Water is flowing in a pipe with a circular cross-section area ,and at all point the water completely fills the pipe.At point 1 ,the cross sectiona are of pipe is 0.070 meter square and the fluid velocity is 3.50 meter per second.
a.) what is the fluid speed at point in the pipe where the cross-sectional area is 0.105 meter square?
b.) Calculate the volume of the water discharge from the open end of the pine in 1000 hour.
Law of continuity: Mass rate in = mass rate out, or
Area1*density1*Velocity1=area2*velocity2*density2
densities are equal, so
.07m^2*3.5m/sec=.105m^2*V solve for velocity V
volume/hr=V*3600sec/hr
in 1000 hrs, multiply that by 1000
To calculate the fluid speed at point B, where the cross-sectional area is 0.105 m²:
Given:
- Cross-sectional area at Point A (A1) = 0.070 m²
- Fluid velocity at Point A (v1) = 3.50 m/s
We can use the principle of continuity, which states that the mass flow rate of a fluid is constant at all points along the pipe.
The equation for the principle of continuity is:
A1 * v1 = A2 * v2
Where:
A2 = Cross-sectional area at Point B
v2 = Fluid velocity at Point B
Now, plug in the given values and solve for v2:
0.070 m² * 3.50 m/s = 0.105 m² * v2
v2 = (0.070 m² * 3.50 m/s) / 0.105 m²
v2 = 2.333 m/s
Therefore, the fluid speed at point B is 2.333 m/s.
To calculate the volume of water discharge from the open end of the pipe in 1000 hours:
Given:
- Fluid velocity at the open end of the pipe (v_out) = v2 = 2.333 m/s
The volume flow rate (Q) is given by the equation:
Q = A_out * v_out
Where:
A_out = Cross-sectional area at the open end of the pipe
To find A_out, we need to know the shape of the cross-section at the open end of the pipe. If it is circular like the rest of the pipe, you can use the same formula as before.
Assuming it is circular and using the given value of A1 and v1, we can solve for A_out:
0.070 m² * 3.50 m/s = A_out * 2.333 m/s
A_out = (0.070 m² * 3.50 m/s) / 2.333 m/s
A_out = 0.105 m²
Now, we can calculate the volume of water discharged in 1000 hours:
Q = A_out * v_out
Q = 0.105 m² * 2.333 m/s
Q = 0.244 m³/s
To convert this to the volume discharged in 1000 hours, we multiply it by the time:
Volume discharged = Q * time
Volume discharged = 0.244 m³/s * 1000 hours
Note that we need to convert hours to seconds as follows:
1 hour = 3600 seconds
Volume discharged = 0.244 m³/s * 1000 hours * 3600 s/hour
Volume discharged = 878,400 m³
Therefore, the volume of water discharged from the open end of the pipe in 1000 hours is 878,400 cubic meters.
To find the fluid speed at point 2 where the cross-sectional area is 0.105 m^2, we can use the principle of conservation of mass. According to this principle, the mass flow rate of fluid is constant at all points along the pipe.
a) Since the density of water is constant, we can say that the mass flow rate (m_dot) at point 1 is equal to the mass flow rate at point 2:
m_dot_1 = m_dot_2
The mass flow rate is equal to the product of the density (ρ) and the volume flow rate (Q):
m_dot = ρ * Q
The volume flow rate (Q) is equal to the product of the cross-sectional area (A) and the fluid speed (v):
Q = A * v
Plugging these equations together, we get:
ρ * A1 * v1 = ρ * A2 * v2
Since the density cancels out, we get:
A1 * v1 = A2 * v2
Now we can plug in the given values:
A1 = 0.070 m^2 (cross-sectional area at point 1)
v1 = 3.50 m/s (fluid velocity at point 1)
A2 = 0.105 m^2 (cross-sectional area at point 2)
v2 = ?
Solving for v2, we get:
v2 = (A1 * v1) / A2
v2 = (0.070 m^2 * 3.50 m/s) / 0.105 m^2
v2 = 2.333 m/s
Therefore, the fluid speed at point 2 where the cross-sectional area is 0.105 m^2 is approximately 2.333 m/s.
b) To calculate the volume of water discharged from the open end of the pipe in 1000 hours, we can again use the principle of conservation of mass.
The volume flow rate (Q) is equal to the product of the cross-sectional area (A) and the fluid speed (v):
Q = A * v
The volume discharge (V) can be calculated by multiplying the volume flow rate (Q) by the time (t):
V = Q * t
Let's plug in the given values:
A = 0.105 m^2 (cross-sectional area)
v = 2.333 m/s (fluid speed)
t = 1000 hours
First, we need to convert the time from hours to seconds:
t = 1000 hours * 3600 seconds/hour = 3,600,000 seconds
Now we can calculate the volume discharge:
V = (A * v) * t
V = (0.105 m^2 * 2.333 m/s) * 3,600,000 seconds
V ≈ 88,060,000 m^3
Therefore, the volume of water discharged from the open end of the pipe in 1000 hours is approximately 88,060,000 cubic meters.