Chapter 7. Energy and Energy Balance
Introduction
Energy is expensive….
Effective use of energy is important task for chemical engineers.
Topics of this chapter
Energy balance
Energy and energy transfer
Forms of energy : Kinetic / Potential / Internal Energy
Energy transfer : Heat and Work
table) Using tables of thermodynamic data (steam ( steam table)
Mechanical energy balances
Energy Consumption
Typical problems
Power requirement for a pump
Pure mechanical energy balance
Heat / Work calculation for a desired change
Removal of heat from reactor
Combustion problem
Requirement of energies for each apparatus
Terminology Associated with Energy Balance
System : The quantity of matter or region chosen for study enclosed by boundary
Surroundings : Everything outside the system
Boundary : The surface that separates the system from the surroundings. It may be a real or imaginary surface, either rigid or movable. Boundary
System
Surroundings
Terminology Associated with Energy Balance
Adiabatic System A system does not exchange heat with surroundings during a process. Isothermal System A system in which the temperature is invariant during a process Isobaric System A system in which the pressure is constant during a process Isochoric System A system in which the volume is invariant during a process
Q
ΔT
ΔP
ΔV
Terminology Associated with Energy Balance
State Variable (State Function)
Any variable (function) whose value depends only on the state of the system and not upon its previous history.
Path Variable (Path Function)
Any variable (function) whose value depends on how the process take place, and differ for different histories.
7.1 Forms of Energy – The First Law of Thermodynamics
Forms of energy
mv 2
Kinetic energy : due to the motion of the system
EK =
Potential energy : due to the position of the system
Ep = m
Internal energy : due to the motion of internal molecules
Expressed as Temperature
U : from thermodynamic calculation
Forms of energy transfer
Heat (Q) : energy flow due to temperature difference
Work (W) : energy flow due to the driving force other than temperature difference (force, torque, voltage, …)
2g c g gc
h
Notation : rate of energy / energy transfer (energy/time) ^ : specific properties (energy / mass)
Example ) Kinetic Energy EK =
& = E K ˆ = E K
mv 2 2g c
& v2 m 2g c
Unit : Joule
Unit : Joule/s
v2 2g c
Unit : Joule/kg
Unit Conversion
Force
Pressure
1 atm = 1.01325 bar = 1.01325 ×105 Pa (N/m2) = 101.325 kPa 2 (psi) = 760 mm Hg = 14.696 lb /in f
Energy
1 N = 1 kg · m /s2, 1 dyne = 1 g cm /s 2 1 lbf = 32.174 lbm · ft /s2
1 J = 1 N · m = 107 dyne · cm = 0.23901 cal = 9.486 × 10-4 Btu 1 Btu = 1055 J
Power
1 W = 1 J/s = 1.341 × 10-3 hp
Example 7.2-1
Water flows into a process unit through a 2-cm ID pipe at a rate of 2.00 m3 /h. Calculate Ek for this stream in J/s.
& = E K
&v m
2
선 속도와 질량 유속을 알아야 함.
2 gc v=
2m h
& = E K
1
(100 cm)
π (1 cm)
2
2 m 3 1000 kg
&= m mv
3
h 2
2gc
=
1m
3
(1 m) 1h 3600 s
2
2
2
s
3600 s
= 1.77 m/s
= 0.556 kg/s
0.556 kg/s (1.77 m) 2 2
1h
1 N2 1 kg ⋅ m/s
2
= 0.870 N ⋅ m/s = 0.870 J/s
Example 7.2-2
Crude oil is pumped at a rate of 15.0 kg/s from a point 220 meters below the earth’s surface to a point 20 meters above ground level. Calculate the attendant rate of increase of potential energy. & = m& E p
g
h
gc & = m& g Δh Δ E p gc
Δh = 220 + 20 = 240 m
& = m& Δ E p
g gc
Δh =
15kg 9.81 N 240 m s
kg
= 35300 N ⋅ m/s = 35300 J/s
7.3 Energy balance on closed systems
Balance equation (Final System Energy) – (Initial System Energy) = (Net Energy Transfer)
Initial System Energy Final System Energy Net Energy Transfer
U i + E pi + E ki U f + E pf + E kf Q+W
The first law of thermodynamics for closed systems
ΔU + ΔE p + ΔE k = Q + W Text definition: work is done by the surroundings on the system
Important points
U depends on composition, state, temperature of the system. Nearly independent of pressure for ideal gases, liquids, solids.
