Answer:
Q= 4.6 × 10⁻³ m³/s
actual velocity will be equal to 8.39 m/s
Explanation:
density of fluid = 900 kg/m³
d₁ = 0.025 m
d₂ = 0.05 m
Δ P = -40 k N/m²
C v = 0.89
using energy equation
[tex]\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g} = \dfrac{P_2}{\gamma}+\dfrac{v_2^2}{2g}\\\dfrac{P_1-P_2}{\gamma}=\dfrac{v_2^2-v_1^2}{2g}\\\dfrac{-40\times 10^3\times 2}{900}=v_2^2-v_1^2[/tex]
under ideal condition v₁² = 0
v₂² = 88.88
v₂ = 9.43 m/s
hence discharge at downstream will be
Q = Av
Q = [tex]\dfrac{\pi}{4}d_1^2 \times v[/tex]
Q = [tex]\dfrac{\pi}{4}0.025^2 \times 9.43[/tex]
Q= 4.6 × 10⁻³ m³/s
we know that
[tex]C_v =\dfrac{actual\ velocity}{theoretical\ velocity }\\0.89 =\dfrac{actual\ velocity}{9.43}\\actual\ velocity = 8.39m/s[/tex]
hence , actual velocity will be equal to 8.39 m/s
