First law of thermodynamics:
It is based on the conservation principles ( mass and energy ). It gives the effect of interactions between the system and surrounding on the total energy of the system.
First law of thermodynamics for control mass:
Conservation of mass for a control mass:
it states that total mass of control mass is always constant. It is given by: dm = 0
Conservation of volume for a control mass:
it states that the change in total energy of control mass is equal to the heat supplied to the control mass minus work produced by control mass. It is given by:
First law of thermodynamics for control mass undergoes a cyclic process:
It states that the total heat transferred to the control mass is equal to the net work done by the control mass. And given by :
Application of first law of thermodynamics for non – flow process:
Constant volume or isochoric process : For a constant volume there is no displacement so work transfer is zero. So it is given by:
Hence, heat transferred to the control mass is equal to the change in internal energy when volume is constant.
Constant pressure or isobaric process: for constant pressure process , work transfer is given by:
Specific heats of ideal gas:
- Specific heat at constant volume for an ideal gas is defines as the change in specific internal energy per degree change in temperature during constant volume.
Specific heat at constant pressure for an ideal gas is defines as the change in enthalpy per degree change in temperature during constant pressure.
First law of thermodynamics for control volume:
Conservation of mass for control volume:
it states that change in mass in control volume is always equal to difference between mass enter and mass exit in control volume. It is given by:
Mass flow rate : let us consider a fluid flowing through the port as shown in figure. Having a cross section area A and small displacement dS and velocity of fluid is V.
conservation of volume for control volume:
it states that” The total change in energy in control volume is net energy transport by the fluid in control volume plus heat transferred by fluid minus work done by the control volume.”
This is shown schematically in Figure. Let E , termed the energy rate, denote the energy transported with the fluid at the inlets and outlets. The energy within the control volume is denoted by , so the conservation of energy is
Heat transfer occurs due to the temperature difference between system and surrounding so
In control volume difference types of work transfer are includes that is flow work, shaft work, expansion or compression work etc but in case of flowing fluid flow work is occurs so the relation of flow work is given below:
Consider a fluid is flowing in port then take a small distance then fluid force is F in a control volume then work flow is given by:
Where F is force acting on a fluid particles which is given as F = PA where P= pressure of fluid and A= cross section area of tube. Substituting F= PA then
The rate of work done in the control volume formulation is the sum of the shaft and flow work plus the general forms.
Control volume analysis:
Control volume can be analyzed either with reference to space or with reference to time. Control volume can be classified as uniform system or non- uniform system with respect to space and steady state and unsteady state with respect to time.
Steady state analysis:
In this state the mass and energy should not change with respect to time. Mathematically, for mass
Unsteady state analysis:
In this state the mass and energy is change with respect to time. Mathematically it is given below
Difference between steady state and unsteady state:
|Steady state||Unsteady state|
|The energy and mass is not change with respect to the time.||The energy and mass is change with respect to the time.|
|The properties at the boundaries of the control volume is not change with time.||The properties of the boundaries of the control volume is change with time.|
|The heat and work interactions between the system and surrounding is not change with time.||The heat and work interactions between the system and surrounding is change with time.|
|They continue indefinitely.||They start and stop over the finite time.|
|Eg: pumps, boilers, nozzle, heat exchanger, turbine, compressor etc||Eg: cooking gas cylinder, start and stop time of turbine, compressor etc.|
Control volume application
It can divides into four groups that is steady state work applications , steady state flow application, unsteady state work application, unsteady flow application. They can be describe below:
Steady state work applications
The device which produce and consumes the work is called steady state work applications. Some steady state work applications are given below:
- Turbine: it is device which converts the fluid energy into mechanical work. Energy equation for turbine is given below:
- Compressor: it is device which converts work to the fluid energy. Energy equation for compressor
- Pump: it converts work into potential energy of fluid. Energy equation for pump
- Fan: it converts the work into kinetic energy of fluid. The energy equation for the fan
Steady state flow application:
The device which are not produce and consume the work are called steady state flow application. Some application are given below:
- Nozzle and diffuser: the device which is used to increase the velocity of fluid is called nozzle. And device which is used to decrease the velocity is called diffuser. The equation for nozzle and diffuser are
- Heat exchanger, evaporator, condenser throttling valve: heat exchanger is used to transfer heat from one devices to another devices. Evaporator converts liquid to vapor by absorbing the heat from surrounding but condenser converts the vapor to liquid by rejecting the heat to the surrounding. And throttling valve reduces the pressure of fluid without performing work. The equation is given below for control volume A is
Unsteady state work applications:
Turbine, compressor ,pump etc are acts as unsteady state work application during the start up and shut down period. For unsteady state work applications as shown in figure below:
Then applying the mass and energy conservation equation for the unsteady state work application:
Unsteady state flow applications
Consider a cooking gas cylinder shown in fig during the cooking the food gas is consuming and mass of the system is decreasing but it does not produces the work.
Perpetual motion machine of the first kind is not possible:
Perpetual motion or continuously moving machine is not possible because of friction. Hence, we can not take continuous useful output effect without corresponding supply of input energy.