Pipe wall stress for 3 s valve closure

The valve closure time is compared with the critical time $2L/c$. If the closure is slower than $2L/c$, the pressure rise is found from the gradual-closure formula $\Delta p=\rho L V/t_c$. Hoop (circumferential) stress in the wall is then $\sigma=pD/2t$ using the total pressure.

Given data

$D=0.25$ m, $L=1500$ m, $Q=0.075$ m³/s, $t=6$ mm, $t_c=3$ s, static head = 200 m, $K=2.2\times10^9$ N/m², $E=2.1\times10^{11}$ N/m².

Step 1: Velocity

$$A=\frac{\pi}{4}(0.25)^2=0.04909\ \text{m}^2,\qquad V=\frac{Q}{A}=\frac{0.075}{0.04909}=1.528\ \text{m/s}$$

Step 2: Wave celerity in elastic pipe

$$c=\sqrt{\frac{K/\rho}{1+\dfrac{KD}{Et}}}=\sqrt{\frac{2.2\times10^9/1000}{1+\dfrac{2.2\times10^9\times0.25}{2.1\times10^{11}\times0.006}}}=1237.5\ \text{m/s}$$

Step 3: Type of closure

$$\frac{2L}{c}=\frac{2\times1500}{1237.5}=2.42\ \text{s}<t_c=3\ \text{s}$$

So the closure is gradual (slow), and the Joukowsky value $\rho cV$ does not apply.

Step 4: Pressure rise

$$\Delta p=\frac{\rho L V}{t_c}=\frac{1000\times1500\times1.528}{3}=763944\ \text{N/m}^2=0.764\ \text{MPa}\ (=77.9\ \text{m of water})$$

Step 5: Stress in pipe wall

Static pressure $p_s=\rho g h=1000\times9.81\times200=1.962$ MPa.

Total pressure $p=p_s+\Delta p=1.962+0.764=2.726$ MPa.

$$\sigma=\frac{pD}{2t}=\frac{2725944\times0.25}{2\times0.006}=56.79\ \text{MPa}$$

The stress due to the static head alone is 40.88 MPa; the water hammer adds 15.92 MPa.

Answer: pressure rise = 0.764 MPa (77.9 m); maximum hoop stress in pipe wall = 56.8 MPa (about 57 N/mm²).