Machine Parts

The ordinary or commercial efficiency of a transformer is given by the ratio –(output in watts)/(Input in watts). But there are certain type of transformer whose performance can not judged by this efficiency. Distribution transformers have their primaries energized all the 24 hours,although their secondaries supply little or no load most of the time during the day except the   housing lighting period.It means that whereas core loss occurs only when the transformers are loaded.This is why the transformer’s all day efficiency is always less than the commercial efficiency.

The ordinary or commercial efficiency of a transformer is given by the ratio                          Output in watts                         Input in watts Again, for all-day efficiency, ¶=Output in Kwh/input in Kwh (For 24 hours) Whereas core loss occurs throughout the day, the Cu loss occurs only when the transformers are loaded. Hence it is considered a good pracice to design such transformers so that Core loss is very low. The Cu losses are relatively less important, because they depend on the load. The performance of such a transformer should be judged by all day efficiency. Math: The energy output is 5*08*6=24 Kw=hr (operates at rated KVA)                                         ½*5*.5*12=15 Kw-hr (one-half rated KVA)   Total Output = 39 KW-hr The energy loss due to core loss is 35*24=840 W-hr=0.84 Kw-hr The copper loss is    40*(1)2=40*6=0240 Kw-hr                               ¼*40*42=120 Kw-hr =0.120 Kw-hr Total Copper energy loss = (0

(I).    High load carrying capacity. (II). High rigity. (III).                       Longer life (IV).                      Easy heat removal by oil leakage. (V).                         No variation of friction force because there is no metal to metal contact. (VI).                      Dust and metallic microelements are flashed away by oil forces. <![endif]--> Determine the actual diameter for drawing   a cup of 2- inch diameter having 2- inch depth and corner radius 1/6 – inch. If the material is 1/32 –inch thick also determine radius of Punch and Die.        Solution : Given data, d=2˝, h=2˝, r=1/10˝, t=1/32˝ The ratio   =   =32>20 D= 2 +4dh) = 2 +4*2*2) =4.47 ˝ According to the general rule for adding 1/8 to the blank diameter for each inch of cup diameter 2*1/8 =1/4  So, D=4.47+0.25=4.72˝ The radius is 4 times of the material th