Question

Turpentine flows by gravity from tank 1 to tank 2 through 24 m of 1.5-inch diameter commercial steel pipe. Both tanks are open to atmosphere. The tanks are large, so assume that the level of turpentine in both tanks remains the same with time. The level of tank 1 is 15 m higher than the level of tank 2 (indicated on the schematic as z1 and z2). There are two standard 90° elbows and one open gate valve in the line. Calculate the volumetric flow rate in the line. Note: ρturpentine = 863 kg/m3 ; µturpentine = 0.0016 Pa s

Answer 1

In both the tanks pressure is same(atmospheric). As the height in Tank 1 is 15 m higher than the Tank 2. This head will be responsible for the flow of Turpentine.

let assume that frictional loss to be minor. After solving we will take a look into it.

Apply bernoulli equation between tank 1 and tank 2.

K_f = loss coefficients

K_f ( Standard elbow 90⁰ ) = 0.75   K_f ( Gate valve Open) = 0.17

P₁ = 1 atm P₂ = 1 atm Since both the tanks are open to atmosphere

Z₁ - Z₂ = 15 m ; The level of tank 1 is 15 m higher than the level of tank 2.

K_f (overall) = 2*K_f (Standard elbow 90⁰) + K_f ( Gate valve Open) =2*0.75 +0.17 = 1.67

Pipe dimensions ; Diameter = 1.5 inch = 0.0381 m Length = 24 m

Cross sectional area of Pipe

Volumetric flow rate

Checking for frictional loss:

Turbulent flow Re = 2.15*10⁵

Calculating friction factor for turbulent flow;

Again iterating for friction factor

Again iterating for friction factor

Here we can stop because velocity is nearly coming constant.

Volumetric flow rate;

You can also check solution by putting value in equation (a).