91Ó°ÊÓ

Mass flow rate is essential in thermodynamics. It describes the mass of a substance passing through a unit area per unit time. For liquid water, given the density (\rho) is approximately 1000 kg/m³, we utilize the relationship: \ \text{mass flow rate} = \rho \times Q, \( Q \) being the volumetric flow rate. If we know how much volume is flowing through an area in a certain time and we multiply by the density, we get the mass flow rate. Here we used the following data:

• For the 80 mm pipe, with a cross-sectional area of 0.0050 m², the water velocity is 0.5 m/s, making the volumetric flow rate 0.0025 m³/s.
• For the 120 mm pipe, with a cross-sectional area of 0.0113 m² and water velocity of 0.1 m/s, the volumetric flow rate becomes 0.00113 m³/s.

By multiplying each volumetric flow rate by the density of water:

• Mass flow rate in the 80 mm pipe: 1000 kg/m³ * 0.0025 m³/s = 2.5 kg/s.
• Mass flow rate in the 120 mm pipe: 1000 kg/m³ * 0.00113 m³/s = 1.13 kg/s.

The same method is applied to find the mass flow rate at the inlet.

Volumetric flow rate is the volume of fluid flowing through a cross-sectional area per unit time. For any pipe, it can be calculated using: \ \text{volumetric flow rate} = Area \times Velocity, \ where Area (\text{A}) is the cross-sectional area of the pipe, and Velocity (\text{V}) is the fluid velocity in the pipe.
In the given exercise,

• For the second pipe (80 mm diameter): Thus, using the formula: \( Q_2 = A_2 \times V_2 \) results in: 0.0050 m² * 0.5 m/s = 0.0025 m³/s.
• For the third pipe (120 mm diameter): Again, applying the formula: \( Q_3 = A_3 \times V_3 \) results in: 0.0113 m² * 0.1 m/s = 0.00113 m³/s.

The total volumetric flow rate through the inlet can be determined using the principle of conservation of mass. Therefore, the sum of the volumetric flow rates in the exit pipes should equal the inlet. Hence, \( Q_1 = Q_2 + Q_3 = 0.0025 + 0.00113 = 0.00363 m³/s \).

The continuity equation is a principle derived from the conservation of mass in fluid dynamics. It states that the mass flow rate into a system must equal the mass flow rate out of the system when it is at a steady state. This means that in an unchanged control volume under steady-state conditions, the amount of mass entering must equal the amount exiting. \( \text{Mass in} = \text{Mass out} \).
In the context of our problem, the continuity equation can be expressed as: \ \( \rho_1 \times Q_1 = \rho_2 \times Q_2 + \rho_3 \times Q_3 \). \ For water, assuming relatively incompressible fluid, the density remains constant, simplifying the equation to: \ \( Q_1 = Q_2 + Q_3 \). Therefore, we used: \ \( Q_1 = 0.0025 m³/s (from pipe 1) + 0.00113 m³/s (from pipe 2) = 0.00363 m³/s \). \ This allows us to cross-verify our volumetric and mass flow rate calculations.