Momentum Equation & Forces

When water rushes through a fire-hose nozzle and the firefighters lean back to hold it, they are fighting the momentum of the jet. When a rocket lifts off, it is the change of momentum of the exhaust gases that drives it skyward. The momentum equation is how engineers turn "fluid in motion" into "force on a structure."

In this lesson, we build the linear momentum equation from the Reynolds transport theorem, then apply it to pipe bends, jets striking vanes, and propulsion. We finish with the angular momentum principle that underlies every pump and turbine.

Reynolds Transport Theorem (RTT)

Newton's laws apply to a fixed mass (a system), but in fluid mechanics we prefer a fixed region in space (a control volume, CV). The RTT bridges the two. For any extensive property $B$ with intensive value $b = dB/dm$:

$$\frac{dB_{sys}}{dt} = \frac{\partial}{\partial t}\int_{CV} \rho\, b \, dV + \int_{CS} \rho\, b \,(\vec{V}\cdot\hat{n})\, dA$$

The first term is the rate of change inside the control volume; the second is the net flux across the control surface (CS). Setting $b = \vec{V}$ (so $B$ is momentum) and using $dB_{sys}/dt = \sum \vec{F}$ gives the momentum equation.

Linear Momentum Equation

For a control volume:

$$\sum \vec{F} = \frac{\partial}{\partial t}\int_{CV} \rho\, \vec{V} \, dV + \int_{CS} \rho\, \vec{V}\,(\vec{V}\cdot\hat{n})\, dA$$

For steady flow with uniform properties at each inlet/outlet, this simplifies to the form we use most:

$$\sum \vec{F} = \sum_{out} \dot{m}\,\vec{V} - \sum_{in} \dot{m}\,\vec{V}$$

where the mass flow rate is $\dot{m} = \rho A V$. The forces $\sum \vec{F}$ include pressure forces, body forces (gravity), and the reaction force from solid surfaces. This is a vector equation — always resolve into $x$ and $y$ components.

Force on a Pipe Bend

Consider a reducing bend that turns the flow through angle $\theta$. Applying momentum in $x$ and $y$, the force the fluid exerts on the bend has components:

$$F_x = p_1 A_1 - p_2 A_2 \cos\theta + \dot{m}(V_1 - V_2\cos\theta)$$

$$F_y = -p_2 A_2 \sin\theta - \dot{m}\,V_2\sin\theta$$

The resultant magnitude and direction are:

$$F_R = \sqrt{F_x^2 + F_y^2}, \qquad \alpha = \tan^{-1}\!\left(\frac{F_y}{F_x}\right)$$

Python: Force on a Reducing Bend

import numpy as np

# Reducing 45-degree bend, water
rho   = 1000.0          # kg/m^3
theta = np.radians(45)  # bend angle
D1, D2 = 0.30, 0.20     # inlet/outlet diameters (m)
p1    = 250e3           # inlet gauge pressure (Pa)
Q     = 0.30            # flow rate (m^3/s)

A1 = np.pi/4 * D1**2
A2 = np.pi/4 * D2**2
V1 = Q / A1
V2 = Q / A2
mdot = rho * Q

# Estimate p2 from Bernoulli (neglect losses, horizontal bend)
p2 = p1 + 0.5*rho*(V1**2 - V2**2)

Fx = p1*A1 - p2*A2*np.cos(theta) + mdot*(V1 - V2*np.cos(theta))
Fy = -p2*A2*np.sin(theta) - mdot*(V2*np.sin(theta))

FR    = np.hypot(Fx, Fy)
alpha = np.degrees(np.arctan2(Fy, Fx))

print(f"V1 = {V1:.2f} m/s, V2 = {V2:.2f} m/s")
print(f"p2 = {p2/1e3:.1f} kPa")
print(f"Fx = {Fx/1e3:.2f} kN")
print(f"Fy = {Fy/1e3:.2f} kN")
print(f"Resultant = {FR/1e3:.2f} kN at {alpha:.1f} deg")

V1 = 4.24 m/s, V2 = 9.55 m/s
p2 = 213.9 kPa
Fx = 12.71 kN
Fy = -6.92 kN
Resultant = 14.47 kN at -28.6 deg

The anchor block holding this bend must resist about 14.5 kN — roughly the weight of 1.5 tonnes.

Jet on a Vane

A free jet (area $A$, velocity $V$) striking a surface delivers a force because its momentum is redirected.

$$F = \rho A V^2 = \dot{m}\,V$$

$$F_x = \rho A V^2 (1 - \cos\theta)$$

For a moving vane travelling at speed $u$ in the jet direction, only the relative velocity $(V-u)$ matters, and the mass actually striking the vane is $\rho A (V-u)$:

$$F_x = \rho A (V-u)^2 (1 - \cos\theta)$$

The power delivered is $P = F_x\, u$, maximized (for a series of vanes on a wheel) near $u = V/2$.

