 Hello students, myself Ganesh Biajlawe working as an assistant professor in Department of Mechanical Engineering, Walchain Institute of Technology, Singapore. In this fifth session of extended surfaces or fins, we will see few numericals. Learning outcome. At the end of this session, students will be able to evaluate fin performance. Now, we will see first numerical. The statement is, it is required to heat oil to about 300 degree Celsius for frying purpose. A ladle is used in the frying. The section of the handle is 5 mm by 18 mm. The surrounding are at 30 degree Celsius. The conductivity of the material is 205 watt per meter Kelvin. If the temperature at a distance of 380 mm from the oil should not reach 40 degree Celsius. Determine the convective heat transfer coefficient. Now, as per the statement, we should see the given data. A ladle is given, so it is required to heat oil to about 300 degree Celsius for frying purpose. A ladle is used in the frying. The section of the handle is 5 mm by 18 mm. The surrounding temperature T infinity temperature is 30 degree Celsius. The conductivity of the material is 205 watt per meter Kelvin. If the temperature at a distance, so suppose this is the origin and consider this as 380 mm distance and the given temperature is 40 degree Celsius, 40 degree Celsius. Then they are asked to find the convective heat transfer coefficient, convective heat transfer coefficient. Now, we can use the condition of, we can use the equation theta by theta naught is equal to e raise to minus mx. Now, this theta is nothing but T minus T infinity by T naught minus T infinity, which is equal to e raise to m and x is 0.380 meter. Now, here we can substitute T T infinity, T naught T infinity. Then we will get m as we know m is nothing but square root of HP by KA and this value we will get as 8.673. Now, we can calculate perimeter area, KA is also given. So, after substituting the values, we will get the edge as 30.17 watt per meter Kelvin. So, this is the convective heat transfer coefficient, this is the answer. Now, we will move to the next number, the statement is a pinfin 2.5 mm diameter is made of copper, conductivity is given 396 watt per meter Kelvin. It protrudes from a wall maintained at 95 degree Celsius and placed in 25 degree Celsius air. The convective heat transfer coefficient over the fin is 10 watt per meter square Kelvin. Calculate the heat loss for fin for infinite length. Now, we will write the given data, a pinfin of diameter 2.5 mm protrudes from the wall, which is maintained means base temperature becomes 95 degree Celsius and it is placed in the surrounding where temperature is 25 degree Celsius air. The convective heat transfer coefficient between the fin and the surrounding is 10 watt per meter square Kelvin. Now, they asked to find the heat loss, infinite length of the fin, this second case we required to consider. Now, first we will find out the m value, which is equal to square root of H p k a, H is given perimeter pi by 4 d square into L, a pydial perimeter is pydial and air is pi by 4 d square k conductivity is 396, we will obtain m as 6.36, then as we know the rate of heat conduction through the infinite length fin. In last sessions, we have derived the equation, so q is equal to square root of H p k a theta naught, which is equal to square root of H p k a theta naught is a t naught minus t infinity, which is equal to H given is 10, this is pi d L into k is 396 into air is pi by 4 d square. So, this is H p k a, t naught is 95 minus 25, sorry here only pi d, pi d. So, if you substitute here, then you do the now calculation, you will get the answer as 0.861 watt, depending upon the given cases, the rate of heat transfer formulae changes, then the temperature distribution relations also changes. For given numerical, the material given is copper, so we use the conductivity of the copper, given conductivity, we can analyze the problem, same problem we can analyze for different materials. For example, if the material is copper, we got the answer as 0.61, you can alter it, you change the material instead of copper, instead of copper you take the aluminum, find out the rate of heat transfer, for gold, for silver etcetera, you can find out the cube, then this is the homework, you find out, you can plot the rate of heat transfer for different materials. So, you find out the curve, or you can plot the histogram, and from that you can select that particular material, depending upon the budget available for that particular application. For further study, you can refer fundamentals of heat and mass transfer by Incropera David. Thank you.