Energy and Mass Balances for Open Systems in Chemical Engineering Thermodynamics

Energy Balance for Open Systems

Introduction to Open Systems

Open systems, also known as control volumes, are regions in space where mass and energy can cross the boundaries. In chemical engineering thermodynamics, analyzing open systems is essential for understanding processes involving flowing fluids, such as reactors, heat exchangers, and pipelines.

• Open system: A system where mass and energy can enter or leave.

• Control volume: The region of space selected for analysis.

Measures of Flow

Flowing streams in open systems are characterized by several measurable quantities. These include mass flowrate, volumetric flowrate, molar flowrate, and velocity.

• Mass flowrate ($\\dot{m}$): The amount of mass passing through a cross-section per unit time. $\\dot{m} = M \\dot{n}$ $\\dot{m} = \\rho v A$

• Volumetric flowrate ($q$): The volume of fluid passing through a cross-section per unit time. $q = vA$ $q = \\frac{\\dot{m}}{\\rho}$

• Molar flowrate ($\\dot{n}$): The number of moles passing through a cross-section per unit time. $\\dot{n} = \\frac{\\dot{m}}{M}$

• Velocity ($v$): The speed at which the fluid moves through the system.

Definitions:

• M: Molar mass

• A: Cross-sectional area

• \\rho: Density (specific or molar)

Example: Blood Flow in an Artery

This example demonstrates the calculation of velocity and mass flowrate in a bifurcating artery.

• Given: Diameter upstream = 5 mm, flowrate = 5 cm3/s, density = 1.06 g/cm3. Downstream: two vessels, each 3 mm diameter.

• Velocity upstream: $v_{up} = \\frac{q}{A} = \\frac{5 \\text{ cm}^3 \\text{ s}^{-1}}{(\\pi/4)(0.5 \\text{ cm})^2} = 25.5 \\text{ cm} \\text{ s}^{-1}$

• Velocity downstream: $v_{down} = \\frac{2.5 \\text{ cm}^3 \\text{ s}^{-1}}{(\\pi/4)(0.3 \\text{ cm})^2} = 35.4 \\text{ cm} \\text{ s}^{-1}$

• Mass flowrate upstream: $m_{up} = q \\rho = (5 \\text{ cm}^3 \\text{ s}^{-1})(1.06 \\text{ g cm}^{-3}) = 5.30 \\text{ g s}^{-1}$

• Mass flowrate downstream: $m_{down} = q \\rho = (2.5 \\text{ cm}^3 \\text{ s}^{-1})(1.06 \\text{ g cm}^{-3}) = 2.65 \\text{ g s}^{-1}$

Example: Flow of Liquid n-Hexane

This example illustrates the calculation of volumetric flowrate, molar flowrate, and velocity for liquid n-hexane in pipes of different diameters.

• Given: $\\dot{m} = 0.75 \\text{ kg/s}$, $D = 5 \\text{ cm}$, $\\rho = 659 \\text{ kg/m}^3$, $M = 86.177 \\text{ g/mol}$

• For $D = 5$ cm:

  • Volumetric flowrate: $q = \\frac{\\dot{m}}{\\rho} = \\frac{0.75 \\text{ kg/s}}{659 \\text{ kg/m}^3} = 0.00114 \\text{ m}^3/ ext{s}$

  • Molar flowrate: $\\dot{n} = \\frac{\\dot{m}}{M} = \\frac{0.75 \\text{ kg/s} \\times 10^3 \\text{ g/kg}}{86.177 \\text{ g/mol}} = 8.703 \\text{ mol/s}$

  • Area: $A = \\frac{\\pi}{4} D^2 = \\frac{\\pi}{4} (5 \\times 10^{-2} \\text{ m})^2 = 0.00196 \\text{ m}^2$

  • Velocity: $v = \\frac{q}{A} = \\frac{0.00114 \\text{ m}^3/ ext{s}}{0.00196 \\text{ m}^2} = 0.582 \\text{ m/s}$

• For $D = 2$ cm (same $q$ and $n$):

  • Area: $A = \\frac{\\pi}{4} D^2 = \\frac{\\pi}{4} (2 \\times 10^{-2} \\text{ m})^2 = 0.000314 \\text{ m}^2$

  • Velocity: $v = \\frac{q}{A} = \\frac{0.00114 \\text{ m}^3/ ext{s}}{0.000314 \\text{ m}^2} = 3.63 \\text{ m/s}$

Mass Balance for Open Systems

Control Volume Concept

A control volume is a defined region in space for which mass and energy balances are performed. The boundaries of the control volume are called the control surface.

• Control volume: Region of space identified for analysis.

• Control surface: The boundary through which mass and energy can flow.

General Mass Balance Equation

The mass balance for a control volume accounts for the rate of change of mass within the volume and the net flow of mass across its boundaries.

• General mass balance: $\\frac{dm_{cv}}{dt} + \\Delta(\\dot{m})_{fs} = 0$ $\\Delta(\\dot{m})_{fs} = \\dot{m}_3 - \\dot{m}_1 - \\dot{m}_2$

• Continuity equation: $\\frac{dm_{cv}}{dt} + \\Delta(\\rho v A)_{fs} = 0$

Steady-State Flow Processes

In steady-state processes, the conditions within the control volume do not change with time. The mass of fluid in the control volume remains constant, so the accumulation term is zero.

• Steady-state mass balance: $\\frac{dm_{cv}}{dt} = 0$ $\\Delta(\\rho v A)_{fs} = 0$

• Key point: Steady-state does not necessarily mean constant flowrates, but the inflow of mass equals the outflow of mass.

Energy Balance for Open Systems

Forms of Energy in Flowing Streams

Each unit mass of a flowing stream carries energy in internal, potential, and kinetic forms. The total energy per unit mass is given by:

• Total energy per unit mass: $U + \\frac{v^2}{2} + zg$

• Energy transport rate: $\\dot{m}(U + \\frac{v^2}{2} + zg)$

Energy Conservation in Control Volumes

The rate of change of energy within the control volume equals the net rate of energy transfer into the control volume, including heat and work.

• General energy balance: $\\frac{dU_{cv}}{dt} = Q + W + \\text{(energy in by flow)} - \\text{(energy out by flow)}$

• Work: Includes shaft work, work associated with moving streams, and expansion/contraction of the control volume.

• Enthalpy: $H = U + PV$

Energy Balance for Closed Systems

For closed systems (no mass flow across boundaries), the energy balance simplifies:

• Energy balance equation: $\\frac{dU_{cv}}{dt} = Q + W$ $\\Delta U_{cv} = Q + W$

Summary Table: Key Flow Quantities

Quantity	Symbol	Formula	Units
Mass flowrate	$\\dot{m}$	$\\dot{m} = \\rho v A$	kg/s
Volumetric flowrate	$q$	$q = vA$	m3/s
Molar flowrate	$\\dot{n}$	$\\dot{n} = \\frac{\\dot{m}}{M}$	mol/s
Velocity	$v$	$v = \\frac{q}{A}$	m/s

Additional info:

• These principles are foundational in chemical engineering thermodynamics and are applicable to a wide range of processes, including those involving organic chemicals, but the content is not specific to organic chemistry reactions or mechanisms.

• Reference: Smith, J.M., Van Ness, H.C., Abbott, M.M. (2008). Introduction to Chemical Engineering Thermodynamics, 7th ed. NY: McGraw-Hill.