Mass and Energy for Open System

This lecture started off with an example, here are some important notes from that example:

• For the Moving Boundary Work and Polytropics equation, you can substitute in the polytropic equation and solve the integral to get an expression that you can plug values in to solve
• If not given a u, you can use the specific heat equation from Moving Boundary Work and Polytropics to solve for U

1st Law in an Open System

When mass enters and leaves an open system, it carries its energy with it
$m_{in}^{\prime }e$ goes in and out carrying:
$e=u+KE+PE$

Conservation of Mass

For a control volume, mass going in = mass going out

More on the Conservation of Mass

1. At steady state (things changing with time don’t)
2. Mass going in is equal to mass going out for one inlet and one outlet
3. Mass flow rate for flow in one dimensions across a cross sectional area (of a pipe maybe) is equal to $m^{\prime }=\rho AV=\frac{AV}{v}$ Where V in this case is boundary and v is specific volume

For the change in energy of an open system $\frac{d{E}_{cv}}{dt}=Q^{\prime }-W^{\prime }+m_{in}^{\prime }{\left(u+\frac{v^2}{2}+gz\right)}_{in}-m_{out}^{\prime }{\left(u+\frac{v^2}{2}+gz\right)}_{out}$
If F = PA then the time rate of energy transfer by “flow work”
$flowwork=PAV$
Where V in this case is velocity

$w_{tot}^{\prime }=w_{cv}^{\prime }+PAV$
Where w_cv is all other energy transfer by work
Using the mass flow rate equation above
$w_{tot}^{\prime }=w_{cv}^{\prime }+m^{\prime }PV$
Substituting into the change of energy of an open system equation
$\frac{d{E}_{cv}}{dt}=Q^{\prime }-W_{cv}^{\prime }+m_{in}^{\prime }{\left(h+\frac{v^2}{2}+gz\right)}_{in}-m_{out}^{\prime }{\left(h+\frac{v^2}{2}+gz\right)}_{out}$

Intrinsic vs Extrinsic

When you switch to the intrinsic form, you divide the above equation by mass

When Q’=w’=delta(u)=dE/dT=0…

$\frac{p}{\rho }+\frac{v^2}{2}+gz=const$

General Strategy

1. Apply mass balance equation to get m’ (might need to use ideal gas law (PV=RT))
2. Apply energy balance equation (might need to use tables to get values to solve, or using a solved value to get solution)

Steady Flow Devices

Common Assumptions

• Adiabatic (Q’=0)
• No work (W’=0)
• No change in potential or kinetic energy
  

Nozzles and Diffusers

Nozzles

• A nozzle increases the velocity of a fluid at the expense of pressure Diffusers
• A diffuser increases the pressure of a fluid by slowing it down

Typically for questions that use these processes

• Adiabatic
• No work
• Change in PE = 0 and change in KE is NOT 0
  

Turbines

• Usually insulated/adiabatic ($Q^{\prime }=0$)
• Change in PE = change in KE = 0
  

Compressors

Use work input to increase the pressure of a gas

• Negligible heat transfer unless there is intentional cooling $Q^{\prime }=0$)
• Change in PE = change in KE = 0
  

Pumps

Use work input to increase the pressure of a liquid

• Negligible heat transfer unless there is intentional cooling $Q^{\prime }=0$)
• Change in PE = change in KE = 0
  

Throttles

Induces a sudden drop in pressure by passing the fluid through an obstruction Typically

• Adiabatic
• No work
• Change in PE = change in KE = 0
• ”Isenthalpic is h1=h2” Used for flow control or to induce a phase change
  

Heat Exchanger

Used to transfer heat between two fluids

Assumptions:

• Adiabatic outside (change in heat is 0)
• No work (change in work is 0)
• Change in PE = change in KE = 0
• $m_A^{\prime }{\left(h_{in}-h_{out}\right)}_A=m_B^{\prime }{\left(h_{out}-h_{in}\right)}_B$