Linear momentum: forces on bends, nozzles and jets

Where the 600 mm transmission main turns 90° at the foot of the reservoir ridge, a massive concrete anchor block surrounds the bend. Jun Park asked why a slow-moving water main needs so much concrete. The answer comes from Newton's second law applied to a control volume: whenever a flow changes direction or speed, or the pressure in it acts on a changing area, the pipe must push on the water, and the water pushes back on the pipe. Without the anchor block, the bend would be shoved off its joints the first time the main was pressurized. This lesson develops the linear momentum equation and applies it to bends, reducers, nozzles and jets.

The momentum equation for a control volume

For a system, Newton's second law says that the net force equals the rate of change of momentum. Moving to a fixed control volume with the Reynolds transport theorem, as we did for mass in Lesson 6, gives the linear momentum equation. The net force on the fluid inside the control volume equals the rate of increase of momentum inside plus the net rate at which momentum flows out across the control surface.

For steady flow with uniform velocities at each inlet and outlet, the first term vanishes, and for a single inlet and outlet the equation reduces to \(\sum \mathbf{F} = \dot{m}(\mathbf{V}_2 - \mathbf{V}_1)\). It is a vector equation, applied separately in each coordinate direction, and the velocities are vectors: a flow that turns without changing speed still changes its momentum.

The forces on the fluid in the control volume are of three kinds. Pressure forces act on every part of the control surface where fluid crosses it; at an inlet or outlet of a pipe, this is the pressure times the area, pushing inward. Body forces, mainly gravity, act on the mass inside. And the solid boundaries, such as the pipe wall and the bend, exert a force on the fluid, which is usually the unknown we want. The force on the pipe from the water is equal and opposite.

Choosing the control volume and using gauge pressure

Good control volumes cut across inlets and outlets perpendicular to the flow, where velocity and pressure are known, and enclose the object whose force we want. Draw the control volume, mark the coordinate axes, and draw every force: pressure forces at each cut, the weight if gravity has a component along an axis, and the unknown reaction force from the solid. Then write the momentum equation in each direction, taking care that each velocity component has the right sign.

Atmospheric pressure acts all around the outside of a pipe and the anchor block. If we used absolute pressure at the inlet and outlet cuts, we would also have to include atmospheric pressure acting on the outside surfaces. Since a uniform pressure over a closed surface exerts no net force, it is simpler to use gauge pressures at the cuts and ignore atmospheric pressure entirely. The result is the force the anchor must provide beyond what the surrounding air already balances.

The program below repeats the calculation for 45° and 90° bends and separates the pressure and momentum contributions. In water mains, pressure dominates; the momentum terms matter more in high-velocity jets and in low-pressure, high-velocity flows such as fan ducts. A bend in a gravity sewer, flowing at low pressure, needs little thrust restraint, while the same bend in a pressurized force main needs a block.

# Linear momentum: anchor force on a reducing bend and force of a jet on a plate
import math
rho = 998.0

def bend_force(Q, D1, D2, p1g, p2g, theta_deg, W=0.0):
    """Force the anchor block must exert on a horizontal-plane reducing bend (x along the inlet)."""
    A1, A2 = math.pi * D1**2 / 4, math.pi * D2**2 / 4
    V1, V2 = Q / A1, Q / A2
    th = math.radians(theta_deg)
    mdot = rho * Q
    # sum F = mdot (V_out - V_in), pressures in gauge
    Rx = mdot * (V2 * math.cos(th) - V1) - p1g * A1 + p2g * A2 * math.cos(th)
    Ry = mdot * (V2 * math.sin(th)) + p2g * A2 * math.sin(th)
    return Rx, Ry, V1, V2

Q = 0.45
p1 = 520e3                                 # Pa gauge upstream
D1, D2 = 0.600, 0.500
V1, V2 = Q / (math.pi * D1**2 / 4), Q / (math.pi * D2**2 / 4)
p2 = p1 + rho * (V1**2 - V2**2) / 2        # Bernoulli across the bend, losses neglected
for th in (45, 90):
    Rx, Ry, V1, V2 = bend_force(Q, D1, D2, p1, p2, th)
    R = math.hypot(Rx, Ry)
    print(f"{th:3d} deg bend: V1 = {V1:.2f}, V2 = {V2:.2f} m/s, Rx = {Rx/1000:7.1f} kN, Ry = {Ry/1000:6.1f} kN, |R| = {R/1000:6.1f} kN")

