Interactions between heat and work
Consider a system consisting of a cylinder and a piston:
Assume that the system is in equilibrium, i.e. PA = Force due to weight.
Now if we remove the small weight, the finite unbalanced force will cause the system to pass through non-equilibrium states. This is not desired: Thermodynamic state cannot be defined?!
So we assume that process is "quasi-static", that is, the force is infinitesimal and so the state is infinitesimally near a thermodynamic equilibrium. This is called fully resisted expansion.
But AdL = dV, change in volume of the gas, so:
The total work done at the moving boundary can be performed by integrating this equation, i.e;
A number of points should be made here:
i) P is the pressure at the systems' boundary and not necessarily the system pressure!
ii) dV is the differential of the volume swept out by the system boundary and not in general the differential of the system volume
iii) The total work done by a system is not equal to integral of pdV since other forms of work such as electrical, magnetic etc might be present!
The relationship between P&V
To integrate pdV, we need to know the relationship between P and V.
Graphical approach- consider the compression of air in a cyclinder:
At the start of the process, the piston is at position 1, the pressure being relatively low. At the end of the process the piston is at 2. Assume the process is quasi-static so that we know the state of the system from 1-2 and so can draw a full line. The work done on the air during the compression process =
= Area under the curve 1-2, or (1-2-b-a)
Possible to go from 1 to 2 along many different quasi-static paths, such as B & C. Since the work done is equal to the area underneath the curve, we can see that the amount of involved in each case is a fraction not only of the end-states of the process, but in addition is dependant on the path that is followed in going from 1-2. For this reason work is called a path function, as opposed to thermodynamics properties which are point functions.
Special cases of P&V relationships
1.The constant volume process
Area underneath curve = 0
2. Constant pressure process
3. PV = constant process
This simple algebraic formula is in fact satisfied by many substances when expanding in fully resisted expansion at constant temperature, so called isothermal process.
4. The process PVn = k
Also known as the polytropic process. This represents a family of processes which are very useful approximations to real processes.
Consider a rigid vessel divided by a light diaphragm:
When diaphragm gives way under pressure the gas expands to fill the whole vessel. If we consider system S1:
(since dV =0)
The pressure on the moving boundary, i.e. The face of the boundary exposed to vacuum is always zero while that part of the boundary is in motion.
Consider S2: Initial 0, finally volume is the same, so W = 0
Consider a block being pushed along a rough horizontal surface by an agent which applies a force F.
Assume that we want to calculate the work done on the lower part of the block by the system consisting of the agent and the upper part of the block. A shear exists along the part of S in the block tending to make the upper half of the block slide relative to the lower half. If this shear stress has uniform value and acts over area A, then from static:
or in general:
I.e. when sigma is a function of (P), we consider each element of area dA separately and multiply by the value of sigma prevailing there at the moment in question and by the elementary distance dL.
A very important form of shear work which we are concerned with is shaft work.
This comes across in actions such as power transmission in cars and steam turbines, gas turbines etc. Consider a shaft penetrating a system boundary:
Where the shaft crosses the boundary S, for each element of cross-section area dA there is a shear stress sigma in the tangential direction, tending to cause the relative rotation of the two parts of the shaft on either side of S i.e. the torque dT on the element =
Suppose that shaft turns through a small angle
and as a result each element (both sides of S) moves a distance in the direction of the stress. The shear work done by the element dA on its surroundings during the elementary rotation is given by:
Thus the shear work done by the system on the surroundings is given by:
For steady condition T is constant and independant of , then:
Dividing both sides by dT (time increment), we obtain the shaft power,
In an electric field, electrons in a wire move under the effect of electromotive forces, doing work. When N electrons move through a potential difference V, the electrical work done is:
We = VN(kJ)
Which can be expressed in the rate form as:
Heat and temperature
The thermodynamic definition of heat is:"Heat transfer is the interaction between systems which occurs by virtue of their temperature difference when they communicate". But what is temperature? Temperature is the thermodynamics property that we are most familiar with and yet it is amongst the most difficult properties to define! On the microscopic level temperature is related to the kinetic energy of the molecules. But in thermodynamics we are interested in temperature because it enables us to know whether or not two system are in thermal equilibrium.So we define temperature as:"The temperature of a system is a property that determines whether or not a system is in thermal equilibrium with other systems". The concept of thermal equilibrium is the basis of the Zeroth law of thermodynamics.
Heat transfer is +ve when the surroundings have a higher temperature i.e. heat is transferred to the system
Heat transfer is -ve when the system has a higher temperature, i.e. heat is transferred from the system Heat like work is transitory, i.e. not an observable characteristic of the system. Of course it is a path function.
