Physics Problem: Hydrodynamic Shock
Question 12
Water is continuously flowing at a speed $v = 1.0 \text{ m/s}$ in a steel pipeline of radius $r = 5.0 \text{ cm}$. A valve installed in the pipeline is closed almost instantaneously resulting in an abrupt rise in water pressure. What should the minimum thickness of the pipe be in order to provide a safety factor of $\eta = 5$? Density of water is $\rho = 1.0 \times 10^3 \text{ kg/m}^3$, speed of sound in water $c = 1.5 \times 10^3 \text{ m/s}$ and tensile strength of steel is $\sigma_b = 0.35 \text{ GPa}$.
Part 1: Derivation of Pressure Rise (Joukowsky Equation)
1. Consider a fluid element of length $\Delta x = c \Delta t$ and mass $m = \rho A c \Delta t$.
2. Change in momentum as it stops: $\Delta p = m(0 – v) = – (\rho A c \Delta t) v$.
3. Force causing this is due to pressure difference: $F = -(\Delta P) A$.
4. From $F = \Delta p / \Delta t$:
$$ -(\Delta P) A = \frac{- \rho A c \Delta t v}{\Delta t} \implies \Delta P = \rho c v $$
Part 2: Derivation of Hoop Stress (From Image)
Consider a small section of pipe of length $l$ subtending an angle $d\theta$.
1. Outward Force (Pressure):
$$ F_P = P \times \text{Projected Area} = P \times (r d\theta \times l) $$
2. Inward Force (Tension):
The hoop stress $\sigma$ creates a tension $T$ in the pipe walls.
$$ T = \text{Stress} \times \text{Area} = \sigma \times (t \times l) $$
The radial component of $T$ from both sides balances the pressure.
$$ F_{inward} = 2T \sin(d\theta/2) $$
3. Equilibrium:
$$ 2(\sigma t l) \sin(d\theta/2) = P (r d\theta l) $$
For small angles, $\sin(d\theta/2) \approx d\theta/2$:
$$ 2 \sigma t l \frac{d\theta}{2} = P r l d\theta $$
$$ \sigma t = P r $$
$$ \sigma = \frac{Pr}{t} $$
Part 3: Numerical Calculation
1. Calculate Pressure Rise:
$$ \Delta P = \rho c v = (1000)(1500)(1.0) = 1.5 \text{ MPa} $$
2. Calculate Allowed Stress:
$$ \sigma_{allowed} = \frac{\sigma_b}{\eta} = \frac{350}{5} = 70 \text{ MPa} $$
3. Find Minimum Thickness:
Using $\sigma = \frac{Pr}{t} \implies t = \frac{Pr}{\sigma}$:
$$ t = \frac{(1.5 \times 10^6 \text{ Pa})(0.05 \text{ m})}{70 \times 10^6 \text{ Pa}} $$
$$ t = \frac{0.075}{70} \approx 0.00107 \text{ m} $$