Transcript ENGR-1100 Introduction to Engineering Analysis

Lecture 1
Laws of Thermodynamics
 Thermodynamic state - equilibrium
 Thermodynamic processes
 Laws of thermodynamics
 Absolute Temperature
 Problems 2.5, 2.6, 2.8
Thermodynamic state - equilibrium
 Thermodynamic intensive coordinates are uniform
across the whole system (T, P, ) or across each
macroscopic phase (e.g., water and ice density
density at the melting point.
 All thermodynamic coordinates are time independent
 Mechanical equilibrium, thermal equilibrium and
chemical equilibrium
Macroscopic vs. microscopic state
 Thermodynamic coordinates (T, P, ) define
macroscopic state of equilibrium.
 Microscopic state is defined by atomic positions, and
momenta - many microscopic states are consistent
with a macroscopic state
 Statistical mechanics connects microscopic
description and detail with macroscopic state via
ensemble average
Thermodynamic process  change of the
thermodynamic state
 Infinitesimal process  infinitesimal change of
coordinates, e.g., dT, dV, dP
 Quasi static process  always near equilibrium
 Adiabatic process  no heat
 Reversible process  can be restored to the initial
state without charging surroundings
0th law of thermodynamics
Two or more systems in equilibrium do not exhibit heat
flow among each other, they are at the same temperature
Later we will see that criterion of equilibrium for isolated system,
i.e., const E, V, T is the maximum entropy state
dS(E, V, N) = 0
dS = dS1+dS2 = 0
Allowing only energy exchange between
two isolated systems
dS1 
dS2 
dS1  dS2 
S1
S
S
S
dE1  1 dV1  1 dN1  1 dE1
E1
V1
N1
E1
S1 S2

E1 E 2
S2
S
S
S
dE 2  2 dV2  2 dN 2  2 
dE 2
E 21
V2
N 2
E 2
From conservation of energy
dE 2  dE1
S1
S
dE1  2 dE1
E1
E 2

E1, V1, N1
T
E
S
E2, V2, N2
1st law of thermodynamics - conservation of
energy
dE = dQ-dW
d indicates inexact differential - depends on the
integration path
In a cycle, E = 0
net Q in = net W out
 Work and heat are not state functions
 Energy is a state function
dW = Fdx can be PdV, -dl, -it
2nd law of thermodynamics - entropy
For a reversible process
dQ= TdS
Where S is entropy which a state function, and T is an
absolute temperature
The entropy can by calculated by integrating heat over a
reversible path
dQ
S f  Si  
R T

Absolute temperature
Consider the Carnot cycle
Since entropy is the
state function
Q1
T1
S1  S2
T
T2
Q
Q3
Q2

S
Q1
S1 
T1
Q2
S2 
T2
Q1 Q2

T1 T2

Using reference
T3 = 273.16 K

Q
T  273.16
Q3
Q>0
Problem 2.5
When a system is taken from a to b state along acb 80 joules of heat flows
into the system and the system does 30 joules of work.
a)
How much heat flows into the system along path adb, if the work done
by the systems is 10 joules.
b)
When the system is returned from b to a along the curved path the work
done on the system is 20 joules. Does the system absorb or liberate the
heat? How much?
c)
If Ea = 0 and Ed = 40 joules, find the heat absorbed in process ad and db.
c
b
Answers:
P
a
d
V
a)
Qadb = 60 joules.
b)
Qba = - 70 joules joules
(liberate heat) of the heat
c)
Qad= 50 J, Qdb= 10 J
Problem 2.6
A vessel of volume VB contains n moles of gas at high pressure. Connected to
the vessel is a capillary tube trough which the gas may slowly leak out into
the atmosphere, where P=P0. Surrounding the vessel and capillary is a water
bath, in which is immersed an electric resistor. The gas is allowed to leak
slowly trough the capillary into the atmosphere while, at the same time,
electrical energy is dissipated in the resistor at such a rate that the temperature
of the gas, the vessel, the capillary and the water is kept equal to that of the
surrounding air. Show that, after as much gas is leaked as is possible during
time , the change of internal energy is
E  i  P0 (nv0 VB )
where, v0 = molar volume of gas at P=P0,  = potential on the resistor, and i
is the current.

Problem 2.8
The tension force in a wire is increased quasi statically and isothermally from
1 to 2. If the length, cross-sectional area and isothermal Young’s modulus
(Y) remain practically constant, show that the work done by the wire is:
L
2
2
W 
( 2  1 )
2AY
