TRC2200: Thermo Fluids & Power Systems

LECTURE 17

Mass and Energy Analysis of Open Systems (Control Volumes)

Lecturer: Alpha Agape Gopalai

• Conservation of mass

• Flow work and the energy of a flowing fluid

• Energy analysis of steady-flow systems

• Some steady-flow engineering devices

Lecture Outline

• Conservation of mass: Mass, like energy, is a conserved property, and it cannot be created or destroyed during a process.

• Closed systems: The mass of the system remain constant during a process.

• Control volumes: Mass can cross the boundaries, and so we must keep track of the amount of mass entering and leaving the control volume.

Mass is conserved even in chemical reactions

Conservation of Mass

Mass & Volume Flow Rates

Conservation of Mass Principle

Conservation of Mass Principle

Total Energy of a Flowing and Non-Flowing Fluid

Total Energy of a Flowing and Non-Flowing Fluid

9

Flow work and the energy of a flowing fluid

Schematic for flow work.

Flow work, or flow energy: The work

(or energy) required to push the mass into

or out of the control volume. This work is

necessary for maintaining a continuous

flow through a control volume.

10

Mass and Energy balances for a steady-flow process

Mass balance

Energy balance

11

• Nozzles and Diffusers

• Turbines and Compressors

•Throttling valve

• Mixing chambers

• Heat exchangers

• Pipe and duct flow

Case studies

12

Nozzle and Diffuser

A nozzle is a device that increases the velocity of

a fluid at the expense of pressure.

A diffuser is a device that increases the pressure

of a fluid by slowing it down.

Nozzles and diffusers are commonly utilized in jet

engines, rockets, spacecraft, and even garden hoses.

13

Nozzles and Diffusers

Energy balance for a nozzle or

diffuser:

14

Compressor

Compressors, as well as pumps and fans, are devices used to increase the

pressure of a fluid. Work is supplied to these devices from an external source

through a rotating shaft.

Energy balance for the compressor in

this figure:

15

Turbine

A turbine is a rotary engine that extracts energy from a fluid flow and converts it into useful work.

16

Throttling valves

Throttling valves are any kind of flow-restricting devices that cause a significant

pressure drop in the fluid.

What is the difference between a turbine and a throttling valve?

The pressure drop in the fluid is often accompanied by a large drop in temperature,

and for that reason throttling devices are commonly used in refrigeration and air-

conditioning applications.

17

Throttling valves

The temperature of an ideal gas does not

change during a throttling (h =

constant) process since h = h(T).

Energy balance

18

Mixing chambers

In engineering applications, the section where the mixing process takes place is commonly referred to as a mixing chamber.

Energy balance for the adiabatic

mixing chamber in the figure is:

19

Heat exchangers

Heat exchangers are devices where two moving fluid streams exchange heat without mixing. Heat exchangers are widely used in various industries, and they come in various designs.

A heat exchanger can be as simple

as two concentric pipes.

20

Pipe and duct flow

The transport of liquids or gases in pipes and

ducts is of great importance in many

engineering applications. Flow through a pipe

or a duct usually satisfies

the steady-flow conditions.

Heat losses from a hot fluid flowing through an uninsulated

pipe or duct to the cooler environment may be very significant.

Pipe or duct flow may involve more than

one form of work at the same time.

Example

21

Air enters the compressor of a gas-turbine plant at 100 kPa and 25°C with a low velocity and exits at 1 MPa and 347°C with a velocity of 90 m/s. The power input to the compressor is 250 kW and it is cooled at 1500 kJ/min. (a) Determine the mass flow rate of air through the compressor. Take the molecular mass of air as 29 kg/kmol and allow for variation in specific heat of air as a function of temperature (in Kelvin) as given by the equation:C pa = 28.11 + 0.1967 x 10-2T +0.4802 x 10-5T2 -1.966 x 10-9T3 kJ/kmol.K

Example:

You may begin by considering the general energy equation given below:

outoutinin WQgZ

ChmWQgZ

Chm 2

2

221

2

11

22

P1=100kPaT1=25oC

C1=0

P2=1MPaT2=347oC

C2=90m/s

1500 kJ/min

250 kWGiven:

Molecular mass of air 29 kg/kmol

Compressor power input Win = 250 kW

Heat rejection Qout = 1500 kJ/min

22

P1=100kPaT1=25oC

C1=0

P2=1MPaT2=347oC

C2=90m/s

250 kW

1500 kJ/min

The potential energy change, input heat, inlet velocity and output mechanical work are negligible in the case of the compressor

Determine:

a) Mass flow rate of air

2

2

2121

2Q

CmWTCTCm papa

C pa = 28.11 + 0.1967 x 10-2+0.4802 x 10-5T2 -1.966 x 10-9T3 kJ/kmol.K

At T1 = 298K Cpa = 29.07 kJ/kmol K

= 29.07/29 kJ/kgK

= 1.004 kJ/kgK

h=CpT

Example

23

At T2 = 620K Cpa = 30.71kJ/kmol K

= 30.71/29 kJ/kgK

= 1.0588 kJ/kgK

2

2

2121

2Q

CmWTCTCm papa

skgm

xmxxm

/62.0

60

1500

10002

902506200588.1298004.1

2

P1=100kPaT1=25oC

C1=0

P2=1MPaT2=347oC

C2=90m/s

1500 kJ/min

250 kW

Example

• Conservation of mass

• Mass and volume flow rates

• Mass balance for a steady-flow process

• Mass balance for incompressible flow

• Flow work and the energy of a flowing fluid

• Energy transport by mass

• Energy analysis of steady-flow systems

• Some steady-flow engineering devices

• Nozzles and Diffusers

• Turbines and Compressors

• Throttling valves

• Mixing chambers and Heat exchangers

• Pipe and Duct flow

Lecture Summary

Attendance Code