 Myself, Mr. Akshay Kumar Suvde, Assistant Professor, Mechanical Engineering Department. Today, we will study Harnel Governor, Learning Outcome. At the end of this session, student will be able to derive and determine the stiffness of the spring, centrifugal force at minimum and maximum as well as intermediate position of the balls. So, Harnel Governor is a spring-controlled governor as shown in the figure. It consists of two bell crank levers, pivoted at 0.00 and attached to the frame. The frame is fitted on the spindle of the governor and the frame also rotates with the spindle. Each lever carries a ball at the end of vertical arm OB as well as a roller at the end of the horizontal arm. A helical compression spring is provided on the spindle which will provide an equal downward force on the roller through the collar on the sleeve. This compression of the spring can be adjusted by tightening or loosening the nut at the top of the spindle. Now, consider the minimum and maximum position of the Harnel Governor, where small m is the mass of the ball in kg, capital letter m for mass of the sleeve in kg, r 1 and r 2, minimum and maximum radius of rotation of the ball in meters, omega 1 and omega 2, angular speed of the governor at minimum and maximum radius in radians per second respectively. S1 and S2 spring force exerted on the sleeve at omega 1 and omega 2 in Newton respectively, f c 1 and f c 2 centrifugal force at a speed of omega 1 and omega 2 respectively. Small letter S represents the stiffness of the spring or the force required to compress the spring in by 1 mm. S is the length of the vertical arm of the ball arm of the lever and y is the length of the horizontal or sleeve arm of the lever in meter. R is the distance of the fulcrum, O from the axis of the governor or the radius of rotation of the governor is in mid position in meter. Now, we will consider the two positions that is as shown in the figure minimum position and maximum position. So, consider the forces acting on one bell crank lever of the Harnel Governor. So, let us consider H is the compression of the spring due to change in the radius of rotation from R1 to R2. So, let us consider the minimum position of the governor. So, when the radius of rotation changes from R2 to R1, what will be the compression of the spring that we will determine. So, H1 is the compression due to the change in radius of rotation R2, R1. Hence from the figure A, we can calculate the compression of the spring H1 divided by y is equal to A1 divided by x. So, A1 is nothing but R minus R1 and therefore, H1 upon y is equal to R minus R1 divided by x and therefore, compression of the spring at minimum position it is R minus R1 divided by x into y. This is your equation number 2. Similarly, for the maximum speed position, the radius of rotation changes from R to R2 where the compression of the spring is represented by H2 and therefore, for maximum position H2 divided by y is equal to A2 divided by x and therefore, H2 divided by y is equal to A2 is R2 minus R1. So, R2 minus R divided by x and therefore, H2 is equal to R2 minus R divided by x into y. So, this is your equation number 2. So, the total compression of the spring is nothing but summation of equation 1 and 2. So, H is equal to H1 plus H2 and therefore, H is equal to R2 minus R1 into y by x. This is equation number 3. Now, we will take the moment about the pivot point of the fulcrum where y1, x1 these are the obliquity effect of the arms of the bell crank lever. So, taking the moment about point O, we will get mg plus S1 taking the moment about point O mg plus S1 divided by 2 into perpendicular distance y1 then mg into A1. So, mg into A1 is equal to centrifugal force acting on the fly ball fc1 into x1. So, mg plus S1 divided by 2 is equal to fc1 into x1 minus mg into A1. So, I can rewrite this equation mg plus S1 is equal to 2 by y1 in bracket fc1 into x1 minus mg into A1. This is my equation number 4 and similarly taking the moment about point O at maximum position I will get mg plus S2 divided by 2 into y2 is equal to fc2 into x2 minus plus mg into A2 and therefore, mg plus S2 is equal to 2 by y2 into fc2 into x2 plus mg into S2 minus S1 is equal to 2 divided by y2 into fc2 multiplied by x2 plus mg into A2 minus 2 upon y1 into fc1 into x1 minus mg into A1. Now, as we know the stiffness of the spring is defined as it is the spring force required to compress the spring by 1 mm and therefore, S2 minus S1 is equal to S into H and therefore, stiffness of the spring is difference between the spring force divided by the compression of spring and hence forth it is equal to S2 minus S1 divided by R2 minus R1 multiplied by x upon y. So, this is your equation number 5. Now, if the obliquity effect of the arms of the bell crank lever is not considered or neglected then x1 is equal to x2 is equal to x, y1 is equal to y2 is equal to y and moment due to the weight of the ball is also neglected then taking the moment about the pivot point O at minimum position then mg plus S1 divided by 2 into y is equal to fc1 into x and therefore, mg plus S1 is equal to 2 times fc1 into x divided by y. Therefore, for maximum position taking the moment about point O mg plus S2 divided by 2 into y is equal to fc2 into x and hence forth mg plus S2 is equal to 2 times fc2 into x divided by y. So, from equation 6 and equation 7 if we deduct equation 7 7 minus 6 therefore, S2 minus S1 is equal to 2 times fc2 minus 2 times fc1 into x upon y and therefore, you will get 2 into fc2 minus fc1 into x upon y and therefore, stiffness of the spring small s is equal to S2 minus S1 divided by h. Hence, S2 minus S1 that will be S2 minus S1 divided by R2 minus R1 into x upon y. So, putting the value of S2 minus S1 you will get S is equal to 2 times fc2 minus fc1 divided by R2 minus R1 into x upon y bracket square. This is your equation number 8. References the references are taken from theory of machines by R. S. Kurmi. Thank you.