Can a company actually actually become efficient when using Quality management Standards>?. (i.e. 9001:2015,)
An increase in efficiency cannot be achieved by just an examination and subsequent confirmation. In practice this is quite different. It means that if you are already certified with so many standards but this doesn't apply to the process, this is all useless.
A billet is a parallelepiped bar used as raw material for the production of reinforcing steel bars in Davao City Philippines. How much Energy is transferred if a block of Billet with a dimension of 150mm x 150mm x 12m is heated from 30°C to 1120°C considering the specific heat of the billet is 0.12 kcal/kg-°C and a density of 8,050 kg/m3?
First, we need to determine what is ask in the problem. In this case, the Energy is ask. Next, we need to determine what formula will be use. Q=mcp∆T. Q is the Energy transferred in Joules(J), m=is the mass in kg. , cp=is the specific heat and delta T (∆T) =is the change in temperature in °C. In the Given, Mass is not given, but density and Volume can be solve, and use the formul of density to find the mass. (density=mass/volume, so mass=density*volume) Volume=0.15*0.15*12 Volume=0.27m3 Density=8,050kg /m3 Mass=(0.27m3)*(8,050kg/m3) Mass=2173.5kg. After that, substitute all the values above. m=2173.5kg cp=0.12 kcal/kg-°C ∆T=Final Temp-Initial Temp Final Temp=1120°C Initial Temp=30°C ∆T=(1120-30) Q=mcp∆T Q=(2173.5kg)*( 0.12 kcal/kg-°C)*((1120°C-30°C) Q=284,293.8kcal - This is the Energy transferred in a block of Billet.
A combustion motor with a rated capacity of 146 kVA runs 24/7 at 90% power factor in a steel manufacturing plant in Davao City Philippines. Provided the motor runs at full rated capacity constantly, what will be the power consumption in peso(PHP) of the motor in one(1) month considering an 8 peso / Kw-hr electricity cost?
First, we have to determine what is ask in the question. -In the problem, it was ask the power consumption of Electricity in a Motor in a month. In this case, the given was in KVA, so we need to convert it to KW since the Electricity cost was in KW. Therefore, in the Power triangle, we need to multiply KVA with power factor KW=(KVA*p.f.) (KW=146*.9=131.4KW) Then, after getting the KW, multiply it to the time when the machine operates. Since the plant runs at 24/7 constantly, multiply KW to 24hrs. to get the Energy done.. (E=131.4*24=3153.6kw-hr) Lastly, multiply Energy to the cost / energy rate. Since the rate of Electricity is at 8PHP /Kw-hr, multiply 3153.6Kw-hr by 8.. (Cost=3153.6kw-hr*8php /kw-hr.) (Cost=25,228PHP) Therefore, the motor consumption is 25,228PHP in a day but the problem was asking the consumption monthly, so just multiply 25,228 PHP by 31 days. Monthly Consumption=25228 PHP * 31 = 782,092 PHP