2. (Problem 4.60 on main book, diameters different) Water flows steadily through a fire hose and nozzle. The hose is 35 mm diameter and the nozzle tip is 25 mm diameter; water gage pressure in the hose is 510 kPa, and the stream leaving the nozzle is uniform. The exit speed and pressure are 32 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.

Answers

Answer 1

Answer:

coupling is in tension

Force = -244.81 N

Explanation:

Diameter of Hose ( D1 ) = 35 mm

Diameter of nozzle ( D2 ) = 25 mm

water gage pressure in hose = 510 kPa

stream leaving the nozzle is uniform

exit speed and pressure = 32 m/s and atmospheric

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

attached below is the remaining part of the  detailed solution

Inlet velocity ( V1 ) = V2 ( D2/D1 )^2  

= 32 ( 25 / 35 )^2

= 16.33 m/s

2. (Problem 4.60 On Main Book, Diameters Different) Water Flows Steadily Through A Fire Hose And Nozzle.

Related Questions

Combinations of velocity and acceleration

Answers

Answer:

acceleration=change in velocity/ time

Explanation:

The velocity of an object is its speed in a particular direction. Velocity is a vector quantity because it has both a magnitude and an associated direction. To calculate velocity, displacement is used in calculations, rather than distance.

Fig_Q5
6. A steel rod is stressed by a tension force of 250 N. It is found that the rod has length of 45
m and diameter of 1.5 mm. If the modulus of elasticity of the steel rod is assumed as 2 x 105
MPa, determine the strain of the steel rod due to the applied force.

Answers

Answer:

The strain of the steel rod due to the applied force is 41.93

Explanation:

Modulus of elasticity is equal to stress divided by strain.

And stress is equal to force divided by area

Surface area of cylindrical rod

[tex]2\pi r (r+h)[/tex]

Substituting the given values we get -

[tex]2 *3.14 * \frac{1.5}{1000} * 45 (45 + \frac{1.5}{1000}) = 19.07[/tex]

[tex]2 * 10 ^5 = \frac{250}{19.07 * S=(\frac{\Delta L}{L} )}[/tex]

Hence, strain is equal to

Strain = 41.93

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