Thermodynamics-I

Lecture No.05Akbar Ali Qureshi

Lecturer Email: akbaraliqureshi@bzu.edu.pk

Contact no: 0335-6387138 Mechanical Engineering Department

UCE&T, BZU Multan.

Sequence

• The working fluid• Liquid, vapor and gas• The use of vapor tables• Properties of wet vapor• Problems

The working fluid

• The matter contained within the boundaries of the system is defined as the working fluid.

• In thermodynamic systems the working fluid can be in the liquid, vapor or gaseous phase.

• All substances can exist in any of these phases, but we tend to identify all substances with the phase in which they are in equilibrium at atmospheric temperature and pressure.

• For example, substances such as oxygen and nitrogen are gases; H2O is liquid or vapor (water or steam); H2O can become a gas at very high temperature.

Liquid, vapor and gas .

• The points P, Q and R represents the boiling points of the liquid.

Boiling points plotted on PV diagram

Liquid, vapor and gas...Contd. At points P’, Q’ and R’ the liquid is

completely changed into vapor.

Points of complete vaporization plotted on PV diagram

Liquid, vapor and gas … Contd. Saturation State: A state at which a change of phase may

occur without the change of temperature or pressure. Saturated Liquid line: The boiling points P, Q and R are

saturation states, and series of such boiling points joined up is called the saturated liquid line.

Saturated vapor line: The points P’,Q’ and R’ are saturation states, at which liquid is completely changed into vapor, and the series of such points joined up is called the saturated vapor line.

Wet vapor: The substance existing at a state point inside loop consists of a mixture of liquid and dry vapor and is known as wet vapor.

Liquid, vapor and gas ...Contd. Specific enthalpy of vaporization: The additional heat

supplied which changes the phase of the substance from liquid to vapor at constant temperature and pressure is called specific enthalpy of vaporization.

Liquid, vapor and gas … Contd. Isotherms: The lines of constant temperatures. Superheated: When a dry saturated vapor is heated at constant

pressure, its temperature rises and it becomes superheated.

Isothermals for a vapor plotted on PV diagram

Liquid, vapor and gas ...Contd.

• Dryness fraction: The condition or quality of a wet vapor is most frequently defined by its dryness fraction.

Dryness fraction, x= mass of dry vapor in 1kg of mixture.Note: For dry saturated vapor, x=1 For dry saturated liquid, x=0

The use of vapor tables.

• The tables which will be used in this course are arranged by Rogers and Mayhew.

• The tables are mainly concerned with steam, but some properties of refrigerants are also given.

• For example, the table for wet steam is shown below.

Properties of wet vapor.

• The properties of wet vapor are.

Problems.

Problem1: Calculate the specific volume, specific enthalpy and specific internal energy of wet steam at 18 bar, dryness fraction 0.9.(v=0.0994m3/kg, h=2605.8kJ/kg and u=2426.5kJ/kg)

Problem 2: Calculate the dryness fraction, specific volume and specific internal energy of steam at 7 bar and specific enthalpy 2600kJ/kg.(x=0.921, v=0.2515m3/kg and u=2420kJ/kg)

Sequence

• The properties of superheated vapor• Problems• Interpolation• The perfect gas• Problems

The properties of superheated vapor

• For steam in superheated region, temperature and pressure are independent properties.

• When the temperature and pressure are given for superheated steam then the state is defined and all the other properties can be found.

• For example, steam at 2 bar and 200C is superheated since the saturation temperature at 2 bar is 120.2C, which is less than the actual temperature. The steam in this state has a degree of superheat of 220-120.2=79.8C

Problems.

Problem 1: Steam at 110 bar has a specific volume of 0.0196m3/kg. Calculate the temperature, the specific enthalpy and specific internal energy.

Problem 2: Steam at 150 bar has a specific enthalpy of 3309kJ/kg. Calculate the temperature, the specific volume and specific internal energy

Interpolation. For properties which are not tabulated

exactly in the tables it is necessary to interpolate between the values tabulated.

For example to find the temperature at 9.8 bar, it is necessary to interpolate between the values given in the tables.

The perfect gas. The characteristic equation of state: At temperatures that are considerably in excess of the

critical temperature of a fluid, and also at very low pressures, the vapor of the fluid tends to obey the equation.

• The constant R is called the specific gas constant and the units of R are Nm/kgK or kJ/kgK

• Perfect gas: An imaginary ideal gas which obeys the law is called a perfect gas.

The characteristic equation of state of a perfect gas

The perfect gas ...Contd. For a mass m and occupying a volume V, ……………………(a) Another form of the characteristic equation can be derived by

using the amount of substance (mol). The amount of substance of a system is that quantity which

contains as many elementary entities as there are atoms in 0.012kg of C-12.

The normal unit for the amount of substance is ‘mol’. In SI it is convenient to use ‘kmol’

Molar mass: The mass of any substance per amount of substance is known as the molar mass.

where m is the mass and n is the amount of substance and the units of

molar mass is kg/kmol .

The perfect gas ...Contd. Substituting for m in eq. (a) gives,

Now Avogadro’s hypothesis states that the volume of 1 mol of any gas is same as volume of 1 mol of any other gas, when the gases are at the same temperature and pressure .

Therefore V/n is same for all gases at same value of p and T. i.e. the quantity pV/nT is constant for all gases . this constant is called the molar gas constant. And is given the symbol and the value of R is 8.3145kJ/kmolKR̃

Problems.

Problem1: A vessel of volume 0.2m3 contains nitrogen at 1.013bar and 15C. If 0.2kg of nitrogen is now pumped into the vessel, calculate the new pressure when the vessel has returned to its initial temperature. The molar mass of nitrogen is 28kg/kmol, and it may be assumed to be a perfect gas. (Answer 1.87 bar)

Problem 2: A certain perfect gas of mass 0.01 kg occupies a volume of 0.003m3 at a pressure of 7 bar and temperature of 131C. The gas is allowed to expand until the pressure is 1 bar and the final volume is 0.02m. Calculate the molar mass of the gas and the final temperature. (Answers 16 kg/kmol; 111.5C)

Specific Heat

Specific Heat at Constant Pressure (CP) :

The Energy required to raise the temperature of a unit mass of a substance by 1

degree, as the Pressure is maintained CONSTANT.

Specific Heat at Constant Volume (CV) :

The Energy required to raise the temperature of a unit mass of a substance by 1

degree, as the Volume is maintained CONSTANT.

m = 1 kg∆T = 1 ºC

Sp. Heat = 5 kJ/kg ºC

5 kJ

DEFINITION :

The Energy required to raise the temperature of a

unit mass of a substance by 1 degree.

Specific heat

Specific heat at constant volume Cv

Hence, CV is change in Internal

Energy of a substance per unit

change in temperature at constant

Volume.

Specific Heat at constant Pressure

Hence, CP is change in Enthalpy of a

substance per unit change in temperature

at constant Pressure.

Relation between Specific Heat

The End