Machine Parts

Procedure Here is the graphical output from tensile testing of five materials: Perspex, polypropylene, aluminium, brass and mild steel, along with the dimensions of the specimen. No values are given on the x axis, for extension, to simplify these questions. This is because the scales differ – except for aluminium and mild steel where they are identical. The load in Newtons is given on the y axis.   Use the graphs to answer the questions below. Questions 1.          Find the yield load for each specimen. 2.          Find the maximum load for each specimen. 3.          Which has the larger Young modulus – aluminium or mild steel? Hints 1.          The yield point is the point where the graph stops being linear – i.e. it passes from the elastic region, where load is proportional to extension, into the non-linear plastic region where additional load produces increasingly larger extensions. Mark this and measure off the corresponding load. 2.      

If a 60 Hz transformer is operated in 50 Hz, but at rated voltage, will it be hotter or cooler at full load?   We know , If a transformer rated in KW, it will be meaning less. Because   will change according to load. So transformer is rated in KVA. As seen, Cu loss of a transformer depends on current and   iron loss on voltage. Hence total transformer loss depends on volt-amp (VA) and not on phase angle between voltage and current i.e. it is independent of load power factor. That is why rating transformer is in KVA not in KW.             If rated voltage is applied but a frequency other than rated is used in a transformer, it is seen from equation, , that the flux must change inversely with the frequency change in order to maintain the same induced voltage. The core loss varies almost directly as the square of the flux and only directly with the frequency. A decrease in the supply Frequency will therefore require an increase in flux and consequently

Load connected with the transformer always varries i.e. it may be sometime resistive load or inductive load or capacitive load. As a result p.f. sometime unity, sometime lagging or leading and it is not equal. We know that KW=KVA*COSθ As p.f. varries so KW varries. So on the nameplate of   the transformer KW rating is not appropriate.Again cu loss of a transformer depends on current and iron loss on voltage. Hence, total transformer loss deoends on volt-ampre(VA) and not on phase angle berween voltage and current i.e. it is independent of load power factor. That is why rating of transformer is in KVA and not in KW.