Energy and Energy Balances: Chapters 7,8,9,F&R.  Every chemical process involves the transfer of energy: – Distillation (phase changes)  energy added for volatilization at reboiler and energy removed at condenser – combustion  power generation – reactors  breaking and forming chemical bonds – fluid transport  pumping

 Energy balances are used to: – determine the amount of energy that flows into or out of each process unit, that must be added or removed – calculate the energy requirement (and $) for the process and, along with capital costs, assess equipment alternatives. – assess ways of reducing energy requirements (e.g. exchanges within the process) in order to improve process profitability

CHEE 221 1 Units of Energy

 Energy has units of force times distance (masslength2/time2) Nm (=Joule) SI units dynecm (=erg) CGS system

ftlbf British engineering system

 Energy originally defined as the amount of heat required to raise the temperature of a specified mass of water by one degree at 1 atm

Unit Symbol Mass of H2O Temperature Interval kilocalorie kcal 1 kg C calorie cal 1 g C

British thermal unit Btu 1 lbm F

 Unit conversions found inside the front cover of F&R

CHEE 221 2 Forms of Energy

The total energy of a system has three components:

1. Kinetic Energy (Ek) – energy due to the translational motion of the system as a whole 1 2 2 Ek  mu  J  [kg][m/s] m  mass (kg) 2 u  velocity (m/s) 1 2 E  m u2  W  J/s  [kg/s][m/s] k 2

2. Potential Energy (Ep) – energy due to the position of the system in a potential field (e.g., earth’s gravitational field (g = 9.8 m/s2))

2 E p  mgz  J [kg][m/s ][m] m  mass (kg) h  height of object (m)  2 E p  mgz  W J/s [kg/s][m/s ][m]

CHEE 221 3 Example

Benzene (SG = 0.879) flows into a process unit through a 2 in Schedule  40 pipe at a rate of 1000 L/h. Calculate E k for this stream in joules/second.

The same benzene is now pumped into a holding vessel that is 20 m above the level of the piping. Determine the rate of increase in potential energy.

CHEE 221 4 Forms of Energy cont’d

3. Internal Energy (U) – all energy possessed by system other than kinetic and potential energy, including the energy arising from the: – rotational and vibrational motion of molecules within the system – interactions between molecules within the system – motion and interactions of electrons and nuclei within molecules

 Internal energy (U) is related to enthalpy (H)  H U  PV – U and H are a function of temperature, chemical composition, physical state (solid, liquid or gas) and only weakly a function of pressure

 U and H are relative quantities (relative to a “reference state”) – absolute values are not specified or known – values must be defined with respect to their reference state – this is OK, since we are always interested in changes in U and H

CHEE 221 5 Intensive vs Extensive Variables; Specific Property

Intensive Variables – independent of the size of the system – e.g., temperature, pressure, density, composition (mass or mole fraction)

Extensive Variables – depend on the size of the system – e.g., mass, number of moles, volume (mass or molar flow rate and 1 2 volumetric flow rate), kinetic energy ( E k  mu ) potential energy and 2 internal energy

Specific Property – a quantity that is obtained by dividing an extensive property by the total amount of the material.

– denoted by ‘^’  specific volume ( V ˆ ) units of m3/kg – enthalpy and internal energy commonly reported as specific ˆ values U (kJ/kg), H ˆ (kJ/kg)  Hˆ  Uˆ  PVˆ

CHEE 221 6 Example

The specific internal energy of steam at 165 ºC and 7 bars pressure is 2571 kJ/kg, and the specific volume at the same temperature and pressure is 273 L/kg. Calculate the specific enthalpy of steam at this temperature and pressure, and the rate at which enthalpy is transported by a stream at 165 ºC and 7 bars with a molar flow rate of 20 kg-mol/h.

CHEE 221 7 Transfer of Energy

In a closed system (no mass transferred across the system boundaries (i.e., batch system)), energy can still be transferred between the system and the surroundings in two ways:

1. Heat (Q) – energy that flows due to a temperature difference between the system and its surroundings – always flows from high to low temperature – defined to be positive if it flows to a system (i.e. input)

2. Work (W) – energy that flows in response to any driving force (e.g., applied force, torque) other than temperature – defined as positive if it flows from the system (i.e. output) – in chemical processes, work may come, for example, from a moving piston or moving turbine A system does not possess heat or work. Heat or work only refer to energy that is being transferred to the system. CHEE 221 8 First Law of Thermodynamics

The First Law of Thermodynamics states that energy can neither be created or destroyed (just like total mass)

Accumulation = In – Out + Generation – Consumption

But generation=0 and consumption=0 since energy cannot be created or destroyed so the general balance becomes:

Accumulation = In – Out

      m in(Ek,in, E p,in,U in) m out( E k,out, E p,out,U out)

 Q W

CHEE 221 9 Energy Balances on Closed Systems

Closed System – no material crosses the system boundary over a period of time (e.g., batch process).

General balance equation is: Accumulation = Input – Output

Although no mass crosses the boundaries, energy input0 and energy output0 since energy can be transferred across the boundary. Therefore, the balance becomes: final system initial system net energy –= energy energy transferred

initial system energy  Ui  Eki  Epi U  E k  E p Q W

final system energy  U f  Ekf  Epf 1st Law of Thermodynamics for a Closed System energy transferred  Q W ( = final – initial)

CHEE 221 10 Notes on Energy Balances for a Closed System

U  E k  E p Q W

Possible Simplifications:

 if Tsystem = Tsurroundings, then Q = 0 since no heat is being transferred due to temperature difference  if the system is perfectly insulated, then Q = 0 (system is adiabatic) since no heat is being transferred between the system and the surroundings

 if system is not accelerating, then Ek = 0  if system is not rising or falling, then Ep= 0  if energy is not transferred across the system boundary by a moving part (e.g., piston, impeller, rotor), then W = 0  if system is at constant temperature (system is isothermal), no phase changes or chemical reactions are taking place, and only minimal pressure changes, then U = 0

CHEE 221 11 Examples of Closed Systems

Example 1: Heating water in a Example 2: Compressing a gas sealed container in a cylinder.

