Momentum Equation for Flow Force
Hydrodynamic force problems use the linear momentum equation. Resolve forces by components and assign signs before substituting.
$$\sum F_x=\rho Q(V_{2x}-V_{1x})$$$$\sum F_y=\rho Q(V_{2y}-V_{1y})$$
Forces on Pressurized Pipe Bends
For elbows and nozzles, include pressure forces at inlet and outlet, weight if significant, and momentum change. The required anchor force is opposite the force of water on the pipe.
Water Hammer Pressure
Water hammer is the pressure surge created when moving liquid is stopped or redirected suddenly.
$$\Delta p=\rho c\Delta V$$$$h_w=\frac{c\Delta V}{g}$$
For gradual valve closure, reduce the surge using the closure-time relation required by the problem statement.
Sudden Valve Closure Head
Water velocity in a pipe drops from 2.0 m/s to zero. If wave celerity is 1000 m/s, find water hammer head.
$$h_w=\frac{c\Delta V}{g}=\frac{1000(2.0)}{9.81}=203.9\text{ m}$$
Answer: The surge head is approximately 204 m of water.
Problem: Force on a 90° Pipe Elbow
Water flows at Q = 0.080 m³/s through a 150 mm diameter pipe that turns 90° in the horizontal plane (a right-angle elbow). The gauge pressure at the inlet of the elbow is 200 kPa. Neglect elevation changes and head loss through the elbow. Find the x- and y-components of the force that the water exerts on the elbow, and the magnitude and direction of the resultant force.
Pipe area: $A = \pi(0.15)^2/4 = 0.01767$ m². By continuity (same diameter), inlet and outlet velocities are equal: $V = Q/A = 0.080/0.01767 = 4.53$ m/s. For a 90° horizontal bend, inlet flow is in the +x direction and outlet is in the +y direction (or vice versa — assign inlet as +x, outlet as +y).
$$p_1 = 200{,}000 \text{ Pa}, \quad \text{By Bernoulli (no loss, same elevation, same diameter):} \quad p_2 = p_1 = 200{,}000 \text{ Pa}$$
$$\rho Q = 1000(0.080) = 80 \text{ kg/s}$$
Momentum equation (forces on fluid from pipe = $R_x, R_y$; water exerts equal and opposite forces on the pipe):
$$\sum F_x = \rho Q(V_{2x} - V_{1x}): \quad p_1 A - R_x = \rho Q(0 - V)$$
$$R_x = p_1 A + \rho Q V = 200{,}000(0.01767) + 80(4.53)$$
$$R_x = 3534 + 362 = 3896 \text{ N} \quad(\text{pipe pushes fluid in }+x)$$
$$\sum F_y = \rho Q(V_{2y} - V_{1y}): \quad -p_2 A + R_y = \rho Q(V - 0)$$
$$R_y = p_2 A + \rho Q V = 3534 + 362 = 3896 \text{ N}$$
Force of water on elbow is equal and opposite to $R_x, R_y$ (reaction from fluid on pipe):
$$F_x = -3896 \text{ N}, \quad F_y = -3896 \text{ N (away from bend)}$$
$$F_R = \sqrt{F_x^2 + F_y^2} = \sqrt{2}(3896) = 5510 \text{ N} = 5.51 \text{ kN}$$
$$\theta = 45° \text{ (diagonal, away from the bend)}$$
Answer: Each component of the force on the elbow is 3.896 kN. The resultant is 5.51 kN at 45° from each pipe direction (diagonal outward). Pipe elbows must be anchored against this force — it is the combination of pressure thrust and momentum flux that tends to push the bend off its supports.
Problem: Force on a Pipe Nozzle
A 200 mm diameter pipe (inlet) reduces to a 100 mm diameter nozzle at the outlet. Water flows at Q = 0.060 m³/s and the inlet gauge pressure is 150 kPa. Find the force required to hold the nozzle in place (the axial force on the nozzle-pipe connection). Neglect losses and elevation change.
$$A_1 = \frac{\pi(0.20)^2}{4} = 0.03142 \text{ m}^2, \quad V_1 = \frac{0.060}{0.03142} = 1.910 \text{ m/s}$$
$$A_2 = \frac{\pi(0.10)^2}{4} = 0.007854 \text{ m}^2, \quad V_2 = \frac{0.060}{0.007854} = 7.639 \text{ m/s}$$
Apply Bernoulli (same elevation, no loss) to find $p_2$:
$$\frac{p_1}{\gamma} + \frac{V_1^2}{2g} = \frac{p_2}{\gamma} + \frac{V_2^2}{2g}$$
$$\frac{150{,}000}{9810} + \frac{(1.910)^2}{19.62} = \frac{p_2}{9810} + \frac{(7.639)^2}{19.62}$$
$$15.29 + 0.186 = \frac{p_2}{9810} + 2.975$$
$$\frac{p_2}{9810} = 12.50 \text{ m} \Rightarrow p_2 = 122{,}625 \text{ Pa} \approx 122.6 \text{ kPa}$$
Momentum equation in flow direction (x), let F = bolt force (tension) holding nozzle:
$$p_1 A_1 - p_2 A_2 - F = \rho Q(V_2 - V_1)$$
$$150{,}000(0.03142) - 122{,}625(0.007854) - F = 1000(0.060)(7.639 - 1.910)$$
$$4713 - 963.1 - F = 60(5.729)$$
$$3749.9 - F = 343.7$$
$$F = 3749.9 - 343.7 = 3406 \text{ N} = 3.41 \text{ kN}$$
Answer: The nozzle connection must resist 3.41 kN in tension (the pipe pulling back against the nozzle thrust). The dominant contribution is the pressure force at the inlet; momentum change adds only about 10% of the total load.
