Momentum principle and flow analysis

Given: $d_1 = 0.3$ m, $d_2 = 0.15$ m, deflection $120^\circ$, $Q = 0.23$ m³/s, outlet 1.4 m below inlet, volume of fluid in bend $= 0.085$ m³, $p_1 = 140$ kPa (gauge), $h_L = 0.25V_1^2/2g$.

 inlet 1  --->  [ bend ]
 (horizontal)        \ 120 deg
                      \
                       v  outlet 2 (1.4 m lower)

Take $x$ along the inlet flow and $y$ upward. The outlet velocity direction is $120^\circ$ from the inlet direction, pointing downward: unit vector $(\cos120^\circ, -\sin120^\circ) = (-0.5, -0.866)$.

Velocities and outlet pressure

$$\begin{aligned} A_1 &= 0.070686\ \text{m}^2,\quad A_2 = 0.017671\ \text{m}^2\\ V_1 &= \frac{0.23}{0.070686} = 3.254\ \text{m/s},\quad V_2 = \frac{0.23}{0.017671} = 13.015\ \text{m/s}\\ h_L &= 0.25\times\frac{3.254^2}{19.62} = 0.135\ \text{m} \end{aligned}$$ $$\frac{p_2}{\gamma} = \frac{p_1}{\gamma} + \frac{V_1^2 - V_2^2}{2g} + 1.4 - h_L$$ $$p_2 = 140\,000 + 9810\left(0.5396 - 8.6338 + 1.4 - 0.1349\right) = 73\,005\ \text{Pa}$$

Forces

• $p_1A_1 = 140\,000\times 0.070686 = 9896\ \text{N}$ (in $+x$)
• $p_2A_2 = 73\,005\times 0.017671 = 1290\ \text{N}$ (acts opposite to the outlet flow direction, i.e. along $(+0.5, +0.866)$)
• Weight of fluid $W = 9810\times 0.085 = 834\ \text{N}$ (downward)
• $\rho Q = 1000\times 0.23 = 230$ kg/s

Momentum equation (forces on fluid, $R$ = force of the bend on fluid):

$$\begin{aligned} x:\ \ p_1A_1 + p_2A_2\,(0.5) + R_x &= \rho Q\,(V_2\cos120^\circ - V_1)\\ 9896 + 645 + R_x &= 230(-6.508 - 3.254) = -2245\\ R_x &= -12\,786\ \text{N}\\[4pt] y:\ \ p_2A_2\,(0.866) - W + R_y &= \rho Q\,(V_2\sin(-120^\circ))\\ 1117 - 834 + R_y &= 230(-11.272) = -2593\\ R_y &= -2876\ \text{N} \end{aligned}$$

The force exerted by the water on the bend is the opposite of $R$:

$$F_x = +12\,786\ \text{N},\qquad F_y = +2876\ \text{N}$$ $$F = \sqrt{12\,786^2 + 2876^2} = 13\,106\ \text{N},\qquad \theta = \tan^{-1}\frac{2876}{12\,786} = 12.7^\circ$$

Answer: net force on the bend $\approx 13.1$ kN, acting at $12.7^\circ$ above the horizontal (in the direction of the inlet flow), with $F_x = 12.79$ kN and $F_y = 2.88$ kN upward.

Given: $d_1 = 0.3$ m, $d_2 = 0.15$ m, deflection $120^\circ$ in a vertical plane, inlet horizontal, outlet 1.5 m below the inlet, volume of fluid $= 0.09$ m³, $Q = 300$ L/s $= 0.3$ m³/s, $p_1 = 130$ kPa (gauge). Friction is neglected.

 1 --->  [ bend ]
              \ 120 deg
               \
                v 2 (1.5 m lower)

Take $x$ along the inlet flow, $y$ upward. Outlet direction: $(\cos120^\circ, -\sin120^\circ) = (-0.5, -0.866)$.