If there are no temperature differences, Q = 0 Adiabatic process
If there are no moving parts … W = 0
Potential energy change
due to the changes in height
Example 7.3-1
A gas is contained in a cylinder fitted with a movable piston.
The initial gas temperature is 25 oC. The cylinder is placed in boiling water with the piston held in a fixed position. Heat in the amount of 2.00 kcal is transferred to the gas, which equilibrates at 100 oC (and a higher pressure). The piston is then released, and the gas does 100 J of work in moving the piston to its new equilibrium position. The final gas temperature is 100 oC. Write the energy balance equation for each of the two stages of this process, and in each case solve for the unknown energy term in the equation. In solving this problem, consider the gas in the cylinder to be the system, neglect the change in potential energy of the gas as the piston moves vertically, and assume the gas behaves ideally. Express all energies in joules.
Solution ΔU + ΔE p + ΔE k = Q + W
Q
ΔU = Q
W
Q = −W
ΔU = Q + W
ΔU = 2 kcal = 8368 J
Q = − W = −( −100) J = 100J
7.4 Energy Balances on Open Systems at Steady State
Flow work and shaft work
Flow work : work done on system by the fluid itself at the inlet and the outlet
Shaft work : work done on the system by a moving part within the system
& = W & + W & W s fl & (m / s ) V in 3
2
Pin ( N / m )
& (m / s ) V out 3
Process Unit
& =P V & & W fl in in − Pout Vout
2
Pout ( N / m )
Ws – Shaft work?
Shaft work : work done on the system by a moving part within the system
Components such as turbines, pumps, and compressors – all operate by energy transfer to or from the working-fluid
Energy transfer usually through blades rotating on a shaft
Also fluid dynamics problem…
Specific Properties
Specific properties
(Property) / (Amount (Mass, Mole number,…))
Volume , energy, …
Specific volume, specific energy, …
Example)
Extensive properties
Volume : extensive property
Intensive property
depends on system size
V (cm3 ), U (kJ ),...
Extensive properties
Specific Volume : intensive property
ˆ (cm3 / mol ), U ˆ ( kJ / mol ),... V
independent of system size Intensive properties
Enthalpy
It is convenient to define the following property for the calculation of energy balance for flowing systems.
Enthalpy
H ≡ U + PV
Specific Enthalpy
ˆ ≡ U ˆ + PV ˆ H
Example 7.4-1 Enthalpy Calculation
The specific internal energy of helium at 300 K and 1 atm is 3800 J/mol, and the specific molar volume at the same temperature and pressure is 24.63 l/mol.
Calculate the specific enthalpy of helium at this temperature and pressure
and the rate at which enthalpy is transported by a stream of helium at 300 K and 1 atm with a molar flow rate of 250 kmol/h.
Important Point
ˆ ≡ U ˆ + PV ˆ H
Energy Energy
Unit conversion methods Use of Gas constant R Pressure = Force / Area
Pressure × Volume
Solution
Method 1 : Use of Gas Constant, R R = 0.08206 l.atm /mol.K = 8.314 J/mol.K
1 atm 24.63 liter 8.314 J ˆ ˆ ˆ H ≡ U + PV = 3800 J/mol + = 6295 J/mol mol 0.08206 liter.atm
Method 2 : Pressure = Force / Area ˆ = 3800 J/mol H
+
1 atm 24.63 liter 1.01325 × 105 N/m 2 (= Pa) mol
1 atm
1 m3 1000 liter
= 6295 J/mol
& = n& H ˆ = 250kmol × 6295J / mol = 1.57 ×109 J / h H
The Steady-State Open System Energy Balance ΔU + ΔE p + ΔE k = Q + W
W = W s + W f ˆ −m P V ˆ Wf = m in Pin V in out out out
Δ E K = Δ
mv
2gc
Δ E K = Δm
ˆ ≡ U ˆ + PV ˆ H
2
g gc
h
ˆ ΔU = ΔmU
ΔH + ΔE p + ΔE k = Q + Ws
Example 7.4-2
500 kg/h of stream drives a turbine.
