Question

Water flows steadily through a fire hose and nozzle. The hose is 78 mm inside diameter, and the nozzle tip is 20 mm I.D.; water gage pressure in the hose is 427 kPa, and the stream leaving the nozzle is uniform. The exit speed and pressure are 43 m/s and atmospheric, respectively. Find the force transmitted by the coupling between the nozzle and hose. Indicate whether the coupling is in tension or compression.

Answer 1

Answer:

The reaction force is 2.776kN as the force is negative and on the left of the Control Volume this indicates that the coupling in in tension.

Explanation:

By applying the continuity equation as

$A_1V_1=A_2V_2$

By simplifying the equation

$V_1=V_2(\frac{D_2}{D_1})^2$

Here

V_2=43 m/2

D1=78 mm

D2=20 mm

So the velocity is given as

$V_1=V_2(\frac{D_2}{D_1})^2\\V_1=2.827 m/s$

Now by applying the equation of momentum in x direction

$R_x=-P_{1g}A_1+\rho V_1^2A_1-\rho V_2^2A_2$

Here

• P1g is given as 427 kPa

• A1 is given by $\pi D_1^2/4=\pi (78/1000)^2/4=0.00477 m^2$

• ρ is the density of water given as 1000 kg/m3
• V1 is calculated above as 2.827 m/s
• V2 is given as 42 m/s

• A2 is given by $\pi D_2^2/4=\pi (20/1000)^2/4=0.00031 m^2$

Substituting the values in the equation gives

$R_x=-P_{1g}A_1+\rho V_1^2A_1-\rho V_2^2A_2\\R_x=-(470 \times 10^3)(0.00477)+(1000) (2.827)^2(0.00477)-(1000) (43)^2(0.00031)\\R_x=-2776.96 N$

So the reaction force is 2.776kN as the force is negative and on the left of the Control Volume this indicates that the coupling in in tension.