Problem: Water Hammer — Wave Celerity in an Elastic Pipe
A steel water main has an inside diameter of 400 mm and a wall thickness of 10 mm. The pipe is assumed to be anchored at both ends. The bulk modulus of water is $K = 2.07 \times 10^9$ Pa and the Young's modulus of steel is $E = 200 \times 10^9$ Pa. Determine the wave celerity $c$ for water hammer. Then find the water hammer surge pressure if the initial velocity is 1.5 m/s and the valve closes instantaneously.
The wave celerity in an elastic pipe accounts for both water compressibility and pipe wall elasticity:
$$c = \frac{\sqrt{K/\rho}}{\sqrt{1 + \frac{KD}{Et}}}$$
where $D$ = inside diameter, $t$ = wall thickness, $K$ = bulk modulus of water, $E$ = Young's modulus of pipe material.
$$\frac{K}{\rho} = \frac{2.07 \times 10^9}{1000} = 2.07 \times 10^6 \text{ m}^2/\text{s}^2$$
$$\sqrt{K/\rho} = \sqrt{2.07 \times 10^6} = 1438.7 \text{ m/s (celerity in rigid pipe)}$$
$$\frac{KD}{Et} = \frac{(2.07 \times 10^9)(0.40)}{(200 \times 10^9)(0.010)} = \frac{8.28 \times 10^8}{2.0 \times 10^9} = 0.414$$
$$c = \frac{1438.7}{\sqrt{1 + 0.414}} = \frac{1438.7}{\sqrt{1.414}} = \frac{1438.7}{1.189} = 1210 \text{ m/s}$$
$$\Delta p = \rho c \Delta V = 1000(1210)(1.5) = 1{,}815{,}000 \text{ Pa} = 1815 \text{ kPa}$$
$$h_w = \frac{c \Delta V}{g} = \frac{1210(1.5)}{9.81} = 185.0 \text{ m}$$
Answer: Wave celerity c = 1210 m/s. Surge pressure = 1815 kPa (about 185 m water head). Pipe elasticity reduces celerity below the rigid-pipe value of 1439 m/s and correspondingly reduces the surge — this is why steel pipes are better than rigid cast-iron from a water hammer standpoint.
Problem: Safe Valve Closure Time to Limit Surge
Water flows at 2.5 m/s in a 500 m long pipeline with a wave celerity of 1200 m/s. The maximum allowable surge pressure head is 80 m above the normal operating pressure. Determine: (a) whether instantaneous closure is safe, and (b) the minimum closure time needed for the valve to close "gradually" so that reflected waves can reduce the pressure rise. Use the criterion that closure is "gradual" if the closure time $T > 2L/c$.
$$h_w (\text{instantaneous}) = \frac{c \Delta V}{g} = \frac{1200(2.5)}{9.81} = 305.8 \text{ m}$$
305.8 m >> 80 m allowable — instantaneous closure is NOT safe.
The critical time for pressure wave return trip (pipe period):
$$T_c = \frac{2L}{c} = \frac{2(500)}{1200} = 0.833 \text{ s}$$
For gradual closure ($T > T_c$), the surge head is approximately reduced by the ratio $T_c/T$:
$$h_{w,gradual} \approx \frac{c \Delta V}{g} \cdot \frac{T_c}{T} \leq 80 \text{ m}$$
$$T \geq \frac{305.8(0.833)}{80} = \frac{254.7}{80} = 3.18 \text{ s}$$
Answer: (a) Instantaneous closure produces 305.8 m surge — far above the 80 m limit, so it is unsafe. (b) The valve must close in at least 3.18 seconds (more than 4 pipe periods) to keep surge below 80 m. In practice, valve closure times of 10–30 seconds or longer are used for large water mains to prevent catastrophic pipe failure.
Problem: Force of a Jet on a Stationary Flat Plate
A water jet 50 mm in diameter exits a nozzle at 12 m/s and strikes a flat stationary plate perpendicularly. Assuming the jet spreads radially after impact (velocity remains 12 m/s tangentially, zero normal component after), find: (a) the force on the plate, and (b) the force if the plate moves away from the jet at 4 m/s.
(a) Stationary plate:
$$A_{jet} = \frac{\pi(0.05)^2}{4} = 1.963 \times 10^{-3} \text{ m}^2$$
$$Q = V A = 12(1.963 \times 10^{-3}) = 0.02356 \text{ m}^3/\text{s}$$
$$F = \rho Q \Delta V_{normal} = \rho Q(V_{in} - 0) = 1000(0.02356)(12) = 282.7 \text{ N}$$
(b) Moving plate (u = 4 m/s away from jet): Relative velocity of jet to plate = V − u = 12 − 4 = 8 m/s. The mass flow rate actually intercepted by the moving plate uses the relative velocity:
$$\dot{m}_{rel} = \rho A (V - u) = 1000(1.963 \times 10^{-3})(8) = 15.70 \text{ kg/s}$$
$$F = \dot{m}_{rel}(V - u) = \rho A(V-u)^2 = 1000(1.963 \times 10^{-3})(8)^2 = 125.6 \text{ N}$$
Answer: (a) Force on stationary plate = 282.7 N ≈ 283 N. (b) Force on moving plate = 125.6 N ≈ 126 N. The force drops to (8/12)² = 44% of the stationary-plate value because both the intercepted flow rate and the momentum change per unit mass decrease with plate velocity. Maximum power transfer to the plate occurs when u = V/3.