Python: Jet Force vs Vane Angle

import numpy as np
import matplotlib.pyplot as plt

rho = 1000.0
D   = 0.05                      # jet diameter (m)
V   = 30.0                      # jet velocity (m/s)
A   = np.pi/4 * D**2

theta = np.linspace(0, 180, 181)        # deflection angle (deg)
Fx = rho * A * V**2 * (1 - np.cos(np.radians(theta)))

# Print a few representative values
for ang in (0, 45, 90, 135, 180):
    f = rho * A * V**2 * (1 - np.cos(np.radians(ang)))
    print(f"theta = {ang:3d} deg -> Fx = {f:6.1f} N")

plt.figure(figsize=(9, 5))
plt.plot(theta, Fx, 'b-', linewidth=2)
plt.axhline(rho*A*V**2, color='g', ls='--', label='Flat plate (90 part)')
plt.xlabel('Vane deflection angle (deg)')
plt.ylabel('Force on vane Fx (N)')
plt.title('Jet-on-Vane Force vs Deflection Angle')
plt.grid(True, alpha=0.3); plt.legend()
plt.show()

theta =   0 deg -> Fx =    0.0 N
theta =  45 deg -> Fx =  517.6 N
theta =  90 deg -> Fx = 1767.1 N
theta = 135 deg -> Fx = 3016.7 N
theta = 180 deg -> Fx = 3534.3 N

A full reversal ($\theta = 180^\circ$) doubles the force compared to a flat plate — this is why turbine buckets are deeply cupped.

Thrust: Rockets and Jets

Propulsion is pure momentum exchange. For a jet engine ingesting air at $\dot{m}_a$ and ejecting it at $V_e$ while the craft moves at $V_\infty$:

$$T = \dot{m}_a (V_e - V_\infty) + \dot{m}_f V_e$$

For a rocket in vacuum, there is no incoming air, so:

$$T = \dot{m}\, V_e + (p_e - p_a)A_e$$

The first term is momentum thrust; the second is pressure thrust when the nozzle is not perfectly expanded.

# Static rocket thrust at sea level
mdot = 250.0     # propellant mass flow (kg/s)
Ve   = 2800.0    # exhaust velocity (m/s)
pe   = 90e3      # exit pressure (Pa)
pa   = 101.3e3   # ambient (Pa)
Ae   = 0.8       # exit area (m^2)

T = mdot*Ve + (pe - pa)*Ae
print(f"Momentum thrust = {mdot*Ve/1e3:.1f} kN")
print(f"Pressure thrust = {(pe-pa)*Ae/1e3:.1f} kN")
print(f"Total thrust    = {T/1e3:.1f} kN")

Momentum thrust = 700.0 kN
Pressure thrust = -9.0 kN
Total thrust    = 691.0 kN

The negative pressure thrust shows the nozzle is over-expanded at sea level — exit pressure is below ambient, costing 9 kN.

Angular Momentum: The Euler Turbomachine Equation

For rotating machinery (pumps, turbines, fans) we apply the moment-of-momentum principle. The torque on the rotor equals the rate of change of angular momentum:

$$T = \dot{m}\,(r_2 V_{t2} - r_1 V_{t1})$$

where $V_t$ is the tangential (whirl) component of velocity. Multiplying by angular speed $\omega$ and using blade speed $U = \omega r$ gives the Euler head delivered:

$$H_E = \frac{U_2 V_{t2} - U_1 V_{t1}}{g}$$

This single equation governs the energy transfer in essentially every rotodynamic machine.

g = 9.81
mdot = 50.0      # kg/s
r1, r2 = 0.10, 0.25
Vt1, Vt2 = 0.0, 18.0    # radial inflow, no inlet whirl
omega = 1500*2*np.pi/60  # rad/s

U2 = omega*r2
T  = mdot*(r2*Vt2 - r1*Vt1)
HE = (U2*Vt2 - 0)/g
print(f"Blade speed U2 = {U2:.1f} m/s")
print(f"Torque  = {T:.1f} N.m")
print(f"Euler head = {HE:.1f} m")
print(f"Power   = {T*omega/1e3:.1f} kW")

Blade speed U2 = 39.3 m/s
Torque  = 225.0 N.m
Euler head = 72.1 m
Power   = 35.3 kW

Common Pitfalls

• Forgetting it is a vector equation. Always resolve into $x$ and $y$; momentum flux carries direction.
• Mixing gauge and absolute pressure. Use consistent gauge pressures for the pressure-force terms in a bend.
• Wrong velocity for moving vanes. The striking mass uses relative velocity $\rho A (V-u)$, not $\rho A V$.
• Sign of the reaction. The force on the fluid and the force on the wall are equal and opposite — state which you report.
• Ignoring pressure thrust in nozzle problems when the exit is not perfectly expanded.

Key Takeaways

• The momentum equation comes from the Reynolds transport theorem with $b=\vec{V}$: $\sum\vec{F} = \dot{m}\vec{V}_{out} - \dot{m}\vec{V}_{in}$.
• Pipe-bend forces combine pressure and momentum flux; resolve components and take the resultant.
• Jet-on-vane force scales with $(1-\cos\theta)$ — a $180^\circ$ turn gives the maximum.
• Thrust is momentum exchange; rockets add a pressure-thrust term for imperfect expansion.
• The Euler turbomachine equation turns angular momentum into the head and torque of pumps and turbines.

In the next lesson, we use dimensional analysis to predict full-scale behavior from small models.