Respuesta :
Answer:
P₁ - P₂ = 17.18 MPa
Explanation:
Pressure drop along the horizontal pipe = P₁ - P₂
= [tex]\dfrac{128\mu L Q}{\piD^4}[/tex]
length of the pipe = 60 m
flow rate of propyl alcohol = 15 l/s
= 0.015 m³/s (∵ 1 l/s = 10⁻³ m³/s)
for 4 standard pipe,
diameter of the pipe = 8 mm = 0.008 m
viscosity of propyl alcohol = 0.00192 N-s/m²
[tex]P_1-P_2 =\dfrac{128\mu L Q}{\piD^4}[/tex]
= [tex]\dfrac{128\times 0.00192 \times 60 \times 0.015}{\pi\times 0.008^4}[/tex]
P₁ - P₂ = 17.18 MPa
hence, pressure drop is equal to 17.18 MPa
