Liquid ammonia at −20°c is flowing through a 17-m-long section of a 5-mm-diameter copper tube at a rate of 0.09 kg/s. determine the pressure drop, the head loss, and the pumping power required to overcome the frictional losses in the tube. the density and dynamic viscosity of liquid ammonia at –20°c are rho

Pressure drop = 971.52 KPa, Head Loss = 148.9 m, Pumping Power = 0.131 KW

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

answer:

make you upset

explanation: