Answer:
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
