A garden hose attached with a nozzle is used to fill a 20-gallon bucket. The inner diameter of the hose is 1 inch and it reduces to 0.5 inch at the nozzle exit. If the average velocity in the hose is 8ft/s, determine (a) the volume and mass flow rates of water through the hose (4-point), (b) how long it will take to fill the bucket with water (3-point), and (c) the average velocity of water at the nozzle exit (3-point).

a. i. 0.044 ft³/s ii. 2.75 lb/s

b. 62.3 s

c. 32 ft/s

Explanation:

(a) the volume and mass flow rates of water through the hose

i. The volume flow rate

The volume flow rate, Q = Av where A = cross-sectional area of nozzle at inlet = πd²/4 where r = diameter of nozzle at inlet = 1 inch = 1/12 ft and v = average velocity in the nozzle = 8 ft/s.

So, Q = Av

= πd²v/4

= π(1/12 ft)² × 8 ft/s/4

= 0.175/4 ft³/s

= 0.044 ft³/s

ii. The mass flow rate

The mass flow rate, M = ρQ where ρ = density of water = 62.4 lb/ft³ and Q = volume flow rate = 0.044 ft³/s

So, M = ρQ

= 62.4 lb/ft³ × 0.044 ft³/s

=  2.745 lb/s

≅ 2.75 lb/s

(b) how long it will take to fill the bucket with water

Since we have 20 gallons = 20 × 1 gallon, we convert if to cubic feet. So, 20 gallons = 20 × 1 gallon = 20 × 0.137 ft³ = 2.74 ft³.

Since V = Qt where V = volume, Q = volume flow rate and t = time, making t subject of the formula, we have

t = V/Q

= 2.74 ft³/0.044 ft³/s

= 62.27 s

≅ 62.3 s

(c) the average velocity of water at the nozzle exit

Using the continuity equation,

A₁v₁ = A₂v₂ where A₁ = cross-sectional area of nozzle at inlet = πd₁²/4 where d₁ = diameter of nozzle at inlet = 1 inch = 1/12 ft, v₁ = average velocity in the nozzle = 8 ft/s, A₂ = cross-sectional area of nozzle at inlet = πd₂²/4 where d₂ = diameter of nozzle at inlet = 0.5 inch = 0.5/12 ft, v₂ = average velocity at exit nozzle.

So, A₁v₁ = A₂v₂

(πd₁²/4)v₁ = (πd₂²/4)v₂

πd₁²v₁/4 = πd₂²v₂/4

Dividing through by π/4. we have

So, d₁²v₁ = d₂²v₂

making v₂ subject of the formula, we have

v₂ = d₁²v₁/d₂²

v₂ = (d₁²/d₂²)v₁

v₂ = (d₁/d₂)²v₁

substituting the values of the variables into the equation, we have

v₂ = (1/0.5)² × 8 ft/s

v₂ = (2)² × 8 ft/s

v₂ = 4 × 8 ft/s

v₂ = 32 ft/s

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the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

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