Velocities and outlet pressure

$$\begin{aligned} A_1 &= 0.070686\ \text{m}^2,\quad A_2 = 0.017671\ \text{m}^2\\ V_1 &= 4.244\ \text{m/s},\quad V_2 = 16.977\ \text{m/s} \end{aligned}$$

Bernoulli (no loss):

$$p_2 = p_1 + \gamma\left[\frac{V_1^2 - V_2^2}{2g} + 1.5\right] = 130\,000 + 9810\,(0.918 - 14.690 + 1.5) = 9620\ \text{Pa}$$

Forces

• $p_1A_1 = 130\,000\times 0.070686 = 9189\ \text{N}$
• $p_2A_2 = 9620\times 0.017671 = 170\ \text{N}$
• $W = 9810\times 0.09 = 883\ \text{N}$
• $\rho Q = 300$ kg/s

Momentum equation ($R$ = force of the bend on the water):

$$\begin{aligned} x:\ \ 9189 + 170(0.5) + R_x &= 300\,(16.977\times(-0.5) - 4.244) = -3830\\ R_x &= -13\,104\ \text{N}\\[4pt] y:\ \ 170(0.866) - 883 + R_y &= 300\,(-16.977\times 0.866) = -4411\\ R_y &= -3675\ \text{N} \end{aligned}$$

The force exerted by the water on the bend is $(+13\,104,\ +3675)$ N, with $F_x = 13\,104$ N and $F_y = 3675$ N:

$$F = \sqrt{13\,104^2 + 3675^2} = 13\,600\ \text{N},\qquad \theta = \tan^{-1}\frac{3675}{13\,104} = 15.7^\circ$$

Answer: the force of water on the bend $\approx 13.6$ kN, acting at about $15.7^\circ$ above the horizontal in the direction of the inlet flow ($F_x = 13.10$ kN, $F_y = 3.68$ kN upward).

Given: jet $d = 50$ mm, $V = 50$ m/s, $a = \dfrac{\pi}{4}(0.05)^2 = 1.9635\times 10^{-3}$ m². The plate is hinged at the top, and the jet strikes it normally at its centre.

Force required to hold the plate vertical

Force of the jet on the plate (normal to a flat plate at rest):

$$F = \rho aV^2 = 1000\times 1.9635\times 10^{-3}\times 50^2 = 4908.7\ \text{N}$$

This acts at the centre (distance $L/2$ from the hinge, where $L$ is the plate length). Taking moments about the hinge, with the horizontal force $P$ applied at the bottom edge (distance $L$):

$$P\times L = F\times\frac{L}{2}\;\Rightarrow\; P = \frac{F}{2} = 2454.4\ \text{N}$$

Angle of inclination when the force is removed

If the force $P$ is removed, the plate swings out by an angle $\theta$ from the vertical. Equilibrium requires the moment of the jet force about the hinge to equal the moment of the plate weight $W$ (acting at the plate's centre of gravity, $L/2$ from the hinge).

The jet strikes at a distance $\ell$ along the plate from the hinge, with $\ell\cos\theta = L/2$ (jet at fixed height). The normal force on an inclined plate is $F\cos\theta$ (the jet makes angle $90^\circ - \theta$ with the plate).

$$F\cos\theta\times\frac{L}{2\cos\theta} = W\times\frac{L}{2}\sin\theta\;\Rightarrow\;\sin\theta = \frac{F}{W} = \frac{\rho aV^2}{W}$$

The weight $W$ of the plate is needed and is not given. For example, if $W = 10$ kN: $\sin\theta = 4908.7/10000 = 0.4909$, so $\theta = 29.4^\circ$. If the plate weight is neglected, nothing resists the jet and the plate would swing up towards the horizontal ($\theta \to 90^\circ$).

Answer: applied force $P = F/2 = 2454 \approx 2.45$ kN; the angle follows from $\sin\theta = \rho aV^2/W = 4908.7/W$ (e.g. $29.4^\circ$ for $W = 10$ kN).

