undefined | unit 4 properties of pure substances thermodynamics of vapour cycle

THERMO

UNIT 4

Properties of Pure substances & Thermodynamics of Vapour

Q1) Explain the various steps and heats involved in formation of steam.

A1) Formation of steam

The above diagram shows the phase change occurring when water from and at 0°C is heated till is converted to steam and further to superheat it.

Temperature is plotted as ordinate and Enthalpy as abscissa.

Phase Change

During part A, water at 0°C is heated till it is converted into saturated liquid at 100°C, ie, the boiling point of water. During this process, water does not undergo any phase change. Hence, the heat supplied is used only for sensible heating of water. It is denoted by hf and is called as enthalpy of liquid.

During part B, water at 100°C is being heated to convert it into steam. Complete conversion of water into steam with no water droplets presence takes time. The heat transfer taking place during this process is utilized to change the phase of water from liquid to gaseous state. This is denoted by hfgand is called as latent heat of vapourisation.

During parts A and B, the total heat transferred is denoted by hg and it is given by hg = hf + hfg

During part C, the steam is already dry saturated. Any further heating results in superheating of steam. The heat supplied during this process is denoted by hsup and it is given by

hsup= hg + Cp (tsup - tsat)

tsup = temperature of superheated steam

tsat = temperature of saturation of steam

Cp = Specific heat of water at constant pressure.

Q2) Explain the critical point of water with the help of T-v diagram.

A2) 1. T-v diagram for pure substance:

Critical Point

As we go on increasing the pressure of the system, the saturation line continues to become smaller and smaller, and it becomes a point when the pressure is 22.06 MPa (220.6 bar) for water.

This point is called the critical point, and it is defined as the point where the saturated liquid and saturated vapor states are same or latent heat of vapourisation becomes 0.

The temperature, pressure, and specific volume of a substance at the critical point are called, respectively, the critical temperature Tcr, critical pressure Pcr, and critical specific volume vcr.

The properties of water at critical-point are Pcr = 22.06 MPa, Tcr = 373.95°C, and vcr = 0.003106 m3/kg.

Q3) Explain triple point of water.

A3) Triple point

Under some special conditions, all 3 phases of a pure substance can co-exist in equilibrium

On P-v or T-v diagrams, these triple-phase states form a line called the triple line.

The states on the triple line of a substance have the same pressure and temperature but different specific volumes.

The triple line appears as a point on the P-T diagrams and, therefore, is often called the triple point.

For water, the triple-point temperature and pressure are 0.01°C and 0.6117 kPa, respectively.

Q4) Define dryness fraction.

A4) Dryness fraction is defined ratio of mass of dry steam to total mass of mixture of wet steam.

It is denoted by ‘x’.

For dry saturated, it is 1.

For wet steam, it is between 0 and 1.

It is very important property of steam and needs to be determined before the steam is fed to the boiler.

Q5) Explain the Simple Rankine Cycle.

A5) Simple Rankine Cycle:

The above diagrams illustrate the various processes of a simple Rankine Cycle applied to a simple steam powered power plant.

Let us the various processes.

Process 1-2 Water is compressed in Pump till intake pressure of boiler. It is an isentropic compression process.

Process 2-3 Water enters boiler at intake pressure, heat (qi) is added to the system to convert water into steam. The entire process takes place at constant pressure. It is an isobaric heat addition process.

Process 3-4 Steam from boiler enters the turbine where it expands to produce useful work. The process is isentropic expansion.

Process 4-1 Exhaust wet steam from the turbine enters the condenser where it is condensed back to water to be used as feed to boiler. It is an isobaric heat rejection process.

Analysis of Rankine Cycle

Assume 1 kg of steam in the cycle and applying SFEE to various processes, we get,

for Process 1-2

Pump work = h2 - h1 = ʃ-V dp

for Process 2-3

Heat supplied in boiler (qi) = h3 - h2

For Process 3-4

Work done in turbine = WT= h3 - h4

For Process 4-1

Heat rejected in condenser (qr) = h4 – h1

Efficiency of Rankine Cycle (η)

Efficiency of Rankine Cycle η = (Shaft work) / (Heat supplied)

= WT – WP / qi

= {(h3 - h4) - (h2 - h1)} / h3 - h2

Q6) Explain the modified Rankine cycle.

A6) 1. Modified Rankine Cycle – Rankine Cycle with Reheat

Analysis of Rankine cycle with reheat

Efficiency of reheat cycle for two stage turbines with one reheater in between the stage can be computed as

WT = WT(HP) + WT(LP)

= (h3 – h4) + (h5 – h6)

Wp= (h2 - h1)

Heat Supplied = Heat supplied in boiler and reheater = (h3 - h2) + (h5 - h4)

Advantages

Improves condition of steam at exhaust of LP turbine, thereby reduces erosion of blade.

Improves thermal efficiency of plant as additional heat is supplied at higher mean temperature increases the net work output of turbine.

Reduces steam rate per kWh.

Disadvantages

Increases cost and size of plant due to inclusion of reheater and long piping

Increases size of condenser based on unit mass flow of steam due to improved quality of steam at exhaust from LP turbine.

Q7) Describe the working of throttling calorimeter.

A7) 1. Throttling Calorimeter

The steam to be analyzed, is taken from the pipe by a tube.

It passes into an insulated container and is throttled through an orifice to atmospheric pressure.

The throttling process is shown on h-s diagram in Fig. by the line 1-2.

If steam initially wet is throttled through a sufficiently large pressure drop, then the steam at state 2 will become superheated.

State 2 can then be defined by the measured pressure and temperature.

The enthalpy, h2 can then be found and hence

h2 = h1 = hf1 + x1hfg1

x1 = (h2 - hf1) / hfg1

Q8) A throttling calorimeter is used to measure the dryness fraction of the steam in the steam main which has steam flowing at a pressure of 8 bar. The steam after passing through the calorimeter is at 1 bar pressure and 115°C. Calculate the dryness fraction of the steam in the main.

Take Cps = 2.1 kJ/kg K.

A8) Condition of steam before throttling :

Condition of steam after throttling :

As throttling is a constant enthalpy process

720.9 + x1  2046.5 = 417.5 + 2257.9 + 2.1(388 – 372.6)

720.9 + 2046.5 x1 = 2707.7

Q9) Calculate the internal energy per kg of superheated steam at a pressure of 10 bar and a temperature of 300°C. Also find the change of internal energy if this steam is expanded to 1.4 bar and dryness fraction 0.8.

Q9) At 10 bar, C. From steam tables for superheated steam.

And corresponding to 10 bar (from tables of dry saturated steam)

To find

Internal energy of superheated steam at 10 bar,

= 2806.2 kJ/kg. (Ans.)

At 1.4 bar. From steam tables :

Enthalpy of wet steam (after expansion)

Internal energy of this steam,

= 2105.49 kJ

Hence change of internal energy per kg

u2 - u1 = 2105.49 – 2806.2

= - 700.7 kJ. (Ans.)

Q10) In a steam power cycle, the steam supply is at 15 bar and dry and saturated. The condenser pressure is 0.4 bar. Calculate the Carnot and Rankine efficiencies of the cycle. Neglect pump work.

A10) Steam supply pressure, p1= 15 bar, x1= 1

Condenser pressure, p2 = 0.4 bar

Carnot and Rankine efficiencies :

At 15 bar :