U  E k  E p Q W U  E k  E p Q W

U  Q U W

CHEE 221 12 Problem 7.9 F&R

Write and simplify the closed-system energy balance for each of the following processes, and state whether nonzero heat and work terms are positive or negative. Begin by defining the system.

(a) The contents of a closed flask are heated from 25C to 80C.

(b) A tray filled with water at 20C is put into a freezer. The water turns into ice at -5C. (Note: When a substance expands it does work on its surroundings and when it contracts the surroundings do work on it.)

(c) A chemical reaction takes place in a closed adiabatic (perfectly insulated) rigid container.

(d) Repeat part (c), only suppose the reactor is isothermal rather than adiabatic and that when the reaction was carried out adiabatically the temperature in the reactor increased.

CHEE 221 13 Energy Balances on Open Systems at Steady-State

Open System – material crosses the system boundary as the process occurs (e.g., continuous process at steady-state).

In an open system, work must be done to push input fluid streams at a  pressure Pin and flow rate V in into the system (“PV” work), and work is  done by the output fluid streams at pressure Pout Vout and flow rate on the surroundings as it leaves the system.

V (m3/s)  3   in Vout (m /s)   Win  PinVin 2 Process Unit 2 Wout  PoutVout Pin(N/m ) Pout (N/m )

     Net rate of flow work done by the system:W fl  Wout Win  PoutVout  PinVin

W  P V  P V For several input and output streams, fl  j j  j j output input streams streams CHEE 221 14 Flow Work and Shaft Work

The total rate of work (W  ) done by a system on its surroundings is divided into to parts, where:    W  Ws W fl where,

 Ws = shaft work – rate of work done by the process fluid on a moving part within the system (e.g., piston, turbine, rotor)

 W fl = flow work – rate of work done by the fluid at the system outlet minus the rate of work done on the fluid at the system inlet

CHEE 221 15 Steady-State Open System Energy Balance

The general balance equation for an open system (i.e., continuous process) at steady-state is: Input = Output

      m in(Ek,in, E p,in,U in) m out E(k,out E, p,out U,out )

 Q W    energy input  Uin  Ek,in  E p,in    energy output  Uout  Ek,out  E p,out    energy transferred  Q  (Ws W fl )          U in  E k,in  E p,in Q  U out  E k,out  E p,out (W s W fl) CHEE 221 16 Steady-State Open System Energy Balance

 If E j is the total rate of energy transport for j input and output energy streams,      E  E  Q (W W ) E j  Q  E j  (Ws W fl )  j  j s fl input output output input streams streams streams streams

E (E E U )     j   k, j  p, j  j E j  (Ek, j  E p, j U j ) input input output output streams streams streams streams

 m (Eˆ  Eˆ Uˆ ) m (Eˆ Eˆ Uˆ )  j k, j p, j j   j k, j  p, j  j input output streams streams

 ˆ ˆ ˆ ˆ  ˆ ˆ ˆ ˆ  m j (Ek, j  E p, j  H j  PjV j )  m j (Ek, j  E p, j  H j  PjV j ) input output streams streams energy transferred Q (W W ) Q (W  P Vˆ  P Vˆ ) s fl s  j j  j j output input CHEE 221 streams streams 17 Steady-State Open System Energy Balance

ˆ PjV j terms cancel

  ˆ  ˆ H  m j H j  m j H j output input streams streams      H  Ek  Ep  Q Ws E m u 2 / 2 m u 2 / 2  k   j j   j j st output input 1 Law of Thermodynamics for an streams streams Open System at Steady-State    ( = output – input) E p  m j gz j  m j gz j output input streams streams CHEE 221 18 Notes on Energy Balances for an Open System

     H  Ek  Ep  Q Ws

Possible Simplifications:

 if Tsystem = Tsurroundings, then Q = 0 since no heat is being transferred due to temperature difference  if the system is perfectly insulated, then Q = 0 (system is adiabatic) since no heat is being transferred between the system and the surroundings  if energy is not transferred across the boundary by a moving part  (e.g., piston, impeller, rotor), then Ws  0   if inflow and outflow streams are of the same velocity, then Ek  0  if there is no large vertical distance between the inlets and outlets of a system, then  E p  0  if system is at constant temperature (system is isothermal), no phase changes or chemical reactions are taking place, and only minimal pressure changes, then H  0

CHEE 221 19 Example F&R 7.4-2

Five hundred kilograms per hour of steam drives a turbine. The steam enters the turbine at 44 atm and 450C at a linear velocity of 60 m/s and leaves at a point 5 m below the turbine inlet at atmospheric pressure and a velocity of 360 m/s. The turbine delivers shaft work at a rate of 70 kW, and the heat loss from the turbine is estimated to be 104 kcal/h. Calculate the specific enthalpy change associated with the process.

CHEE 221 20 Summary on Energy Balances

The First Law of Thermodynamics for a closed (i.e. batch) system is:

U  E k  E p Q W and the First Law of Thermodynamics for an open system at steady- state (i.e., continuous) system is:      H  Ek  Ep  Q Ws

 Changes in kinetic and potential energy can be calculated, but are usually small for chemical systems.  Heat and work inputs are given in the problem, or are what you must solve for.  The major task is calculating changes in U or H: – Ch 7: Using tabulated values (steam tables) – Ch 8: Phase changes with no reaction – Ch 9: Energy balances with reaction

CHEE 221 21