Given: nozzle $d = 2$ cm, standard air $\rho_a = 1.225$ kg/m³, water manometer reading $h = 7$ cm (water $\rho_w = 1000$ kg/m³). The jet is deflected by the fixed vane through $30^\circ$ and leaves at the same speed (no friction).

Speed of the air from the pitot reading

A stagnation pitot tube in the nozzle exit plane (where static pressure is atmospheric) reads the dynamic pressure:

$$p_0 - p = \frac{1}{2}\rho_aV^2 = \rho_wgh$$ $$V = \sqrt{\frac{2\rho_wgh}{\rho_a}} = \sqrt{\frac{2(1000)(9.81)(0.07)}{1.225}} = \sqrt{1121.1} = 33.5\ \text{m/s}$$

Horizontal force on the vane

$$A = \frac{\pi}{4}(0.02)^2 = 3.1416\times 10^{-4}\ \text{m}^2$$ $$\dot m = \rho_aAV = 1.225\times 3.1416\times 10^{-4}\times 33.48 = 0.01288\ \text{kg/s}$$

The jet enters horizontally with velocity $V$ and leaves at $30^\circ$ to the horizontal with the same speed (pressure atmospheric everywhere, so only momentum flux matters). Taking $x$ along the incoming jet:

$$\begin{aligned} -R_x &= \dot m\,(V\cos30^\circ - V)\\ R_x &= \dot mV(1 - \cos30^\circ) = 0.01288\times 33.48\times 0.13397\\ &= 0.0578\ \text{N} \end{aligned}$$

where $R_x$ is the force on the vane in the direction of the incoming jet.

Answer: speed of air $V \approx 33.5$ m/s; horizontal force on the vane $\approx 0.058$ N in the direction of the jet. (Equivalently $R_x = \rho_aAV^2(1-\cos30^\circ) = 0.4315 \times 0.134$.)

Setup: The left tank has water head $h$ above its orifice; the right tank has head $H$ above its orifice (water depth in the right tank is large and constant). Both orifices have equal area $a$. The plate covers the right orifice and is kept in place only by the jet.

Force of the jet

By Torricelli's theorem, the jet speed is $V = \sqrt{2gh}$. The jet strikes the vertical plate normally and loses all its horizontal momentum:

$$F_{jet} = \rho aV^2 = \rho a(2gh) = 2\rho g\,a\,h$$

(If the jet area is $a$ at the vena contracta, the same expression applies for small smooth orifices.)

Force from the water in the right tank

The water pressure acts on the plate covering the orifice, trying to push it away. Taking the pressure at the orifice (small area) as $\rho gH$:

$$F_{water} = \rho g\,H\,a$$

Condition to keep the plate in place

$$F_{jet} \ge F_{water}\;\Rightarrow\; 2\rho g\,ah \ge \rho g\,aH\;\Rightarrow\; h \ge \frac{H}{2}$$

Answer: the minimum height is $h_{min} = H/2$, i.e. the head in the left tank must be at least half of the head in the right tank.

Given: total flow $Q = 10$ L/s $= 0.01$ m³/s entering vertically at the axis; two equal nozzles of $d = 12$ mm; $r_1 = 0.6$ m, $r_2 = 0.2$ m; jets leave horizontally at $\theta = 60^\circ$ to the tangential direction (assumed from the figure). Arm held stationary ($\omega = 0$). Both jets give torque in the same sense.

