Ch. 2: Energy & 1st Law of Thermo
Closed Systems
Ch. 2: Energy & 1st Law of Thermo
Esys
Closed Systems
I. E, Energy (a property)
Mechanical Energy:
Kinetic Energy, Translational
Potential Energy, Gravitational
I. E, Energy
Internal Energy, U:
sum of microscopic forms of energy:
Electron and nuclear forces (atomic level)
Bonds in molecules
Kinetic Energy of molecules
Translation, rotation, vibration
Molecule to molecule forces (eg. H2O)
A change in Temperature (or sometimes pressure) or phase causes a change in U
Sensible heat: change of temperature of substance
Latent heat: change of phase of substance
u, specific internal energy:
I. E, Energy (continued)
Total Energy E:
specific energy e:
Height change z must be relative to the direction of gravity
Scalar: no associated direction
relative to a reference point
A property !!
Energy conversion between KE ad PE (pendulum swings)
I. E, Energy (continued)
Change in Energy:
II. Energy Transfers:
Energy transfer = flow of energy across a boundary of a system during a process
Q, Heat:
W, Work:
II. Energy Transfers, continued
Q and W can flow in and/or out, at several places
Q and W are NOT properties
they are path dependent
Qin Win
Qout
II. Energy Transfer by HEAT: Q
Heat vs. Thermal Energy vs. Temperature
Units (kJ, Btu)
Sign convention
Heat added to the system is positive
Heat removed from the system is negative
Adiabatic process: no heat transfer
Types of heat transfer modes Conduction
Convection
Radiation
II. Energy Transfer by HEAT: Q
2
1
2
1
2 2
Q: amount of energy transfer by heat (J, BTU)
: rate of heat transfer (W=J/s, BTU/h)
: heat flux, heat transfer rate per unit area (W/m , BTU/h.ft )
t
t
t
t
Q
q
Q Qdt
Q qdA
II. Energy Transfer by HEAT: Q
Heat is not a property, it is path dependent
2
2 1
1
The amount of energy transfer by heat for a process
Heat is not a property
Q Q Q Q
II. Energy Transfer by WORK: W
Units (kJ, Btu, ft-lbf)
Work is not possessed by a system-only measured as it crosses the system boundary
Sign convention:
positive if done by a system,
negative if done on a system
Many types of Work Mechanical
Electrical
Expansion/compression
II. Energy Transfer by Work: W
2
1
2
1
Work is not a property
= . (J, BTU)
: power (W=J/s, BTU/h)
=
s
s
t
t
W F ds
W
W Wdt
II. Work is path dependent
Work is not a property
II. Work is path dependent
III. 1st Law of Thermodynamics
Closed System
(integrated over time):
Qin Wout E
Example 2.1:
A 1 kg metal weight that is initially 1 m above the ground is connected by a cable
through a frictionless pulley to a smaller
mass. The 1 kg mass is dropped from rest
and does 5 Joules of work as it lifts the
smaller weight.
Determine the speed at which the 1 kg mass hits the ground.
III. 1st Law of Thermodynamics, cont.
Closed System (Instantaneous):
Qin Wout E
IV. Cycles
Cycle: a series of
processes that begin and
end at the same state
IV. Cycles, cont.
1st Law Energy Balance
Equation for a cycle:
0 cycleE
cyclecycle WQ
or
Example 2.3:
A closed system (stationary) goes through a 3-process cycle beginning with U1 = 100 kJ.
1500 J of heat is added to the system during process 1-2 until
the internal energy increases to U2 = 200 kJ.
Process 2-3 is adiabatic as 10 kJ work is done on the system.
No work is done during process 3-1.
Find U, Q, and W for each process.
