frac{p_{A}}{\rho g}
+
\frac{V_{A}^2}{2g}
+h
$
$
h=
\frac{p_{atm}-p_{A}}{\rho g}
-
\frac{V_{A}^2}{2g}
$
$
\text{At }h_{max},\quad p_{A}=p_{vap}
$
$
h_{max}
=
\frac{p_{atm}-p_{vap}}{\rho g}
-
\frac{1}{2g}
\left(\frac{4Q}{\pi D^2}\right)^2
$
$
\boxed{
h_{max}
=
\frac{p_{atm}-p_{vap}}{\rho g}
-
\frac{8Q^2}{g\pi^2D^4}
}
$
# [[ME 308 HW 05.pdf#page=5|Problem 5]]
$
\dot{m}_{in}=\dot{m}_{out}
$
$
\rho bD_{1}V_{1}=\rho bD_{2}V_{2}
\to
V_{1}D_{1}=V_{2}D_{2}
\to
V_{2}=\frac{V_{1}D_{1}}{D_{2}}
$
$
F_{p1}=\int_{0}^{D_{1}}\rho gyb\,dy=\frac{1}{2}\rho gbD_{1}^{2}
$
$
F_{p2}=\int_{0}^{D_{2}}\rho gyb\,dy=\frac{1}{2}\rho gbD_{2}^{2}
$
$
\sum F_{x}=\dot{m}(V_{2}-V_{1})
$
$
\frac{1}{2}\rho gb(D_{1}^{2}-D_{2}^{2})
=
\rho bD_{1}V_{1}(V_{2}-V_{1})
$
$
\frac{1}{2}g(D_{1}^{2}-D_{2}^{2})
=
D_{1}V_{1}(V_{2}-V_{1})
$
$
\frac{1}{2}g(D_{1}^{2}-D_{2}^{2})
=
D_{1}V_{1}
\left(
\frac{V_{1}D_{1}}{D_{2}}-V_{1}
\right)
$
$
gD_{2}(D_{1}^{2}-D_{2}^{2})
=
2V_{1}^{2}D_{1}(D_{1}-D_{2})
$
$
gD_{2}(D_{1}-D_{2})(D_{1}+D_{2})
=
2V_{1}^{2}D_{1}(D_{1}-D_{2})
$
$
gD_{2}(D_{1}+D_{2})=2V_{1}^{2}D_{1}
$
$
D_{2}^{2}+D_{1}D_{2}-\frac{2V_{1}^{2}D_{1}}{g}=0
$
$
D_{2}
=
\frac{-D_{1}+\sqrt{D_{1}^{2}+\frac{8V_{1}^{2}D_{1}}{g}}}{2}
$
$
\boxed{
D_{2}
=
\frac{D_{1}}{2}
\left[
\sqrt{1+\frac{8V_{1}^{2}}{gD_{1}}}-1
\right]
}
$
$
D_{1}=0.6m,\qquad V_{1}=5m/s
$
$
\boxed{D_{2}=1.47m}
$
# [[ME 308 HW 05.pdf#page=6|Problem 6]]
$
\sum_{CS}v_{x}'(\rho\vec{v}'\cdot d\vec{A})=-Ma_{rf}
$
$
\dot{m}(V-U)\cos120^\circ-\rho A(V-U)^2=-M\frac{dU}{dt}
$
$
\dot{m}=\rho A(V-U)
$
$
\rho A(V-U)^2\left(\cos120^\circ-1\right)
=
-M\frac{dU}{dt}
$
$
-\frac{3}{2}\rho A(V-U)^2
=
-M\frac{dU}{dt}
$
$
\frac{3}{2}\rho A(V-U)^2
=
M\frac{dU}{dt}
$
$
\frac{dU}{dt}=a=2.5m/s^2
$
$
V-U
=
\sqrt{\frac{2Ma}{3\rho A}}
$
$
\boxed{V-U=12.91m/s}
$
$
U(t)=at=2.5t
$
$
V(t)=U(t)+(V-U)
$
$
\underline{V(t)=12.91+2.5t}
$
$
\boxed{ V(0)=12.91m/s }
$
$
\boxed{ V(2.5)=19.16m/s }
$
$
\boxed{ V(5)=25.41m/s }
$