Answer:
Pressure drop = 971.52 KPa, Head Loss = 148.9 m, Pumping Power = 0.131 KW
Step-by-step explanation:
T = -20 C, L = 17 m, Diameter = 5 x 10⁻³m, Mass (flow rate) = 0.09 kg/s
From table of properties of saturate ammonia at T = -20 C, ρ(density) = 665.1 kg/m³, μ(dynamic viscosity) = 2.36 x 10⁻⁴ kg/ms, Roughness of Copper tube)ε = 1.5 x 10⁻⁶
Calculate Area of cross sectional flow = (π/4) (D)² = (π/4) (5 x 10⁻³)² = 19.635 x 10⁻⁶ m²
Velocity of liquid = Mass (flow rate)/(ρ x Area cross sectional) = 0.09/665.1 x 19.635 x 10⁻⁶ = 6.89 m/s
Flow Rate (V) = Mass (flow rate)/ρ = 0.09/665.1 = 1.35 x 10⁻⁴ m/s
Reynold's Number (Re) = ρ x velocity x D/μ = 665.1 x 6.89 x 5 x 10⁻³/ 2.36 x 10⁻⁴ = 97087.7
The Re is greater than 4000, so the flow is said to be turbulent
For turbulent flow, use the Cole- brook equation
(1/√f) = -2.0㏒[{(ε/D)/3.7} + {2.51/(Re x √f)}]
(1/√f) = -2.0㏒[{(1.5 x 10⁻⁶/5 x 10⁻³)/3.7} + {2.51/(97087.7 x √f)}]
(1/√f) = -2.0㏒[{2.027 x 10⁻¹⁰ + {2.59 x 10⁻⁵/√f)}]
solve for f which is equal to 0.01810
Pressure Drop = f x (L/D) x (ρ x velocity²/2)
ΔP(l) = 0.01810x (17/5 x 10⁻³) x (665.1x 6.59²/2) = 971522.6 Pa = 971.52 KPa
Head Loss = ΔP(l)/ρ x g, where g is the acceleration due to gravity
H(l) = 971522.6/(665.1 x 9.81) = 148.9 m
Pumping Power Required = V x ΔP(l)
W(pump) = 1.35 x 10⁻⁴ x 9715522.6 = 131.16 W = 0.131 KW