A fluid moves in a steady flow manner between two sections ina flow line. At section 1: A1=10ft2, v1=100ft/min, specificvolume1= 4ft3/lb. At section 2: A2=2ft2, density 2= 0.20 lb/ft3.Calculate the mass flow rate and speed at section 2.
To calculate the mass flow rate and speed at section 2, we first need to understand the principles of steady flow and the equations involved.
1. Mass Flow Rate (ṁ):
The mass flow rate is the amount of mass passing through a given area per unit time. It can be calculated using the equation:
ṁ = ρ * A * v
Where:
ṁ = mass flow rate (lb/min)
ρ = density (lb/ft³)
A = cross-sectional area (ft²)
v = velocity (ft/min)
2. Speed at Section 2 (v2):
The speed at section 2 can be obtained using the continuity equation, which states that the mass flow rate remains constant along a flow line.
Since mass flow rate is constant, we can equate the mass flow rate at section 1 to the mass flow rate at section 2:
ṁ1 = ṁ2
Therefore, we can write:
ρ1 * A1 * v1 = ρ2 * A2 * v2
Now we can substitute the given values and solve the equations:
Given:
A1 = 10 ft²
v1 = 100 ft/min
specific volume1 = 4 ft³/lb
A2 = 2 ft²
density2 = 0.20 lb/ft³
First, let's calculate the density at section 1:
density1 = 1 / specific volume1 = 1 / 4 ft³/lb = 0.25 lb/ft³
Next, calculate the mass flow rate (ṁ) at section 1:
ṁ1 = density1 * A1 * v1 = 0.25 lb/ft³ * 10 ft² * 100 ft/min = 250 lb/min
Now, using the continuity equation, we can solve for v2:
ρ1 * A1 * v1 = ρ2 * A2 * v2
Rearranging the equation for v2:
v2 = (ρ1 * A1 * v1) / (ρ2 * A2)
Substituting the given values:
v2 = (0.25 lb/ft³ * 10 ft² * 100 ft/min) / (0.20 lb/ft³ * 2 ft²)
= 1250 ft/min
So, the mass flow rate at section 2 is 250 lb/min, and the speed at section 2 is 1250 ft/min.
To calculate the mass flow rate, we can use the formula:
ṁ = ρ * A * v
Where:
ṁ is the mass flow rate
ρ is the density of the fluid
A is the cross-sectional area
v is the velocity of the fluid
Given values:
A1 = 10 ft^2
v1 = 100 ft/min
Specific volume 1 (v1) = 4 ft^3/lb
A2 = 2 ft^2
Density 2 (ρ2) = 0.20 lb/ft^3
First, let's calculate the mass flow rate at section 1:
Using the specific volume (v1) at section 1, we can calculate the density (ρ1) using the formula:
ρ1 = 1 / v1
ρ1 = 1 / 4
ρ1 = 0.25 lb/ft^3
Now we can calculate the mass flow rate (ṁ1) at section 1:
ṁ1 = ρ1 * A1 * v1
ṁ1 = 0.25 * 10 * 100
ṁ1 = 250 lb/min
Next, let's calculate the mass flow rate (ṁ2) at section 2. Since the fluid is in a steady flow, the mass flow rate remains constant along the flow line.
ṁ1 = ṁ2
ṁ2 = ṁ1
ṁ2 = 250 lb/min
Finally, let's calculate the speed (v2) at section 2:
v2 = ṁ2 / (ρ2 * A2)
v2 = 250 / (0.20 * 2)
v2 = 625 ft/min
So, the mass flow rate at section 2 is 250 lb/min and the speed at section 2 is 625 ft/min.