Water at a pressure of 72 $kN/m^2$ flows through a horizontal pipe of diameter 360 mm at the rate of 300 $\mu s$. The direction of water is changed through $120^\circ$ by a vertical bend whose exit diameter is 240 mm. The volume of the bend is 0.14 $m^3$. The exit of the bend is 2.4 m above the inlet. Find the magnitude and direction of the resultant force on the bend due to water. Neglect friction and minor losses.

a) According to Reynold's Transport theorem (RTT)

$\frac{dN}{dt}=\frac{\partial}{\partial t}\int \int \int_{cv} \rho \cdot \eta \cdot dt+\int \int _{c.s} \eta \cdot \rho \cdot \overline \nu \cdot \overline {dA}$.............(1)

Mass flow rate equation:

-The system form of conservation of mass is $\frac{dm}{dt}=0$ which states that mass 'm' within system remains constant in time.

-In equation (1), let N=m, then $\eta$ is mass per unit mass

$\therefore \eta =1$

$\therefore \text{equation (1) becomes}$

$0=\frac{\partial}{\partial t}\int \int \int _{c.v} \rho \cdot dt+\int \int _{c.s} \rho \cdot \overline \nu \cdot \overline {dA}$...........(2)

-Consider cylindrical tube as shown below.

• Flow enters the tube at section (1) and exit at (2)
• Now flow is permitted through solid surface comprising the tube.

-In application of conservation of mass some assumptions are made as follows.

-Stream lines at inlet and outlet are parallel to boundary of the c.v. at inlet and outlet respectively in such a way that inlet and exit velocities are perpendicular to respective areas.

-Assuming steady flow

$\therefore \frac{\partial}{\partial t}\int \int \int _{c.v.}\rho \cdot dt = 0$

$\therefore \text{Equation (2) becomes}$

$0=\int\int _{c.s}\rho \overline \nu \cdot dA$...........(3)

above equation is applied to control surface where mass entering or leaving

$\int \int_{c.s(1)} \rho _1 \overline \nu _1 \cdot dA_1+\int \int _{c.s(2)}\rho _2 \overline \nu _2 \cdot dA_2=0$

-If the inlet and outlet velocity vectors are at inlet and outlet perpendicular to their respective areas then all outflow integral dot products are evaluated as

$\rho _2\overline \nu _2 dA_2=\rho _2 \nu _2 dA_2$

and inflows are evaluated as

$\rho _1\overline \nu _1 dA_1=\rho _1 \nu _1 dA_1$

$\therefore$ equation becomes

$\int_{c.s(1)}\rho _1 \nu _1 dA_1=\int_{c.s(2)}\rho _2 \nu _2 dA_2$..............(4)

-If $\rho _1$ & $\rho _2$ do not vary across the inlet and outlet then areas becomes

$\rho _1\int_{c.s(1)} \nu _1 dA_1=\rho _2\int_{c.s(2)} \nu _2 dA_2$

For one dimensional representation

$v_1A_1=\int \int _{c.s(1)}=v_1.dA_1$ and $v_2A_2=\int \int _{c.s(2)}=v_2.dA_2$

$\therefore$ equation becomes

$\rho_1 A_1 v_1=\rho_2 A_2 v_2=m$...........(A)

Here m is mass flow rate in kg/sec

Momentum equation

Selecting linear momentum as property of system

$\therefore N=m.\overline v$ and $\eta=\overline v$

$\therefore$ equation (1) becomes

$\frac{d}{dt}(m\overline v)=\frac{\partial}{\partial t}\int \int \int _{cv} \rho \cdot \overline \nu \cdot dt+\int \int _{c.s} \overline \nu \cdot \rho (\overline \nu \cdot \overline {dA})$.............(B)

But $\frac{d}{dt}(m\overline v)=F_{ext}$

Assuming steady flow

$\therefore \frac{\partial}{\partial t}\int \int \int _{cv} \rho \cdot \overline \nu \cdot dt=0$

Equation (5) becomes

$F_ext=\int \int _{c.s} \rho \cdot \overline \nu \cdot (\overline \nu \cdot \overline {dA})$

but $\rho \cdot \overline \nu \cdot \overline{dA}=m$

Also outlet and inlet velocities changes are $v_2$ and $v_1$

$\therefore$ equation becomes

$F_{ext}=m(v_2-v_1)$

$\therefore F_{ext}=\rho Q[v_2-v_1]$............(B)

Above equation is momentum equation and can be used to solve control volume problems.

b) Data:-

$\rho _1=72 kN/m^2$

$P_1=360 mm=0.36 m$

$Q=300 lps=0.3 m^3/s$

$D_2=240 mm=0.24 m$

$Z_2=2.4 m$

$Z_1=0(Datum)$

$h_F$ and $h_L=0$

To find: $F_{water}=?$

$A=\frac{\pi}{4}D^2$

$\therefore A_1=0.1018 m^2$

$\therefore A_2=0.04524 m^2$

Also, $Q=A_1v_1=A_2v_2$.........continuity equation

$v_1=\frac{Q}{A_1}=\frac{0.3}{0.1018}=2.95 m/sec$

$v_2=\frac{Q}{A_2}=\frac{0.3}{0.04524}=6.63 m/sec$

Now applying bernoulis equation between inlet and outlet

$\frac{P_1}{w_1}+\frac{v_1^2}{2g}+Z_1=\frac{P_2}{w}+\frac{v_2^2}{2g}+Z_2+h_L$

$\frac{72}{9.81}+\frac{2.95^2}{2\times 9.81}+0=\frac{P_2}{9.81}+\frac{6.6313^2}{2\times 9.81}+2.4$

$P_2=30.82 kN/m^2$

weight of water in bend

$wt=W_w\times \text{Volume of bend}=9.81\times 0.14$

$=1.3734 kN$

Applying equation (B) to control volume in x-direction

$\therefore P_1A_1+P_2A_2cos 60+f_x=\frac{wQ}{g}[-v_2cos 60-v_1]$

$\because \rho=\frac{w}{g}$

$(72\times 0.1018)+[30.8201\times0.04524cos 60]+F_x=1\times 0.3[-6.6313cos 60-2.95]$

$\therefore F_x=-9.9064 (\leftarrow)$

Applying equation (B) to control volume in y-direction

$0-P_2A_2sin 60-wt+F_y=\frac{wQ}{g}[v_2sin 60-0]$

$(-30.8201\times 0.04524 sin 60)-1.3734+F_y=1\times 0.3[6.6313\times cos 60]$

$\therefore F_y=3.5756 kN(\uparrow)$

Force exerted by the bend

$F_{bend}=\sqrt{f_x^2+F_y^2}$

$=\sqrt{(-9.9064)^2+(3.5756)^2}$

$F_{bend}=10.532 kN$

$Inclination \Rightarrow \phi =\tan ^{-1}\frac{F_y}{F_x}=\tan ^{-1}(\frac{3.5756}{(9.9064})$

$=19.85^\circ$

According to Newtons $3^{rd}$ law

$F_{bend}=F_{water}$

$\therefore F_{water}=10.352 kN$