The stream enters the turbine at 44 atm and 450 oC 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 10 4 kcal/h
Calculate the specific enthalpy changes associated with the process
Solution Q = -104 kcal/h W = -70 kW
ΔH + ΔE p + ΔE k = Q + Ws
500 kg/h
ΔH = Q + Ws − ΔE p − ΔE k
44 atm, 450oC 60 m/s 5m 500 kg/h
& = m
1 atm 360 m/s
ΔE k =
& m 2
2 2
2 1
(u − u ) =
& g ( z 2 − z1 ) = ΔE p = m
Q=
− 10 4 kcal h
500kg / h 3600s / h
0.139 kg / s
1N
( 260 2 − 60 2 ) m 2
2
1kg ⋅ m / s 2
s2
1N ⋅ m / s 10 3 W
1kW
= −6.81× 10 −3 kW
0.139kg / s 9.81N ( −5) m
1J
2 1h
kg 1kW
0.239 × 10 −3 kcal 3600s 10 3 J / s
3
10 N ⋅ m / s
= −11.6kW
1W
1kW
= 0.139kg / s
= 8.75kW
Ws = −70kW
& −W & =Q & − ΔE& − ΔE& = −90.3kW ΔH s p k ˆ −H ˆ ) & =m & (H ΔH 2 1
− 90.3kJ / s ˆ −H ˆ = ΔH & / m & H = = −650 kJ / kg 2 1 0 139 kg /
7.5 Tables of Thermodynamic Data
U, H, S, V,… Thermodynamic function
Tables of Thermodynamic Data
Tabulation of values of thermodynamic functions ( U, H, V ,..) at various condition (T and P )
It is impossible to know the absolute values of U , H for process materials Only changes are important ( ΔU , ΔH ,…)
Reference state
Choose a T and P as a reference state and measure changes of U and H from this reference state
tabulation
Steam Tables
Compilation of physical properties of water (H, U, V)
Reference state: liquid water at triple point (0.01 oC, 0.00611 bar) Table B.7 Properties of Superheated Steam
Example 7.5-3
Steam at 10 bar absolute with 190 oC of superheat is fed to a turbine at a rate m = 2000 kg/h. The turbine operation is adiabatic, and the effluent is saturated steam at 1 bar.
Calculate the work output of the turbine in kW, neglecting kinetic and potential energy changes.
ΔH + ΔE p + ΔE k = Q + Ws ΔH = Ws 초기와 최종조건의 엔탈피 변화 = 한 일의양
Table B.7 Properties of Superheated Steam
370
10 bar, saturated T = 179.9
T = 179.9 + 190 = 369.9
Solution
Interpolation (and Extrapolation) of steam table
M = M1 + (
X − X1 X 2 − X1
H = 3159 + (
)( M 2 − M1 )
370 − 350 400 − 350
)(3264 − 3159) = 3201
ˆ = ( 2000kg / h ) × (2675 − 3201) kJ / kg × (1h / 3600s) Ws = ΔH = mΔH
= −292 kJ/s = − 292 kW
7.6 Energy Balance Procedures
Solve material balance
Determine the specific enthalpies of each stream
Get all the flow rate of streams
components
Using tabulated data
Calculation (using heat capacity, C p(T), – Ch.8)
Construct energy balance equation and solve it. ΔH + ΔE p + ΔE k = Q + Ws
Example 7.6-1
Two stream of water are mixed to form the feed to a boiler. Process data are as follows:
Feed stream 1 : 120 kg/min @ 30 oC
Feed stream 2 : 175 kg/min @ 65 oC
Boiler pressure : 17 bar (absolute)
The exiting steam emerges from the boiler through a 6-cm ID pipe. Calculate the required heat input to the boiler in kJ/min if the emerging steam is saturated at the boiler pressure.
Neglect the kinetic energies of the liquid inlet streams.
