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A fluid moves in a steady flow manner between two between two sections

in a flow line. At section 1; 𝐴1= 10𝑓𝑑2
, 𝑣1 = 100 fpm, Ρ΅ = 4 𝑓𝑑3
𝑙𝑏
. At section 2;
𝐴2= 2 𝑓𝑑2
, 𝜌2= 0.20 lb/𝑓𝑑3
. Calculate (a) the mass flow rate and (b) the
speed at section 2.

To solve this problem, we will use the continuity equation and the mass flow equation which states:

Continuity Equation: A1 * v1 = A2 * v2
Mass Flow Equation: m_dot = rho * A * v

(a) Calculate mass flow rate at section 1:
rho1 = 4 ft^3 / lb
m_dot1 = rho1 * A1 * v1

m_dot1 = (4 ft^3 / lb) * (10 ft^2) * (100 fpm)
m_dot1 = 4000 ft^3/min*lb = 4000 lb/min

The mass flow rate is constant along the flow line, so m_dot1 = m_dot2. Therefore, the mass flow rate at section 2 is also 4000 lb/min.

(b) Calculate speed at section 2 using the continuity equation (A1 * v1 = A2 * v2):

A1 = 10 ft^2
v1 = 100 fpm
A2 = 2 ft^2

v2 = (A1 * v1) / A2
v2 = (10 ft^2 * 100 fpm) / 2 ft^2
v2 = 500 fpm

The speed at section 2 is 500 fpm.

To calculate the mass flow rate and speed at section 2, we can use the equation:

π‘šΜ‡ = 𝜌 Γ— 𝐴 Γ— 𝑣

where π‘šΜ‡ is the mass flow rate, 𝜌 is the density, 𝐴 is the cross-sectional area, and 𝑣 is the velocity.

Given:
At section 1:
𝐴1 = 10 ft²
𝑣1 = 100 fpm
𝜌1 = 4 ft³/lb
At section 2:
𝐴2 = 2 ft²
𝜌2 = 0.20 lb/ft³

(a) To calculate the mass flow rate:
Plug in the values for section 1:
π‘šΜ‡1 = 𝜌1 Γ— 𝐴1 Γ— 𝑣1
π‘šΜ‡1 = 4 ftΒ³/lb Γ— 10 ftΒ² Γ— 100 fpm
π‘šΜ‡1 = 4000 lb/min

So, the mass flow rate at section 1 is 4000 lb/min.

(b) To calculate the speed at section 2:
Plug in the values for section 1 and section 2:
π‘šΜ‡1 = π‘šΜ‡2 (mass flow rate is constant)
π‘šΜ‡1 = 𝜌2 Γ— 𝐴2 Γ— 𝑣2
4000 lb/min = 0.20 lb/ftΒ³ Γ— 2 ftΒ² Γ— 𝑣2
𝑣2 = 4000 lb/min / (0.20 lb/ftΒ³ Γ— 2 ftΒ²)
𝑣2 = 4000 lb/min / (0.4 lb/ftΒ²)
𝑣2 = 10000 ft/min

So, the speed at section 2 is 10000 ft/min.

To calculate the mass flow rate, we need to use the equation:

π‘šΜ‡ = 𝜌1 Γ— 𝐴1 Γ— 𝑣1

where π‘šΜ‡ is the mass flow rate, 𝜌1 is the density of the fluid at section 1, 𝐴1 is the cross-sectional area of section 1, and 𝑣1 is the velocity of the fluid at section 1.

Given:
𝐴1 = 10𝑓𝑑^2
𝑣1 = 100 fpm
𝜌1 = 4 𝑓𝑑^3/𝑙𝑏

To calculate π‘šΜ‡, substitute the given values into the equation:

π‘šΜ‡ = (4 𝑓𝑑^3/𝑙𝑏) Γ— (10𝑓𝑑^2) Γ— (100 fpm)

Simplifying the units:

1 𝑙𝑏 = 7.48 gallons = 7.48 Γ— (144) 𝑓𝑑^3

π‘šΜ‡ = (4 𝑓𝑑^3/𝑙𝑏) Γ— (10𝑓𝑑^2) Γ— (100 fpm)
= (4/7.48) Γ— (10𝑓𝑑^2) Γ— (100 fpm)
= 4.79 Γ— 10^2 𝑙𝑏/π‘šπ‘–π‘›

Therefore, the mass flow rate is 4.79 Γ— 10^2 𝑙𝑏/π‘šπ‘–π‘›.

To calculate the speed at section 2 (𝑣2), we can use the equation of continuity, which states that the mass flow rate is constant along a flow line. Mathematically, it can be expressed as:

π‘šΜ‡ = 𝜌2 Γ— 𝐴2 Γ— 𝑣2

Given:
π‘šΜ‡ = 4.79 Γ— 10^2 𝑙𝑏/π‘šπ‘–π‘›
𝐴2 = 2 𝑓𝑑^2
𝜌2 = 0.20 lb/𝑓𝑑^3

Substituting the given values into the equation:

4.79 Γ— 10^2 𝑙𝑏/π‘šπ‘–π‘› = (0.20 lb/𝑓𝑑^3) Γ— (2 𝑓𝑑^2) Γ— 𝑣2

Simplifying the units:

1 𝑙𝑏 = 16 oz

4.79 Γ— 10^2 𝑙𝑏/π‘šπ‘–π‘› = (0.20 lb/𝑓𝑑^3) Γ— (2 𝑓𝑑^2) Γ— 𝑣2
= (0.20 Γ— 16) π‘œπ‘§/𝑓𝑑^3 Γ— (2 𝑓𝑑^2) Γ— 𝑣2
= 6.4 π‘œπ‘§/𝑓𝑑^2 Γ— 𝑣2

To find 𝑣2, divide both sides of the equation by 6.4 π‘œπ‘§/𝑓𝑑^2:

𝑣2 = (4.79 Γ— 10^2 𝑙𝑏/π‘šπ‘–π‘›) / (6.4 π‘œπ‘§/𝑓𝑑^2)
= (4.79 Γ— 10^2) / (6.4) ft/min
= 7.48 Γ— 10^1 ft/min

Therefore, the speed at section 2 is 7.48 Γ— 10^1 ft/min.