Jet velocity

Nozzles are equal, so each nozzle carries $q = Q/2 = 0.005$ m³/s:

$$\begin{aligned} a &= \frac{\pi}{4}(0.012)^2 = 1.131\times 10^{-4}\ \text{m}^2\\ V &= \frac{q}{a} = \frac{0.005}{1.131\times 10^{-4}} = 44.21\ \text{m/s} \end{aligned}$$

Torque

Water enters along the axis with zero angular momentum. Angular momentum leaving: $\rho q\,(r_1V\cos\theta + r_2V\cos\theta)$. Torque needed to hold the arm $=$ rate of change of angular momentum:

$$\begin{aligned} T &= \rho\,q\,V\cos\theta\,(r_1 + r_2)\\ &= 1000\times 0.005\times 44.21\times\cos60^\circ\times(0.6 + 0.2)\\ &= 1000\times 0.005\times 44.21\times 0.5\times 0.8 = 88.4\ \text{N·m} \end{aligned}$$

Individually: jet 1 gives 66.3 N·m and jet 2 gives 22.1 N·m.

Answer: torque required to hold the sprinkler stationary $\approx 88.4$ N·m. (This is the torque the sprinkler would exert if free; if the angle $\theta$ were measured from the arm, replace $\cos60^\circ$ by $\sin60^\circ$.)

Notation: jet velocity $U$, area $A$, density $\rho$. The kinetic power available in the jet is $\dfrac{1}{2}\rho AU^3$.

(i) Plate at rest

$$\begin{aligned} F &= \rho AU^2\\ \text{Power developed} &= F\times 0 = 0\\ \eta &= 0 \end{aligned}$$

A stationary plate receives a force but does no work, so its efficiency is zero (the jet's energy is carried away as kinetic energy of the deflected sheet).

(ii) Plate moving with velocity u in the direction of the jet

Only the relative velocity $(U - u)$ hits the plate, and the mass flow striking the plate is $\rho A(U - u)$:

$$\begin{aligned} F &= \rho A(U - u)^2\\ P &= Fu = \rho A(U - u)^2u\\ \eta &= \frac{P}{\frac{1}{2}\rho AU^3} = \frac{2u(U - u)^2}{U^3} \end{aligned}$$

Condition for maximum efficiency:

$$\frac{d\eta}{du} = 0\;\Rightarrow\; (U - u)^2 - 2u(U - u) = 0\;\Rightarrow\; U - 3u = 0\;\Rightarrow\; u = \frac{U}{3}$$ $$\eta_{max} = \frac{2(U/3)(2U/3)^2}{U^3} = \frac{8}{27} = 29.6\ \%$$

(iii) Series of plates (e.g. wheel with many vanes)

With plates arriving in quick succession, the whole jet discharge $\rho AU$ is used, not just $\rho A(U - u)$:

$$\begin{aligned} F &= \rho AU(U - u)\\ P &= Fu = \rho AU(U - u)u\\ \eta &= \frac{2u(U - u)}{U^2} \end{aligned}$$ $$\frac{d\eta}{du} = 0\;\Rightarrow\; U - 2u = 0\;\Rightarrow\; u = \frac{U}{2},\qquad \eta_{max} = \frac{2(U/2)(U/2)}{U^2} = 0.5 = 50\ \%$$

Answer: (i) $F = \rho AU^2$, $P = 0$, $\eta = 0$; (ii) $F = \rho A(U-u)^2$, $\eta = 2u(U-u)^2/U^3$, maximum $8/27 = 29.6\%$ at $u = U/3$; (iii) $\eta_{max} = 50\%$ at $u = U/2$.

Given: pipe $d_1 = 0.10$ m, nozzle exit $d_2 = 0.05$ m, $Q = 2400$ L/min $= 0.04$ m³/s, volume of assembly $= 18.2$ L $= 0.0182$ m³, head loss in bend $0.5V^2/2g$ and in nozzle $2V^2/2g$ ($V$ = velocity in the pipe).