Solution 120 kg H2O/min 30 oC H = 125.7 kJ/kg
295 kg H2O/min
175 kg H2O/min
17 bar, saturated steam (204 oC) H = 2793 kJ/kg (V = 0.1166m3) 6cm ID pipe
65 oC H = 271.9 kJ/kg Q kJ/min
ΔH + ΔE p + ΔE k = Q + Ws
ΔH =
∑ m Hˆ − ∑ m Hˆ i
i
outlet
i
= (295 × 2793) − (120 × 125.7 + 175 × 271.9) = 7.61× 10 6 kJ / min
inlet
v = V / A =
ΔE K =
i
Q = Δ H + Δ E k
295kg 1 min 0.1166m 3 min
mΔv 2 2g c
60s
kg
1 2
π × 0.03 m
2
= 202m / s
= 6.02 × 103 kJ / min
Q = ΔH + ΔE K = (7.61×10 5 + 6.02 × 103 ) kJ / min = 7.67 × 105 kJ / min
Example 7.6-3 Material balance + Energy balance
Saturated steam at 1 atm is discharged from a turbine at a rate of 1150 kg/h. Superheated steam at 300 oC and 1 atm is needed as a feed to a heat exchanger.
To produce it, the turbine discharge stream is mixed with superheated steam available from a second source at 400 oC and 1 atm.
The mixing unit operates adiabatically.
Calculate the amount of superheated steam at 300 oC produced and the required volumetric flow rate of the 400 oC steam.
Solution 1150 kg H2O/hr 1 atm, sat (100 o C) H = 2676 kJ/ kg
m2 (kg H2O/ hr) 1 atm, 300 o C H = 3074 kJ/ kg
m1 (kg H2O/ hr) 1 atm, 400 o C H = 3278 kJ/ kg
Material Balance 1150 + m1 = m2 Two equation Two unknown
Energy Balance
1150×2676 + m1×3278 = m2×3074 m1=2240 kg/h m2=339z0 kg/h
The specific volume of steam at 400 oC and 1 atm is 3.11 m3/kg (Table B.7)
& = V
2240 kg 3.11m 3 h
kg
= 6980m 3 / h
7.7 Mechanical Energy Balances ΔH + ΔE p + ΔE k = Q + Ws
Chemical equipment (Reactor, Distillation column, Evaporator, Heat exchanger,…) Heat flow, internal energy changes (enthalpy change) are most important Shaft work, kinetic energy, potential energy changes are negligible Δ H ≈ Q
Mechanical equipment (Pump, Reservoir, Pipes, Wells, Tanks, Waste Discharge,…) Heat flow, internal energy changes are negligible Shaft work, kinetic energy, potential energy changes are most important ΔE p + ΔE k = W
Mechanical Energy Balances ΔU + ΔE p + ΔE k = Q + W ˆ = V ˆ = 1 / ρ V in out
Δ P Δv 2 g ˆ − Q / m) = W / m + + Δ z + ( ΔU s 2g c g c ρ ˆ − Q / m ( friction loss ) F = ΔU
Δ P Δv 2 g + + Δz + F = Ws / m ρ 2g c g c F = 0, W s = 0
ΔP ρ
+
Δv 2 2gc
+
g gc
Δ z = 0
Bernoulli Equation Important equation for the calculation of equipments consist of pipes, tanks and pumps
Example 7.7-1 The Bernoulli equation
Water flows through the system shown here at a rate of 20 l/min. Estimate the pressure required at point 1 if friction losses are negligible.
50 m
(1) 0.5 cm ID pipe 20 liter /min H2O P1 = ?
(2) 1 cm ID pipe P2 = 1 atm
Solution (2) 1 cm ID pipe P2 = 1 atm
50 m
(1) 0.5 cm ID pipe 20 liter /min H2O P1 = ?
v1 = v(m / s ) = V / A v2 =
1 m3
20 liter min
2
π (0.25cm) 100liter
1 m3
20 liter min
2
π (0.5cm) 100liter
Δv 2 = ( 4.24 2 − 17.0 2 ) = −271.0 m 2 / s 2
(100 cm) 2 1 min 1m
2
60s
(100 cm) 2 1 min 1m
2
60s
= 17.0 m/s = 4.24 m/s
