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Ch 4 The First Law

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m(u2 u1) = -3.7 kJ (-7.2 kJ) P m (v2- v1) ... Use values of u2 and v2 in the energy balance to see if the equation is satisfied. ... – PowerPoint PPT presentation

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Title: Ch 4 The First Law


1
Ch 4 The First Law
  • Ted Huddleston
  • Dept of Chemical Engineering
  • Univ of South Alabama
  • Mobile, AL

2
First Law of ThermodynamicsClosed Systems
All energy must be accounted for.
Energy can be neither created nor destroyed.
Energy In - Energy out Energy Accumulation
Total Energy entering the system -
Total Energy leaving the system
Change in the total energy of the
system
3
First Law for Closed Systems
4
First Law for Closed Systems
If KE2 KE1 0 and PE2 PE1 0
Energy Balance.
An Accounting Equation
5
Mass Balance Energy Balance Closed System
Mass is constant. DU Q - W
6
DU Q - W
Method One Formal Sign Convention
For heat Q, heat transfer into system has
POSITIVE sign.
For work W, work by a system on surroundings has
POSIIVE sign.

7
DU Q - W
Method Two Intuitive Sign Convention
DU Energy in (heat or work) Energy out (heat
or work)
Think of a checking account. Additions can be
made through cash or check deposits. Account
can be lowered by by ATM withdrawals or by
writing checks.
8
DU Q - W
Example 20 Joules of heat are added to a gas
while at the same time, the gas expands,
performing 15 Joules of work. Find the change
in internal thermal energy.
Method ONE Formal Sign Convention
Q 20 J
W 15 J
DU Q W (20) (15) 5 J
9
DU Q - W
Example 20 Joules of heat are added to a gas
while at the same time, the gas expands,
performing 15 Joules of work. Find the change
in internal thermal energy.
Method Two Intuitive Sign Convention
Qin 20 J
Wout 15 J
DU Qin Wout (20) (15) 5 J
10
DU Q - W
Example 10 Joules of heat are added to a gas
while at the same time, the gas is compressed,
requiring 15 Joules of work. Find the change
in internal thermal energy of the gas.
Method ONE Formal Sign Convention
Q 10 J
W -15 J
DU Q W (10) (-15) 10 15 J 25 J
11
DU Q - W
Example 10 Joules of heat are added to a gas
while at the same time, the gas is compressed,
requiring 15 Joules of work. Find the change
in internal thermal energy of the gas.
Method TWO Intuitive Sign Convention
Qin 10 J
Win 15 J
DU Qin Win (10) (15) 25 J
12
Sign Convention
The formal sign convention Q in
positive W out positive is strongly
(very strongly) recommended.
13
Ex 4-1 pg 171
Initial internal energy is 800 kJ. System loses
500 kJ of heat. Paddle wheel performs 100 kJ of
work on system. Find U2.
Q -500 kJ
W -100 kJ
U2 U1 Q W (-500) (-100) -400 kJ
U2 U1 400 800 400 400 kJ
14
Problem Solving Techniquepage 38
15
Problem Solving Techniquepage 38
  • 1) Problem Statement
  • 2) Schematic
  • 3) Assumptions
  • 4) Physical Laws (M E Balances)
  • 5) Properties (IG, Tables, Z Chart . )
  • 6) Calculations
  • 7) Reasoning, Verification, and Discussion

16
Example 4-2 page 171
Step 1 Problem Statement Saturated water
vapor is in a piston-cylinder device. Electrical
work is performed on the steam and there is some
heat loss. Given 25 g of water vapor.
Initial state is
saturated vapor.
P 300 kPa
Heater is
supplied with 0.2 A at 120 V for 5 min Find
Final temperature of steam.
17
Ex 4-2 Step 2 Schematic
0.2 A 120 V 5 min
Q-3.7 kJ
m 25 g P 300 kPa x1 1
18
Ex 4-2 Step 3 Assumptions
Neglect changes in kinetic energy and potential
energy Piston moves with no friction. Work is
performed. Heat is transferred. The pressure is
constant. Thus, the volume may change, and
probably will change.
19
Ex 4-2 Step 4 Physical Laws
Mass Balance m constant 25 g Energy
Balance DU Q - W
20
Ex 4-2 Step 5 Properties
Use tabulated properties of water Tables A-4,5,6
State 1 P 200 kPa sat vap x1 1 v1
vg 0.8857 m3/kg u1 ug 2529.5 kJ/kg
State 2 P 200 kPa
21
Ex 4-2 Step 6 Calculations
Q -3.7 kJ
22
Ex 4-2 Step 6 Calculations
DU Q W U2 U1 m(u2 u1) -3.7 kJ (-7.2
kJ) Wbw
Wbw P(V2-V1) P m (v2- v1)
m(u2 u1) -3.7 kJ (-7.2 kJ) P m (v2- v1)
23
Ex 4-2 Step 6 Calculations
m(u2 u1) -3.7 kJ (-7.2 kJ) P m (v2- v1)
Everything is known in this equation except u2
and v2. We know P2 P 300 kPa.
When using tabular data, the procedure is trial
and error. One procedure might be
Guess T2
Look up values of u2 and v2.
Use values of u2 and v2 in the energy balance to
see if the equation is satisfied.
Repeat with new values of T2 until equation is
satisfied.
24
Ex 4-2 Step 6 Calculations
m(u2 u1) -3.7 kJ (-7.2 kJ) P m (v2- v1)
Fortunately, in this problem there is an easier
way. Rearrange the energy balance
m(u2 u1) P m (v2- v1) -3.7 kJ (-7.2 kJ)
m(u2 Pv2 u1- Pv1) -3.7 kJ (-7.2 kJ)
m(h2 h1) -3.7 kJ (-7.2 kJ)
Great ! Now there is no trial and error ! !
25
Ex 4-2 Step 6 Calculations
m(h2 h1) -3.7 kJ (-7.2 kJ)
h1 hg 2725.3 kJ/kg m 0.025 kg
At 0.3 MPa and 200 oC h 2865.6 kJ/kg
26
Ex 4-2 Step 7 Discussion
We neglected potential energy. The text verifies
that the numerical change in PE is
insignificant. v2 0.7163 v1 0.6058
m3/kg Thus the system expands in order to keep
the pressure constant.
27
The First Law for Open Systems
Energy and Mass Balances for Control Volumes
Discussion will be restricted to steady-state,
steady-flow open systems.
28
FIGURE 4-22Under steady-flow conditions, the
mass and energy contents of a control volume
remain constant.
4-2
29
Mass Balance
At steady state operation, rate of mass flow in
equals rate of mass flow out. Mass is neither
stored nor depleted in the control volume.
30
Energy Balance (First Law)
At steady-state operation rate of energy entering
the control volume equals the rate of energy
leaving the control volume. There is no energy
accumulation or depletion in the control volume.
Rate of energy in Rate of energy out 0
31
Energy Balance
32
Energy Balance
33
Energy Balance
34
Energy Balance
35
Energy Balance
36
Energy Balance
Subscript 1 refers to input
stream. Subscript 2 refers to output or exit
stream.
37
Mass and Energy BalancesControl Volume at Steady
State
Mass Balance
Energy Balance
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