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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.

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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

[tex]A_1V_1=A_2V_2\\[/tex]

By simplifying the equation

[tex]V_1=V_2(\frac{D_2}{D_1})^2[/tex]

Here

V_2=43 m/2

D1=78 mm

D2=20 mm

So the velocity is given as

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

Now by applying the equation of momentum in x direction

[tex]R_x=-P_{1g}A_1+\rho V_1^2A_1-\rho V_2^2A_2[/tex]

Here

  • P1g is given as 427 kPa
  • A1 is given by [tex]\pi D_1^2/4=\pi (78/1000)^2/4=0.00477 m^2[/tex]
  • 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 [tex]\pi D_2^2/4=\pi (20/1000)^2/4=0.00031 m^2[/tex]

Substituting the values in the equation gives

[tex]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[/tex]

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.

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