# How much of the 90-degree anchor force comes from pressure?
A1, A2 = math.pi * D1**2 / 4, math.pi * D2**2 / 4
print(f"Pressure part of Rx: {-p1*A1/1000:.1f} kN; momentum part: {-rho*Q*V1/1000:.2f} kN")

# Water jet from a fire monitor striking a flat plate normally
d, Vj = 0.040, 30.0
Qj = math.pi * d**2 / 4 * Vj
print(f"Jet force on a fixed plate: {rho*Qj*Vj:.0f} N; on a plate moving away at 10 m/s: {rho*math.pi*d**2/4*(Vj-10)**2:.0f} N")

 45 deg bend: V1 = 1.59, V2 = 2.29 m/s, Rx =   -75.0 kN, Ry =   72.7 kN, |R| =  104.5 kN
 90 deg bend: V1 = 1.59, V2 = 2.29 m/s, Rx =  -147.7 kN, Ry =  102.9 kN, |R| =  180.0 kN
Pressure part of Rx: -147.0 kN; momentum part: -0.71 kN
Jet force on a fixed plate: 1129 N; on a plate moving away at 10 m/s: 502 N

Thrust restraint in practice

Unbalanced forces occur at every bend, tee, reducer, dead end and closed valve in a pressurized pipeline. Pipes with push-on or mechanical joints cannot carry tension across the joints, so the force must be transferred to the soil. Concrete thrust blocks bear against undisturbed soil behind the fitting; their bearing area is the thrust divided by the soil's allowable bearing pressure, which might be 75 to 200 kPa. Where space or soil is poor, restrained joints lock several lengths of pipe together so friction along the pipe and soil resistance carry the thrust instead. A dead end, such as a capped main awaiting extension, sees the full \(pA\) as thrust, which for a 600 mm main at 520 kPa is 147 kN.

Vertical bends add another twist. At a bend that turns upward, the thrust pushes down into the soil and is easily resisted; at a bend that turns downward over a hump, the thrust pulls up, and the block must be heavy enough to hold the bend down by its weight. Hana's crews learned this the hard way when a poorly anchored vertical bend lifted during a pressure test and split its joint.

Nozzles and jets

A nozzle accelerates water, so it needs an axial force to hold it on its pipe. For a fire hose nozzle reducing a 65 mm hose to a 25 mm tip, the jet leaves at high speed and the pressure inside the hose is large; the coupling must hold back the pressure force on the hose area minus the momentum gained. Firefighters feel the reaction of the jet as a push back along the hose, roughly \(\rho Q V_{\text{jet}}\), which for a large handline can exceed 300 N.

A free jet striking a surface exerts a force equal to the rate at which its momentum is changed. A jet hitting a fixed flat plate square-on and spreading sideways loses all its momentum in the original direction, so the force is \(\rho Q V\). If the plate moves away at speed \(u\), the jet strikes it at the relative velocity \(V - u\), and the mass flow reaching the plate is also reduced, so the force becomes \(\rho A (V - u)^2\). A jet turned back on itself by a curved vane changes its momentum by nearly twice as much, which is the principle of the Pelton turbine, whose buckets turn the jet through almost 180°.

Momentum flux correction and the limits of the method

Using the average velocity to compute momentum flux slightly underestimates it, because the velocity varies across the pipe and momentum depends on velocity squared. A momentum flux correction factor \(\beta\) fixes this: about 1.02 to 1.04 for turbulent pipe flow and 4/3 for laminar flow. For most water main calculations the correction is far smaller than the uncertainty in pressure, and it is set to 1.

The momentum equation gives forces from what crosses the control surface, without our needing to know the complicated flow inside. That is its great strength. It cannot by itself give the pressure distribution inside a fitting, and when losses are significant, the outlet pressure must come from the energy equation with a loss term, not from Bernoulli. The anchor block example used Bernoulli for the outlet pressure because the bend is short and smooth; for a sharp-cornered fitting, a measured loss coefficient should be used instead, as Lesson 12 explains.