Water flows steadily through a fire hose and nozzle. The hose is 75 mm inside diameter, and the nozzle tip is 25 mm inside diameter; water gage pressure in the hose is 510 kPa, and the stream leaving the nozzle is uniform. The exit speed 32 m/s and pressure is atmospheric. Determine the force transmitted by the coupling between the nozzle and hose. (25 points)

R = 1804 N

Explanation:

Given:-

- The density of water, ρ = 997 kg/m^3

- The inside diameter of the hose, dh = 75 mm

- The gauge pressure of water in the hose, P1 = 510 KPa

- The exit speed of the water, V2 = 32 m/s

- The inside diameter of the nozzle tip, dn = 25 mm

- The atmospheric pressure (gauge), P2 = 0 KPa ... P = 1 atm (Absolute).

Find:-

Determine the force transmitted by the coupling between the nozzle and hose.

Solution:-

- We will first develop a control surface at the hose-nozzle interface.

- Assuming steady and one dimensional flow - (x-direction).

- Since there are no fictitious unbalanced forces acting on the fluid flow due to roughness of hose any any losses of energy from the fluid are negligible.

- The use of conservation of momentum of fluid flow is valid for an isolated system, where the flow of fluid into the control volume is denoted by (-) and the flow of fluid going out of the control volume is denoted by (+):

- The principle of conservation of momentum, the pair of equal force (Newton's third law) act on the control volume at (nozzle-hose) interface:

                  R = ρ*Q*(V2 - V1) + (P2*A2 - P1*A1)

Where,        Q: Flow rate

                   V1: The velocity of fluid in hose

                   A1: Cross sectional area of the hose

                   A2: Cross sectional area of the nozzle exit

- We see that the reaction force (R) that acts on nozzle-hose interface is due to changes in dynamic and hydrostatic pressures.

- Compute the required quantities Q, A1 and A2 and V1 using the given data:

- The flow rate Q for any flow in the hose can be given, where the cross sectional area of hose (A1)is:  

- The cross sectional area of the nozzle tip with diameter dn = 25 mm is:

- The flow rate (Q) can now be calculated:

- Since, the density of the water does not vary along the direction of flow, the flow rate (Q) remains constant throughout. So from continuity equation we have:

- Now use the calculated quantities and compute the pair of reaction force at the nozzle-hose interface:

                 R = ρ*Q*(V2 - V1) + (P2*A2 - P1*A1)

                 R = (997)*(0.01570)*(32-3.56189) + (0 - 510*0.00441)*1000

                 R = 445.13889 - 2,249.1

                R = - 1803.961 ≈ -1,804 N

- Here the negative sign denotes the direction of in which the force (R) is exerted. Since, (-) denotes into the control volume it acts opposite to the flow of water.

answer:

the answer is d

explanation:

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