Kirchhoff's equation is used in thermodynamics to calculate the increase in enthalpy at different temperatures, since the change in enthalpy does not occur constantly in higher temperature intervals. The German physicist Gustav Robert Kirchhoff was the forerunner of this equation in which he contributed in the scientific field of electrical circuits.
Kirchhoff equation
It starts from the representation of ΔHr and proceeds in relation to the temperature at constant pressure and it results as follows:
But:
So:
If the pressure is constant, we can place the previous equation with total derivatives, and it results like this:
If reordered:
What integrating:
That is to say:
Kirchhoff's laws are two equalities that are based on the preservation of energy and the charge of electrical circuits. These laws are:
- Kirchhoff's first or node law is understood as Kirchhoff's current law and his article describes that, if the algebraic sum of the currents entering or leaving a node is equal to zero at all times. That is, in any node, the sum of all the nodes plus the currents that enter the node is not equal to the sum of the currents that leave.
I = 0 at any node.
- Kirchhoff's second law is understood as the law of voltages, Kirchhoff's law of loops or meshes and his article describes that, if the algebraic sum of the voltages around any loop (closed path) in a circuit, is equal to zero at all times. In every mesh the sum of all the voltage drops is similar to the total voltage supplied, in an equitable way. In every mesh, the algebraic sum of differences in electrical power is equal to zero.
(I.R) on the resistors is zero.
V = 0 in any mesh of the network
For example:
A direction of circulation is selected to circulate in the meshes. It is suggested that they circulate the mesh in a clockwise direction.
If the resistance comes out negative it is considered positive. In generators the electromotive forces (emf) are considered positive when a mesh circulates in the direction of travel that was selected, the negative pole is found first and then the positive pole. If the opposite occurs, the electromotive forces are negative.
M1: 6 (I1 - I2) + 10 (I1 - I 3) - 7 + 7I1 = 0
M2: -4 + (I2) - 6 (I1 - I2) = 0
M3: 1/3 - 25 - 10 (I1 - I3) = 0
Each mesh is solved to obtain the respective equations:
M1: 6I1 - 6I2 + 10I1 - 10I3 - 7 + 7I1 = 0 23I1 - 6I2 - 10I3 = 7 (Equation 1)
M2: -4 + 5I2 - 6I1 + 6I2 = 0 -6I1 + 11I2 = 4 (Equation 2)
M3: 1I3 - 25 - 10I2 + 10I3 = 0 -10I1 + 11I3 = 25 (Equation 3)