Assumed from the figure: water flows upward in the vertical pipe, passes a bend and leaves through a horizontal nozzle into the atmosphere, 5 m above the inlet section.

   exit ====>  (nozzle, d = 5 cm)
   |
   |  5 m
   |
   ^  inlet 1 (pipe d = 10 cm, flow up)

Velocities

$$\begin{aligned} A_1 &= 7.854\times 10^{-3}\ \text{m}^2,\quad A_2 = 1.9635\times 10^{-3}\ \text{m}^2\\ V_1 &= 5.093\ \text{m/s},\quad V_2 = 20.372\ \text{m/s} \end{aligned}$$

Inlet pressure (Bernoulli with losses)

$$\frac{p_1}{\gamma} + \frac{V_1^2}{2g} = 0 + \frac{V_2^2}{2g} + 5 + (0.5 + 2)\frac{V_1^2}{2g}$$ $$\begin{aligned} \frac{p_1}{\gamma} &= 21.15 + 5 + 2.5(1.322) - 1.322 = 28.13\ \text{m}\\ p_1 &= 276\,009\ \text{Pa} \approx 276\ \text{kPa (gauge)} \end{aligned}$$

(using $V_2^2/2g = 21.15$ m, $V_1^2/2g = 1.322$ m).

Momentum equation

• $p_1A_1 = 276\,009\times 7.854\times 10^{-3} = 2168\ \text{N}$ (up, on the inlet face)
• $W = 9810\times 0.0182 = 178.5\ \text{N}$ (down)
• $\rho Q = 40$ kg/s. Inlet velocity $= (0, +V_1)$; exit velocity $= (+V_2, 0)$.

Forces on the fluid ($R$ = force of the assembly on the fluid):

$$\begin{aligned} x:\ \ R_x &= \rho Q\,V_2 = 40\times 20.372 = 815\ \text{N}\\ y:\ \ p_1A_1 - W + R_y &= \rho Q\,(0 - V_1)\\ 2168 - 178.5 + R_y &= -203.7\\ R_y &= -2193\ \text{N} \end{aligned}$$

The hydrodynamic force exerted by the water on the assembly is the opposite of $R$:

$$F_x = -815\ \text{N},\qquad F_y = +2193\ \text{N}$$ $$F = \sqrt{815^2 + 2193^2} = 2339\ \text{N},\qquad \theta = 110.4^\circ\ \text{from the +x direction (the direction of the exit jet)}$$

Answer: the hydrodynamic force on the system $\approx 2.34$ kN, directed upward and backwards (against the jet), at $69.6^\circ$ above the horizontal, with components 0.815 kN opposite to the jet and 2.19 kN upward.

When a jet strikes a series of vanes, the whole jet discharge ($\rho aV$) is used on the moving vanes, because new vanes keep replacing the old ones. The efficiency is the work done on the vanes divided by the kinetic energy supplied by the jet. For semicircular vanes it reaches a maximum of 1 (100%) when the vane speed is $u = V/2$.

Setup

Let $V$ = jet velocity, $u$ = vane velocity (same direction), $a$ = jet area. Take the vanes as smooth, so the relative speed is unchanged.

   jet V -->   ____
 ============>/    \   vane moves
              \____/ -> u

• Relative velocity at inlet: $V_{r1} = V - u$.
• The semicircular vane turns the jet through $180^\circ$, so the relative velocity at exit is $V_{r2} = V - u$ in the opposite direction.
• Absolute exit velocity along the jet: $V_{2x} = u - (V-u) = 2u - V$.

Force and work

Mass flow striking the series of vanes = $\rho a V$.

$$\begin{aligned} F &= \rho a V\,[V - (2u - V)] = 2\rho a V (V-u) \\ \text{Work done per second} &= F u = 2\rho a V (V-u)\,u \end{aligned}$$

Efficiency

Input power (kinetic energy of the jet per second) $= \tfrac12 \rho a V \cdot V^2$.

$$\eta = \frac{2\rho a V (V-u)\,u}{\tfrac12 \rho a V^3} = \frac{4u(V-u)}{V^2}$$

Condition for maximum

$$\frac{d\eta}{du} = \frac{4}{V^2}(V - 2u) = 0 \;\Rightarrow\; u = \frac V2$$

The second derivative is $-8/V^2 < 0$, so this is a maximum.

$$\eta_{max} = \frac{4\,(V/2)(V/2)}{V^2} = 1$$

Hence the maximum efficiency is 1 (100%) at $u = V/2$. At this speed the exit absolute velocity is $2u - V = 0$, so the jet leaves with no kinetic energy and all of its energy is given to the vanes.