Process U Q W
1-2
2-3
3-1
II. Types of Work and Power
Mechanical Work:
Mechanical Power:
.mechW F ds
?dt
dsFWmech
FV
2
2
1VACF fddrag
WeightfFrolling *rollingdrag FFF where
II. Types of Work and Power
Rotational Mechanical Power (Shaft Power):
rotW
sec
min
rev
rad
min
rev
60
1
1
2 RPM
II. Types of Work and Power
Electrical Power:
E* i
elecW
dtWWt
AmpVoltWatt 1*11
II. Types of Work and Power
Fluid Power:
Recall: Relating Power and Work :
VpW fluid *
dtWWt
t
WWaverage
II. Types of Work and Power
Expansion/Compression Work:
II. Expansion / Compression Work:
- Common form of work for a
gas in a piston/cylinder device
V, p
y
cylinder
piston
gas
II. Expansion / Compression Work
To move the piston up (expand):
FdsW
?FwithF
II. Expansion / Compression Work
pAdsW
?AdsandF
pAF
II. Expansion / Compression Work
pdVW
dVAds
II. Expansion / Compression Work
W on p-V diagram
Work = ?
pdVWV
P
Work =
area under curve
V = constant
pV = constant
II. pdV Work: 4 cases
p = constant
pVn = constant (polytropic)
pVn = constant (polytropic)
II. calculating pdV work for 4 cases
V = constant
pV = constant
p = constant
Example 2.4:
0.41 lb of air in a piston/cylinder device goes through a constant pressure (p = 20
lbf/in2) heat addition process as the
volume changes from 5 ft3 to 6.52 ft
3.
Find the work during this process.
Example 2.5:
Nitrogen, which behaves as an ideal gas, is compressed in a piston cylinder device
as temperature is held constant at 27C.
The work required during compression is 7000 J.
The initial pressure and volume are 100 kPa and 0.1 m
3.
Find (a) the final volume and (b) the heat
during this process.
Example 2.6:
0.05 kg of air
expands in a piston cylinder device until
the final volume is 4 times the initial volume.
The initial pressure and volume are 400 kPa and 0.0144 m
3, and
the expansion is polytropic with n = 1.4. Find the work during this process.
IV. Cycles, cont.
Thermal Reservoir: A large mass that can accept or reject heat without
changing temperature
(also called thermal energy
SOURCE or SINK)
at TL
at TH
IV. Cycles, cont.
Thermal Reservoirs:
Thermal cycles operate
between two thermal
reservoirs
IV. Cycles, cont.
1. Power Cycle
Objective:
2a. Refrigeration Cycle
Objective:
2b. Heat Pump Cycle
Objective
IV. Thermal Cycles 3 types
IV. Cycles
1. Power
Cycle
1. Power Cycle
Objective: to produce work
(power), Wnet,out
By using heat added from a high-temperature reservoir
(QH from TH) QH is IN
And rejecting heat to a low-temperature reservoir
(QL to TL) QL is OUT
Power
Cycle
1. Power Cycle
- Power Cycle Energy Balance
Power
Cycle
- Power Cycle Performance
Refrigeration
Objective: to keep LOW
temperature reservoir
cool by removing QL
R
High temp Reservoir
Low temp Reservoir
Driven by Work (Power) put
into the cycle
(compressor)
Refrigeration
R
High temp Reservoir
Low temp Reservoir
- Refrigeration Cycle Performance
- Refrigeration Cycle Energy Balance
Heat Pump
Objective: to keep HIGH
temperature reservoir
warm by adding QH
High temp Reservoir
Low temp Reservoir
R or
HP
HP
Driven by Work (Power) put into
the cycle (compressor)
Heat Pump
High temp Reservoir
Low temp Reservoir
R or
HP
HP
- Heat Pump Cycle Energy Balance
- Heat Pump Cycle Performance
1. Power Cycle
2a. Refrigeration Cycle
2b. Heat Pump Cycle
Example 2.7:
A 600 MW steam power plant is cooled by water from a nearby river.
The thermal efficiency of the plant is 40%. Find the rate of heat rejection from the plant to
the river.
Example 2.8:
A household refrigerator has a coefficient of performance of 1.2.
Heat transfer to the refrigerated space through the insulation and due to opening of the
refrigerator doors is 60 kJ/min.
Find
the electric power consumed by the refrigerator and
the rate of heat transfer to the kitchen air from the refrigerator.
Example 2.9:
A heat pump is used to maintain the air in a home at 69F when
the outside temperature is just below freezing.
Heat loss from the home through doors, windows, the roof, and the walls is 36,000
Btu/hr.
What is the coefficient of performance of the heat
pump if it consumes electrical power at the rate of
1 kW.