Some notes on the definition of heat:1. Heat is not "that which inevitably causes a temperature rise". For example if a system comprising of ice and water is heated, no temperature rise occurs at least until the ice has melted.2. Heat is not "that which is always present when a temperature rise happens". E.g. from below, water can be heated by work only.
How heat transfer occurs
Conduction is a mode of heat transfer involving solid media due to the impact of adjacent molecules vibrating about mean position. Some solids such as metal rods, copper and silver are very good conductors, while others like wood, cork are very poor conductors and are used as insulators.
This is a mode of heat transfer in a fluid media in motion, e.g. hot wall heats a mass of air which moves to heat a cold wall
Thermal communication which involves no matter of media, e.g. sun radiates heat to warm the earth. The rate of radiative heat transfer increases rapidly with temperature but is negligible at room temperature.
If all the above three modes of heat transfer are absent then Q=0, and we have an adiabatic process.
Heat and Work
One important concept in thermodynamics is the relationship between heat and work. However, there is no logical relationship between heat and work and so it had to be found experimentally. The experiments were carried out by J P Joule between 1840 to 1849. The important experiment involved a system consisting of a well-lagged cylinder containing water going through a cycle composed of two processes.
In the first process, shear work was done on the system when a paddle turned as a weight was lowered, as shown below.
As a result of this process the water temperature was found to rise steadily. When the temperature reached a certain value, the paddle wheel was stopped and the work done noted (historically in foot pounds-force).
In the second process the system was brought into contact with a cold body so that heat was transferred from the system, as shown below.
This process was then cut-off when the system reached its original state and again the amount of heat transfer was noted down (in British Thermal units or calorie).
After making the measurements for a variety of systems and for various amounts of work and heat, it was found that the amounts of work and heat were always proportional. These observations were then formalised into the First Law of Thermodynamics which is stated as:
"If any system is carried through a cycle (the end state being precisely the same as the initial state) then the net work is proportional to the net heat transfer".
Where J is a constant known as the Mechanical Equivalent of Heat. In British units J is 778 ft lbf per Btu and in SI units its 1. Nm per Joules i.e.
The Property Enthalpy
Now that by considering the first law for process in which there is a change of state we have managed to define a new property called internal energy, we can define another useful property called enthalpy. For this purpose consider a constant pressure process e.g. the gas in a leak proof cylinder with weights on it. Assume there are no changes in kinetic energy and potential energy. and the only work done is displacement work. Thus from First Law:
That is heat transfer during a constant pressure process is given in terms of the change in the quantity (U + PV) between the initial and final states. Since all thermodynamic properties are functions of the end-states only, then (U + PV) must be a property. Thus we define enthalpy, H, an extensive property:
H = U + PV
The use of H is not restricted to constant pressure process as we shall show later.
Enthalpy is an extensive property which can be intensive by dividing it by mass, i.e. specific enthalpy.
h = u + pv
Specific heats, internal energy and enthalpy of an ideal gas
An ideal gas not only satisfies the equation of state pv = RT, but its specific heats are constant also. For real gases, these vary appreciably with temperature, and a little with pressure.
The properties of a substance are related by:
Tds = du + pdv (1)
Ds = du/T + pdv/T (2)
The internal energy u is assumed to be a function of T and v, i.e.:
From equations 1 and 2;
Again, let s = f(T,v)
Comparing equations 4 and 5;
Differentiating equation 6 with respect to v when T is constant
Differentiating equation 7 with respect to T when v is constant
From equations 8 and 9;
For an ideal gas, pv = RT
From equations 10 and 11;
Therefore, u does not change when v changes at constant temperature.
Similarly, if u = f(T,p) it can be shown that
Therefore, u does not change when v changes at constant temperature.
U does not change unless T changes.
Then: u = f(T) (13) only for an ideal gas. This is known as Joule's law.
If u = f(T,v)
Since the last term is zero by equation 12, and by definition
The equation du = cv dT holds good for an ideal gas for any process, whereas for any other substance it is true for a constant volume process only.
Since cv is constant for an ideal gas;
The enthalpy of any substance is given by: h = u + pv
For an ideal gas, h = u +RT
Therefore, h = f(T) (15)
Only for an ideal gas.
Now dh = du + Rdt
Since R is a constant,
Since h is a function of T only, and by definition:
From equations 16 and 17,
cp = cv + R
cp -cv = R (18)
The equation dh = cpdT holds good for an ideal gas, even when pressure changes, but for any other substance, this is true only for a constant pressure change.
The ratio of cp/cv is of importance in ideal gas computations, and is designated by the symbol, ie.
From equation 18,
If is substituted into equation 19;
and are the molar or molal specific heats at constant volume and constant pressure respectively.
It can be shown by the classical kinetic theory of gases that the values of
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