Water is flowing at 3 m/s in a horizontal pipe under a pressure of 200 kP a. The pipe narrows to half its original diameter.
(a) What is the speed of flow in the narrow section?
(b) What is the pressure in the narrow section?
(c) How do the volume flow rates in the two sections compare? (Assume laminar non-viscous flow)
To find the answers to these questions, we can apply the principle of conservation of mass and Bernoulli's equation.
(a) To find the speed of flow in the narrow section, we can use the principle of conservation of mass, which states that the mass flow rate of a fluid through a pipe is constant. In this case, since the density of water is constant, the mass flow rate is also constant throughout the pipe.
The formula for mass flow rate is given by:
Mass flow rate = density * cross-sectional area * velocity
Since the cross-sectional area and density are constant, we can write:
Mass flow rate = A1 * v1 = A2 * v2
Where:
- A1 is the cross-sectional area of the pipe in the initial section
- v1 is the velocity of the water in the initial section
- A2 is the cross-sectional area of the pipe in the narrow section (which is half of A1)
- v2 is the velocity of the water in the narrow section (the value we want to find)
Given that A2 = (1/2)*A1 and v1 = 3 m/s, we can substitute these values into the equation and solve for v2:
A1 * v1 = A2 * v2
A1 * 3 = (1/2)*A1 * v2
3 = (1/2)*v2
v2 = 6 m/s
Therefore, the speed of flow in the narrow section is 6 m/s.
(b) To find the pressure in the narrow section, we can use Bernoulli's equation, which states that the total energy of a fluid flowing in a pipe is constant along a streamline (assuming laminar non-viscous flow).
The equation is given by:
P + (1/2)ρv^2 + ρgh = constant
Where:
- P is the pressure
- ρ is the density of the fluid
- v is the velocity of the fluid
- g is the acceleration due to gravity
- h is the height of the fluid above a reference point (we can assume it is constant since the pipe is horizontal)
We can assume h = 0 since the pipe is horizontal and there is no vertical displacement. Rearranging the equation, we get:
P + (1/2)ρv^2 = constant
Since the total energy is constant, the pressure in the initial section is 200 kPa. Therefore, substituting the values into the equation:
200 kPa + (1/2)*(density of water)*3^2 = P2 + (1/2)*(density of water)*6^2
Simplifying the equation, we have:
200 kPa + (1/2)*ρ*(9) = P2 + (1/2)*ρ*(36)
200 kPa + (1/2)*ρ*(9) = P2 + (1/2)*ρ*(36)
Substituting the known values, we get:
200 kPa + (1/2)*1000 kg/m^3 * 9 m^2/s^2 = P2 + (1/2)*1000 kg/m^3 * 36 m^2/s^2
Simplifying further, we have:
200,000 Pa + 4,500 Pa = P2 + 18,000 Pa
P2 = 200,000 Pa + 4,500 Pa - 18,000 Pa
P2 = 186,500 Pa
Therefore, the pressure in the narrow section is 186,500 Pa, or 186.5 kPa.
(c) The volume flow rate can be calculated using the formula:
Volume flow rate = cross-sectional area * velocity
Since the density is constant, we can directly compare the volume flow rates.
In the initial section, the cross-sectional area is A1 and the velocity is v1.
In the narrow section, the cross-sectional area is A2 and the velocity is v2.
Using the known values, A1 = (1/2)*A2 and v1 = 3 m/s, we can substitute into the equation:
Volume flow rate in initial section = A1 * v1
Volume flow rate in narrow section = A2 * v2 = (1/2)*A1 * v2
Therefore, the volume flow rate in the narrow section is half of the volume flow rate in the initial section, assuming laminar non-viscous flow.