Respuesta :
Answer:
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:
[tex]A1 =\pi\frac{d_h^2}{4} = \pi\frac{0.075^2}{4} \\\\A1 = 0.00441 m^2\\\\\\[/tex]
- The cross sectional area of the nozzle tip with diameter dn = 25 mm is:
[tex]A2 =\pi\frac{d_n^2}{4} = \pi\frac{0.025^2}{4} \\\\A2 = 0.00049 m^2\\\\\\[/tex]
- The flow rate (Q) can now be calculated:
[tex]Q = A2*V2\\\\Q = (0.00049)*(32)\\\\Q = 0.01570 \frac{m^3}{s}[/tex]
- 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:
[tex]Q = A2*V2 = A1*V1\\\\V1 = \frac{Q}{A1} = \frac{0.0157}{0.00441} \\\\V1 = 3.56189 \frac{m}{s}[/tex]
- 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.
The coupling between the nozzle and hose is -1.81N
This question relates to flow rate of a liquid
Data given:
The density of water = 997kg/m^3
The inside diameter of the hose = 75mm = 0.0075m
The gauge pressure of water in the hose = 510kPa
The exit speed of the water = 32m/s
The inside diameter of the nozzle tip = 25mm = 0.0025m
The atmospheric pressure = 0kPa or 1atm
Let's calculate the inlet velocity
[tex]v_1=v_2=A_2/A_1\\v_1=V_2(\frac{d_2}{d_1})^2\\v_1=32(\frac{25}{75})^2\\v_1=3.50m/s[/tex]
Calculating the force transmitted by coupling between the nozzle and hose
[tex]R_x+p_1gA_1=v_1[-|pv_1A_1|]+v_2[|pv_2A_2|]\\[/tex]
μ[tex]_1[/tex]=[tex]v_1[/tex] and μ[tex]_2[/tex] =[tex]v_2[/tex]
[tex]R_x=-p_1gA_1-v_1pv_1A_1+v_2pv_2A_2\\R_x=-p_1gA+pv_2A_2(v_2-v_1)\\R_x=-510*10^3N/m^3*\frac{\pi }{4}(0.075m)^2+997kg/m^3*32m/s*\frac{\pi }{4} (0.025m)^2(32-3.50)=-1805=-1.81kN[/tex]
The force between the nozzle and hose is -1.81
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