The resultant force is found by applying the Bernoulli equation to get the inlet pressure and then the momentum equation in the $x$ and $y$ directions. The result is about 3.71 kN, acting at about $47^\circ$ to the inlet axis.

Assumptions

The figure does not give the inlet diameter, so I take $d_1 = 20$ cm. The inlet pipe lies along the $x$-axis, nozzle 2 ($d_2 = 12$ cm) continues straight, and nozzle 3 ($d_3 = 10$ cm) leaves at $60^\circ$ to the axis. The plane is horizontal, so weight does not matter. Both nozzles discharge to the atmosphere, so $p_2 = p_3 = 0$ (gauge). $\rho = 1000\ \text{kg/m}^3$.

                    / --> V3 (d3, 60 deg)
   V1, p1  ------->(
                    \ --> V2 (d2, straight)

Discharges and inlet velocity

$$\begin{aligned} A_1 &= 0.03142\ \text{m}^2,\; A_2 = 0.011310\ \text{m}^2,\; A_3 = 0.007854\ \text{m}^2 \\ Q_2 &= A_2V_2 = 0.2262\ \text{m}^3/\text{s},\quad Q_3 = A_3V_3 = 0.1571\ \text{m}^3/\text{s} \\ Q_1 &= Q_2 + Q_3 = 0.3833\ \text{m}^3/\text{s},\quad V_1 = Q_1/A_1 = 12.20\ \text{m/s} \end{aligned}$$

Inlet pressure (Bernoulli, no losses)

$$p_1 = \tfrac12\rho\,(V_2^2 - V_1^2) = \tfrac12(1000)(20^2 - 12.2^2) = 125\,580\ \text{Pa} = 125.6\ \text{kPa}$$

Momentum equation (force of bend on fluid, $F_x$, $F_y$)

$$\begin{aligned} F_x + p_1A_1 &= \rho\,(Q_2V_2 + Q_3V_3\cos 60^\circ - Q_1V_1) \\ F_x &= 1000\,(4.524 + 1.571 - 4.676) - 3945 = -2526\ \text{N} \\ F_y &= \rho Q_3V_3\sin 60^\circ = 1000(0.1571)(20)(0.866) = 2721\ \text{N} \end{aligned}$$

The force exerted on the bend by the water is equal and opposite:

$$R_x = 2526\ \text{N},\quad R_y = -2721\ \text{N}$$ $$R = \sqrt{2526^2 + 2721^2} = 3713\ \text{N},\qquad \theta = \tan^{-1}\frac{2721}{2526} = 47.1^\circ$$

Answer: $R \approx 3.71$ kN, directed $47.1^\circ$ below the inlet axis (away from the branch side), pushing the bend along the flow direction.

For a series of vanes the whole jet discharge is used, so the force follows from the change of momentum in the direction of the jet and perpendicular to it. The resultant force is about 1.62 kN at $21.3^\circ$ to the jet, and the work done is about 18.2 kW.

Given

$d = 5$ cm, $Q = 0.056\ \text{m}^3/\text{s}$, $u = 12$ m/s, deflection $= 135^\circ$, $V_{r2} = 0.9\,V_{r1}$, $\rho = 1000\ \text{kg/m}^3$.

