Thermodynamics Lecture 3 (Jul 13)
Steady flow energy equation with enthalpy introduced:
Conservation of Energy
“Energy cannot be created nor destroyed, One form of
energy can only be converted into Another form”.
-[First Law of Thermodynamics]
Forms of Energy:
Gravitational Energy, P  mg  z2  z1 
1
m  v22  v12 
2
Internal Energy, (per unit mass) u  u2  u1
(per Total mass, U = mu) U  U2 U1
Kinetic Energy, K 
Work for Non-Flow, W 

2
1
pdV
Work for Flow, Wf  p2V2  pV
1 1
Heat, (per unit mass) q  q2  q1
(per Total mass, Q = qm) Q  Q2  Q1
Important Notes:
(1) convention for Work: Work “done on” the system is
negative while Work “done by” the system is positive
(2) convention for heat: heat “added” to a system is
positive while heat “rejected” by a system is negative
Classification of Systems
(1) Closed system – one in which mass does not cross its
boundaries (ex. Closed piston, closed box)
(2) Open system – one in which mass crosses its
boundaries (ex. Pumps, turbines, heat exchangers)
Steady Flow Energy Equation
Characteristics:
(a) There is neither accumulation or diminution of mass
within the system
(b) There is neither accumulation or diminution of energy
within the system
(c) State of working substances at any point in the system
remains constant (ex. Inlet properties, outlet properties)
Energy Balance Equation :
(or Steady Flow Energy Equation)
Energy entering system = Energy leaving system
P1  K1 Wf 1 U1  Q  P2  K2 Wf 2 U2 W
Q  P  K  U  W
Concept of Enthalpy, h = u + pv
Enthalpy, (per unit mass) h  h2  h1
(per Total mass, H = mh) H  H2  H1
P1  K1  H1  Q  P2  K2  H2  W
Q  P  K  H  W
Problems:
1. During a steady flow process, the pressure of the
working substance drops from 200 to 20 psia, the
speed increases from 200 to 1000 fps, the internal
energy of the open system decreases 25 Btu/lb, and
the specific volume increases from 1 to 8 ft3/lb. No
heat is transferred. Sketch an energy diagram.
Determine the work per lb. Is it done on or by the
substance? Determine the work in hp for 10 lb per
min. (Note: 1 hp = 42.4 Btu/min)
2. Steam is supplied to a fully loaded 100-hp turbine at
200 psia with u1=1163.3 Btu/lb, v1=2.65 ft3/lb and
v1=400 fps. Exhaust is at 1 psia with u2=925 Btu/lb,
v2=294 ft3/lb and v2=1100 fps. The heat loss from the
steam in the turbine is 10 Btu/lb. Neglect potential
energy change and determine (a) the work per lb
steam and (b) the steam flow rate in lb/h.
3. An air compressor (an open system) receives 272 kg
per min of air at 99.29 kPa and a specific volume of
0.026 m3/kg. The air flows steady through the
compressor and is discharged at 689.5 kPa and
0.0051 m3/kg. The initial internal energy is 6241
J/kg. The cooling water circulated around the
cylinder carries away 4383 J/kg of air. The change in
kinetic energy is 896 J/kg increase. Sketch an energy
diagram and compute the work.
4. A centrifugal pump operating under a steady flow
condition delivers 2,270 kg/min of water from an
initial pressure of 82,740 Pa to a final pressure of
275,800 Pa. The diameter of the inlet pipe to the
pump is 15.24 cm and the diameter of the discharge
pipe is 10.16 cm. What is the work?
5. A turbine operates under steady flow conditions,
receiving steam at the following state: pressure 1200
kPa, temperature 188degC, enthalpy 2785 kJ/kg,
speed 33.3 m/s and elevation 3m. The steam leaves
the turbine at the following state; pressure 20 kPa,
enthalpy 2512 kJ.kg, speed 100 m/s and elevation
0m. Heat is lost to the surroundings at the rate of
0.29 kJ/s. If the rate of steam flow through the
turbine is 0.42 kg/s, what is the power output of the
turbine in kW?