Thermodynamics 2
Lecturer: MG Moletsane
Lesson 9: Gas Processes – Constant V, P, T
E&McC: Chapter 3 p51-59
Study Guide: Learning Unit 4 p20
Introduction
• We look firstly at the heating/cooling processes with Perfect Gases
(air)
• Engines that uses air as the working fluid are mainly piston-cylinder
systems – i.e. Closed Systems (non-flow processes)
• We will investigate heat transfer and work in/out of a system by
keeping 1 of the 3 major properties (p, V, T) constant
• We will also look at the relations between gas properties for the 3 gas
processes:
• Constant Volume Process [V]
• Constant Pressure Process [P]
• Constant Temperature Process [T] – Isothermal Process
Constant Volume Process [V]
E&McC p51
Work
But dv = 0 → W = 0
Heat
But W = 0 →
(kJ/kg)
(3.1)
(kJ)
(3.2)
for a mass m of the gas:
All the Heat supplied in a [V] process is used to
increase the Internal Energy of the system
Heat transferred at [V] is
which is a proof for Joule’s Law:
Constant Volume Process [V]
E&McC p51
General Gas Equation
Since V1 = V2:
Also, from
Gas equation for a [V] process:
P1 /T1 = P2 /T2
→ P/T = mR/V = Constant
P/T = C
Constant Pressure Process [P]
E&McC p52
Work
because p is constant
Heat
→
for a mass m of the gas:
(kJ/kg)
(3.3)
(kJ)
(3.4)
All the Heat supplied in a [P] process is used to
increase the Enthalpy of the system
Heat transferred at [P] is
Enthalpy,
Constant Pressure Process [P]
E&McC p52
General Gas Equation
Since P1 = P2:
Also, from
Gas equation for a [P] process:
V1 /T1 = V2 /T2
→ V/T = mR/p = Constant
V/T = C
Constant Temperature (Isothermal) Process [T]
E&McC p52
General Gas Equation
Since T1 = T2:
P1 V1 = P2 V2
(3.5)
Also, from
Gas equation for a [T] process:
→ PV = Constant
Constant Temperature (Isothermal) Process [T]
E&McC p55
Work
But
Since c = pv =
→ p = c/v
we can write
(kJ/kg)
(3.6)
(kJ/kg)
Constant Temperature (Isothermal) Process [T]
E&McC p55
Work: Variations on the Formula
since
then
so that
(kJ/kg)
(3.8)
(kJ/kg)
(3.10)
since
then
for a mass m
of the gas:
(kJ)
(3.7)
(kJ)
(3.9)
(kJ)
(3.11)
Constant Temperature (Isothermal) Process [T]
E&McC p55
Heat
from Joule’s Law:
Since T1 = T2:
Thus
this implies that Q = -W
All the Heat supplied in a [T] process is directly
converted into Work Output
NB: Q = -W means that all the previous formulas
derived for Work is the same for Heat (-)
(3.12)
Introduction
• We are still looking at the energy transfer processes with Perfect
Gases (air) in Closed Systems (non-flow processes)
• We will also look what happens to the gas when no heat is
transferred across the boundaries of the system – adiabatic
process [Q = 0]
• Perfect thermal insulation of the system
• Very fast process – no time for heat to be transferred
Adiabatic Process [Q]
E&McC p59, 63
Work
But
from
→
and
(kJ/kg)
(3.23)
Heat
All the Work done on the fluid increases the Internal Energy
Work can be done by the fluid at the expense of the Internal Energy
Adiabatic Process [Q]
E&McC p62
General Gas Equation:
Gas equation for a [Q] process:
(3.19)
Also, from
and from
:
(3.20)
:
(3.21)
End of Lesson 9
Summary
• [V]: Q → ΔU
W=0
P1 /T1 = P2 /T2
• [P]: Q → ΔH
W = p(V2 –V1)
V1 /T1 = V2 /T2
• [T]: Q = -W
• [Q]: W → ΔU
• Study E&McC p51-59
• Look at Ex 3.1(ii), 3.3
• Practice problems 3.1, 3.3, 3.7 on p84