Question

A fluid moves in a steady flow manner between two sections in a flow line. At a section 1: A= 20ft3. "1 = 200 fpm, V1 = 2 ft^3/Ib. At section 2: A2 = 4 ft^2 P2= 0.40 lb/ft^3. Calculate a). the mass flow rate and b). the speed at section 2 Note: Steady flow m1= m2

Answer 1

Alright, let's dive into solving this problem!

a) The mass flow rate is calculated using the formula: mass_flow_rate = A1 * v1 / V1

Where A1 is the area of section 1, v1 is the velocity at section 1, and V1 is specific volume at section 1.

By substituting A1 with 20ft^3, v1 with 200 fpm and V1 with 2 ft^3/lb,
mass_flow_rate = 20ft^3 * 200 fpm / 2 ft^3/lb = 2000 lb/min

So, the mass flow rate is 2000 lb/min.

b) The speed at section 2 (v2) is calculated using the formula: v2 = mass_flow_rate * V1 / A2

Where mass_flow_rate is the rate calculated from part a, V1 is the specific volume at section 1, and A2 is the area of section 2.

By substituting mass_flow_rate with 2000 lb/min, V1 with 2 ft^3/lb and A2 with 4 ft^3,
v2 = 2000 lb/min * 2 ft^3/lb / 4 ft^3 = 1000 fpm

So, the speed at section 2 is 1000 fpm.

I hope these explanations are clear. If you have any questions, don't hesitate to ask!