113k views
2 votes
A solution of ethanol is pumped to a vessel 25 m above a reference level through a 25-mm-inside-diameter steel pipe at a rate of 10 m3 /h. The length of pipe is 30 m and contains two elbows with friction equivalent to 20 diameters each. Compute the power requirements of the pump. Solution properties include density of 975 kg/m3 and viscosity of 4 3 1024 Pa s

User Zabbarob
by
8.8k points

1 Answer

2 votes
Answer:

To compute the power requirements of the pump, we need to determine the head loss and the pump's efficiency. The head loss in the pipeline is given by the Darcy-Weisbach equation:

hL = f (L / D) (V^2 / 2g)

where hL is the head loss, f is the friction factor, L is the length of the pipe, D is the inside diameter of the pipe, V is the average fluid velocity, and g is the acceleration due to gravity.

First, we need to calculate the fluid velocity:

Q = A * V

where Q is the flow rate, A is the cross-sectional area of the pipe, and V is the fluid velocity.

The cross-sectional area of the pipe is:

A = π/4 * D^2

A = π/4 * (0.025 m)^2

A = 4.91 x 10^-4 m^2

So, the fluid velocity is:

V = Q / A

V = 10 m^3/h / (3600 s/h) / (4.91 x 10^-4 m^2)

V = 5.04 m/s

Next, we need to calculate the Reynolds number to determine the friction factor:

Re = (ρVD) / μ

where ρ is the fluid density and μ is the fluid viscosity.

Re = (975 kg/m^3)(5.04 m/s)(0.025 m) / (4.3 x 10^-4 Pa s)

Re = 5.73 x 10^5

Using the Moody chart or a Colebrook equation solver, we can determine the friction factor for the given Reynolds number and roughness of the steel pipe. For simplicity, we will assume a friction factor of 0.02.

The head loss due to friction in the pipe is:

hL = f (L / D) (V^2 / 2g)

hL = 0.02 (30 m / 0.025 m) (5.04 m/s)^2 / (2 x 9.81 m/s^2)

hL = 24.4 m

The head loss due to the two elbows is:

hL = K (V^2 / 2g)

where K is the equivalent length of the elbow in diameters and is equal to 20 diameters each. From a piping handbook, K for a long radius 90° elbow is approximately 30 diameters.

hL = 30 (5.04 m/s)^2 / (2 x 9.81 m/s^2)

hL = 7.82 m

The total head loss is:

hL_total = hL_friction + hL_elbows

hL_total = 24.4 m + 7.82 m

hL_total = 32.2 m

The power required by the pump is:

P = ρQhL_total / η

where η is the pump efficiency.

We will assume a pump efficiency of 75%.

P = (975 kg/m^3)(10 m^3/h)(3600 s/h)(32.2 m)/(0.75)

P = 1.13 x 10^6 W or 1.13 MW

Therefore, the power requirements of the pump are 1.13 MW.
User ScanQR
by
8.5k points

No related questions found