Given that
Density of fluid ρ = 7 × 1 0 2 k g / m 3 ρ=7 \times 10^2 \;kg/m^3 ρ = 7 × 1 0 2 k g / m 3
Inlet hose radius r 1 = 0.024 m r_1 = 0.024 \;m r 1 = 0.024 m
Outlet hose radius r 2 = 0.012 m r_2 =0.012 \;m r 2 = 0.012 m
Pressure difference Δ P = 1.2 × 1 0 3 P a ΔP=1.2 \times 10^3 \;Pa Δ P = 1.2 × 1 0 3 P a
A) From equation of continuity
A 1 v 1 = A 2 v 2 v 1 = A 2 A 1 v 2 v 1 2 = ( π r 2 2 π r 1 2 ) 2 v 2 2 = ( ( 0.012 m ) 2 ( 0.024 m ) 2 ) 2 v 2 2 = ( 0.0325 ) v 2 2 A_1v_1=A_2v_2 \\
v_1= \frac{A_2}{A_1}v_2 \\
v_1^2 =(\frac{\pi r_2^2}{\pi r_1^2})^2 v^2_2 \\
= (\frac{(0.012 \;m)^2}{(0.024 \;m)^2})^2 v^2_2 \\
= (0.0325)v^2_2 A 1 v 1 = A 2 v 2 v 1 = A 1 A 2 v 2 v 1 2 = ( π r 1 2 π r 2 2 ) 2 v 2 2 = ( ( 0.024 m ) 2 ( 0.012 m ) 2 ) 2 v 2 2 = ( 0.0325 ) v 2 2
Applying Bernoulli’s theorem for a hose pipe of the same height
Δ P = 1 2 ρ [ v 2 2 − v 1 2 ] = 1 2 ρ [ v 2 2 − ( 0.0625 ) v 2 2 ] = 1 2 ρ v 2 2 [ 1 − 0.0625 ] v 2 = 2 Δ P ρ ( 1 − 0.0625 ) = 2 × 1200 P a ( 700 k g / m 3 ) × ( 1 − 0.0625 ) = 1.9 m / s ΔP= \frac{1}{2}ρ[v^2_2 -v^2_1] \\
= \frac{1}{2}ρ[v^2_2 -(0.0625)v^2_2] \\
= \frac{1}{2}ρ v^2_2 [1-0.0625] \\
v_2 = \sqrt{ \frac{2ΔP}{ρ(1-0.0625)}} \\
= \sqrt{\frac{2 \times 1200 \;Pa}{(700 \;kg/m^3) \times (1-0.0625)}} \\
=1.9 \;m/s Δ P = 2 1 ρ [ v 2 2 − v 1 2 ] = 2 1 ρ [ v 2 2 − ( 0.0625 ) v 2 2 ] = 2 1 ρ v 2 2 [ 1 − 0.0625 ] v 2 = ρ ( 1 − 0.0625 ) 2Δ P = ( 700 k g / m 3 ) × ( 1 − 0.0625 ) 2 × 1200 P a = 1.9 m / s
b) The flow rate is
Q = A 2 v 2 = π r 2 2 v 2 2 = π × ( 0.012 m ) 2 × ( 1.9123 m / s ) = 8.65 × 1 0 − 4 m 3 / s Q=A_2v_2 \\
= \pi r^2_2 v^2_2 \\
= \pi \times (0.012 \;m)^2 \times (1.9123 \;m/s) \\
= 8.65 \times 10^{-4} \;m^3/s Q = A 2 v 2 = π r 2 2 v 2 2 = π × ( 0.012 m ) 2 × ( 1.9123 m / s ) = 8.65 × 1 0 − 4 m 3 / s
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