Chapter 5

Mass and Energy Analysis of Control Volumes

MAE 2131 - Lecture 10

First Law of Thermodynamics for a Closed System

• Energy Change of a System, ( \Delta E_{system} )

  • Fixed Boundaries:

  • Moving Boundaries:

  • ( E2 - E1 = Q - W )

  • ( \Delta E = \Delta KE + \Delta PE + \Delta U = Q - W )

  • ( \Delta E = \Delta KE + \Delta PE + \Delta H = Q - W )

Objectives

• Develop the conservation of mass principle.

• Apply the conservation of mass principle to various systems including steady- and unsteady-flow control volumes.

• Apply the first law of thermodynamics as the statement of the conservation of energy principle to control volumes.

• Identify the energy carried by a fluid stream crossing a control surface as the sum of internal energy, flow work, kinetic energy, and potential energy of the fluid.

  • Relate the combination of the internal energy and the flow work to the property enthalpy.

• Solve energy balance problems for common steady-flow devices such as:

  • Nozzles

  • Compressors

  • Turbines

  • Throttling valves

  • Mixers

  • Heaters

  • Heat exchangers.

• Apply the energy balance to general unsteady-flow processes with particular emphasis on the uniform-flow process as the model for commonly encountered charging and discharging processes.

Closed System

• Characteristics:

  • Mass: No

  • Energy: Yes

• Examples:

  • Tanks

  • Piston cylinder systems

Open System (Control Volume)

• Characteristics:

  • Mass: Yes

  • Energy: Yes

• Examples:

  • Nozzle/Diffuser

  • Pump

  • Turbine

Conservation of Mass

• Mass, like energy, is a conserved property, and it cannot be created or destroyed during a process.

  • Closed Systems: The mass of the system remains constant during a process.

  • Control Volumes: Mass can cross the boundaries, requiring tracking of the amount of mass entering and leaving the control volume.

• Mass ( m ) and energy ( E ) can be converted to each other according to:

  • ( E = mc^2 )

  • Where ( c ) is the speed of light in a vacuum, approximately ( 2.9979 \times 10^8 \, m/s ).

• Mass is conserved even during chemical reactions.

Mass Flow Rate (One-Dimensional Flow)

• Flow Properties:

  • Flow is normal to the boundary at locations where mass enters or exits the control volume.

  • All intensive properties (Temperature ( T ), Pressure ( P ), Density ( \rho ), etc.) are uniform with position over each inlet or exit area ( A ) through which matter flows.

• Mass flow rate formula:

  • ( \dot{m} = \rho A V )

• Volume flow rate:

  • ( Q = AV = \dot{V} )

  • Where ( \dot{m} ) is mass flow rate, ( \rho ) is density, ( A ) is cross-sectional area, and ( V ) is velocity.

Volumetric Flow Rate

• Units: Cubic meters per second (m³/s)

Example 1: Flow Rate in a Pipe

• Given:

  • Cross-section of pipe: ( 1.5 \, m^2 )

  • Flow velocity: ( 2 \, m/s )

  • Temperature: ( 40 \degree C )

  • Pressure: ( 100 \, kPa )

• Calculations:

  • Volumetric Flow Rate:
    [ V = A V = (1.5 \, m^2)(2 \, m/s) = 3 \, m^3/s ]

  • Mass Flow Rate:
    [ \dot{m} = PAV = (998 \, kg/m^3)(1.5)(2) = 2994 \, kg/s ]

Mass Rate Balance

• Time Rate of Change of Mass:

  • Mass contained within the control volume at time ( t )

  • Rate of flow of mass into and out of the control volume:
    [ \dot{m}{cv} = \frac{d m{cv}}{dt} = \dot{m}i - \dot{m}e ]

Conservation of Mass: Steady-State Summary

• For steady-state control volume:

  • ( \frac{d m{CV}}{dt} + \sum \dot{m}{out} - \sum \dot{m}_{in} = 0 )

  • ( \sum \dot{m}{out} = \sum \dot{m}{in} )

  • ( \dot{m} = \rho A V = A V v )

Example 2: Filling a Garden Hose

• Scenario: A garden hose fills a 10-gal bucket in 50 seconds with a nozzle diameter changing from 2 cm to 0.8 cm.

• Calculations:

  • Determine Volume and Mass flow rates, and average velocity at the nozzle exit and hose.

  • Bucket Volume: ( 10 \text{ gal} = 3.785 \, m^3 )

  • Average velocities computed using area change and steady-flow equations.

Mass Rate Balance (Steady-State Form)

• Steady-state: All properties are unchanging over time, leading to:

  • ( \frac{d m_{cv}}{dt} = 0 )

• For control volume, mass rate in equals mass rate out:

  • ( \sum{i} \dot{m}{i} = \sum{e} \dot{m}{e} )

Example 3: Closed Tank Flow

• Problem: Water flows through a closed tank with diameters ( D1 = 6 \, cm, D2 = 5 \, cm, D_3 = 4 \, cm ).

• Task: Compute flow velocities at each inflow and outflow.

  • Given density: ( \rho_{water} = 998 \, kg/m^3 )

Example 4: Velocity Through a Hose and Nozzle

• Scenario: Water flows at ( 0.5 \, m/s ) through a 2 cm diameter hose; find velocity through a 0.4 cm diameter nozzle.

• Answer Choices: A) 2.5 m/s B) 12.5 m/s C) 0.1 m/s D) 0.02 m/s

• Calculation: Using continuity,

  • ( V{out} = (0.5 \, m/s) \cdot \frac{Dh^2}{D_n^2} = 12.5 \, m/s ) (Correct: B)

  • Confirm using area ratios and flow continuity.

Flow Work and the Energy of a Flowing Fluid

• Flow Work: The energy required for pumping mass in and out of control volumes, expressed as:

  • ( w_{flow} = Pv )

  • Where ( P ) is pressure and ( v ) is specific volume.

• Total Energy of a Flowing Fluid:

  • Total energy consists of:

  • ( e = u + \frac{V^2}{2} + gz + Pv ) (where ( u ): internal energy, ( V ): flow velocity, ( gz ): potential energy).

Energy Analysis of Steady-Flow Systems

• Steady-flow Process: A process in which fluid flows through a control volume steadily.

  • Engineering applications such as power plants operate under steady conditions.

• Under steady-flow conditions, the mass and energy contents of a control volume remain constant.

Upcoming Classes

• February 23: Mass & Energy Analysis of Control Volumes

• February 25: Mass & Energy Analysis of Control Volumes

• March 2: Review

• March 4: ASME Code of Engineering Ethics

• March 9: Spring Break (No Class)

• March 11: Spring Break (No Class)

• March 16: Midterm Exam

Summary

• Conservation of Mass:

  • Steady-State Form: [ \frac{d m{CV}}{dt} + \sum \dot{m}{out} - \sum \dot{m}_{in} = 0 ]

  • [ \sum \dot{m}{out} = \sum \dot{m}{in} ]