Question

Ethanol in room temperature flows in a horizontal, smooth pipe. the pipe diameter isd = 25 cm and its length is l = 10 m .what is the pressure drop (in si), if the flow rate is q = 15 lit/s?

Answer 1

The pressure drop in the pipe can be calculated using Bernoulli's equation. In this case, with a flow rate of 15 L/s, a pipe length of 10 m, and a diameter of 25 cm, the pressure drop is approximately 47.16 kPa.

Step-by-step explanation:

Pressure Drop in a Pipe

To calculate the pressure drop in a smooth pipe, we can use Bernoulli's equation. Bernoulli's equation states that the pressure drop is equal to the product of the flow rate, pipe length, and fluid's density, divided by the pipe diameter.

Pressure Drop = (Flow Rate * Length * Density) / Diameter

In this case, the flow rate is given as 15 L/s, the length is 10 meters, the density of ethanol is approximately 789 kg/m³, and the diameter is 25 cm (or 0.25 m). Plugging these values into the equation, we get:

Pressure Drop = (15 * 10 * 789) / 0.25

Calculating this expression gives a pressure drop of approximately 47160 Pa or 47.16 kPa.