A fire hose 10 cm in diameter delivers water at the rate of 23 kg/s. The hose terminates in a nozzle 3.1 cm in diameter.
(a) What is the flow speed in the hose?
(b) What is the flow speed in the nozzle?
To solve for the flow speed in the hose and nozzle, we can make use of the principle of conservation of mass. According to this principle, the mass flow rate at any given point in a pipe must remain constant.
Given:
Diameter of the hose (Dh) = 10 cm = 0.1 m
Mass flow rate (m) = 23 kg/s
(a) Calculating the flow speed in the hose (Vh):
The cross-sectional area of the hose (Ah) is given by the formula:
Ah = π(Dh/2)^2
Substituting the values, we have:
Ah = π(0.1/2)^2 = π(0.05)^2 = 0.00785 m^2
The flow speed in the hose (Vh) is calculated using the formula:
Vh = m / Ah
Substituting the values, we have:
Vh = 23 / 0.00785 ≈ 2932.49 m/s
Therefore, the flow speed in the hose is approximately 2932.49 m/s.
(b) Calculating the flow speed in the nozzle (Vn):
Given:
Diameter of the nozzle (Dn) = 3.1 cm = 0.031 m
The cross-sectional area of the nozzle (An) is given by the formula:
An = π(Dn/2)^2
Substituting the values, we have:
An = π(0.031/2)^2 = π(0.0155)^2 = 0.00075 m^2
The flow speed in the nozzle (Vn) is calculated using the formula:
Vn = m / An
Substituting the values, we have:
Vn = 23 / 0.00075 = 30666.67 m/s
Therefore, the flow speed in the nozzle is approximately 30666.67 m/s.
To find the flow speed in the hose and the nozzle, we can use the principle of continuity, which states that the volume flow rate of an incompressible fluid remains constant as it flows through a pipe of varying cross-sectional area. The equation for continuity is:
A1v1 = A2v2
where A1 and A2 are the cross-sectional areas of the two points in the pipe, and v1 and v2 are the flow speeds at those points.
(a) First, let's calculate the flow speed in the hose.
Given:
Diameter of the hose (D1) = 10 cm
Radius of the hose (r1) = D1/2 = 5 cm = 0.05 m
Volume flow rate (Q) = 23 kg/s
The cross-sectional area of the hose (A1) can be calculated using the formula:
A1 = π*r1^2
Substituting the values, we get:
A1 = π*(0.05 m)^2
= 0.00785 m^2
Now, using the principle of continuity, we can rearrange the equation to solve for v1:
v1 = (A2/A1) * v2
Since the hose terminates in a nozzle, the cross-sectional area of the nozzle (A2) can be calculated using the diameter of the nozzle (D2):
Diameter of the nozzle (D2) = 3.1 cm
Radius of the nozzle (r2) = D2/2 = 1.55 cm = 0.0155 m
A2 = π*r2^2
= π*(0.0155 m)^2
= 0.00075 m^2
Substituting the values for A1, A2, and v2, we can solve for v1:
v1 = (0.00075 m^2 / 0.00785 m^2) * v2
= 0.09554 * v2
(b) Next, let's calculate the flow speed in the nozzle.
To find v2, we need to use the equation:
Q = A2 * v2
Substituting the values we already know, we get:
23 kg/s = 0.00075 m^2 * v2
Simplifying the equation and solving for v2, we get:
v2 = 23 kg/s / 0.00075 m^2
= 30,666.67 m/s
Finally, substituting the value of v2 in the equation for v1, we get:
v1 = 0.09554 * 30,666.67 m/s
= 2,930.66 m/s
So, the flow speed in the hose is approximately 2,930.66 m/s, and the flow speed in the nozzle is approximately 30,666.67 m/s.
(a) The volume flow rate Q is the mass flow rate divided by the density.
Q = 23/1000 = 0.023 m^3/s
Q = V*A, where A is the cross sectinal area, (pi/4)*D^2 = 10^-2 m^2
and V is the flow speed
V = .023/10^-2 = 2.3 m/s
(b) Q remains contant in the nozzle, but V increases because of the reduction in A. A is reduced by a factor (10/3.1)^2 = 10.4. V increases by the same factor.