Velocities

$$\begin{aligned} V &= \frac{Q}{A} = \frac{0.056}{\frac{\pi}{4}(0.05)^2} = 28.52\ \text{m/s} \\ V_{r1} &= V - u = 16.52\ \text{m/s},\qquad V_{r2} = 0.9 \times 16.52 = 14.87\ \text{m/s} \end{aligned}$$

  Exit velocity triangle
     Vr2 at 135 deg to jet direction
   Vx2 = u + Vr2 cos135 = 12 - 10.51 = 1.49 m/s
   Vy2 =     Vr2 sin135 = 10.51 m/s

(a) Force on the vanes

$$\begin{aligned} F_x &= \rho Q\,(V - V_{x2}) = 1000(0.056)(28.52 - 1.49) = 1514\ \text{N} \\ F_y &= \rho Q\,V_{y2} = 1000(0.056)(10.51) = 589\ \text{N} \\ R &= \sqrt{1514^2 + 589^2} = 1624\ \text{N},\qquad \theta = \tan^{-1}\frac{589}{1514} = 21.3^\circ \end{aligned}$$

The jet is turned upward, so the force on the vanes has a component $F_x$ along the jet and a component $F_y$ of 589 N downward.

Answer (a): $R \approx 1.62$ kN, at $21.3^\circ$ below the direction of the jet.

(b) Work done per second

$$\text{Work} = F_x\,u = 1514 \times 12 = 18\,167\ \text{J/s} \approx 18.2\ \text{kW}$$

Only $F_x$ does work, because the vanes move along the jet direction.

Answer (b): about 18.2 kW.

Apply Bernoulli between the two sections to find $p_2$, then apply the momentum equation in the $x$ and $y$ directions on the water in the bend.

Assumptions

The bend lies in a vertical plane. Inlet flow is along $+x$ (horizontal). "120° bend" means the flow is turned through $120^\circ$, so the outlet velocity makes $120^\circ$ with the inlet direction and points upward ($V_{2x} = V_2\cos120^\circ$, $V_{2y} = V_2\sin120^\circ$). $W = 1500$ N acts downward. $\rho = 1000\ \text{kg/m}^3$.

            ^ y        2 (outlet, 0.4 m higher)
            |         /
            |        /  120 deg from inlet direction
   1 -----> +-------+--------> x

Velocities and pressures

$$\begin{aligned} A_1 &= 0.07069\ \text{m}^2,\; A_2 = 0.03142\ \text{m}^2 \\ V_1 &= 0.30/0.07069 = 4.244\ \text{m/s},\quad V_2 = 0.30/0.03142 = 9.549\ \text{m/s} \\ p_2 &= p_1 + \rho g(z_1 - z_2) + \tfrac12\rho(V_1^2 - V_2^2) \\ &= 210\,000 - 3924 + \tfrac12(1000)(18.01 - 91.19) = 169\,488\ \text{Pa} \end{aligned}$$

Forces: $p_1A_1 = 14\,844$ N, $p_2A_2 = 5325$ N. Momentum flux: $\rho QV_1 = 1273$ N, $\rho QV_2 = 2865$ N.

Momentum equations (forces by bend on fluid)

$$\begin{aligned} x:\;& F_x + p_1A_1 - p_2A_2\cos120^\circ = \rho Q\,(V_2\cos120^\circ - V_1) \\ & F_x = 1000(0.3)(-4.775 - 4.244) - 14\,844 - 2662 = -20\,212\ \text{N} \\[4pt] y:\;& F_y - W - p_2A_2\sin120^\circ = \rho Q\,V_2\sin120^\circ \\ & F_y = 2481 + 1500 + 4611 = 8592\ \text{N} \end{aligned}$$

Resultant

$$F = \sqrt{20\,212^2 + 8592^2} = 21\,962\ \text{N},\qquad \theta = 180^\circ - \tan^{-1}\frac{8592}{20\,212} = 157.0^\circ \text{ from the inlet direction}$$

Answer: $F_x = -20.2$ kN (opposite to inlet flow), $F_y = +8.59$ kN (upward); resultant $\approx 22.0$ kN at $157^\circ$ to the $+x$ axis. The force on the bend is equal and opposite.