I'm not to sure but try soy
Spectrophotometric cell or a cuvette is made of quartz for UV spectrophotometers. These cuvettes are used as sample holders for the spectrophotometric determination of the analytes. The material that makes up the cuvette and the condition of the cuvette is to be taken care of in order to avoid erroneous absorbance readings. The sample holder or the cuvette must be removed from the spectrophotometer in between two successive readings. This is to ensure that the light sensing detector of the instrument is not affected.
Answer:
1) The power needed to process 50 ton/hr is 135.4 HP.
2) The void fraction of the bed is 0.37.
Explanation:
1) For this type of milling operations, we can estimate the power needed for an operation according to the work index (Ei), the passing size of the circuit feed (F80) and the passing size of the product (P80).
We assume the units of Ei are kWh/t.
The equation that relates this parameters and the power is (size of particles in μm):
![W=Ei*(\frac{10}{\sqrt{P80}} -\frac{10}{\sqrt{F80}} )\\\\W=9.45*(\frac{10}{\sqrt{3175\mu m}} -\frac{10}{\sqrt{76200mm}} )\\\\\\W=9.45*(0.1774+0.0362)=2.019 kWh/t](https://tex.z-dn.net/?f=W%3DEi%2A%28%5Cfrac%7B10%7D%7B%5Csqrt%7BP80%7D%7D%20-%5Cfrac%7B10%7D%7B%5Csqrt%7BF80%7D%7D%20%29%5C%5C%5C%5CW%3D9.45%2A%28%5Cfrac%7B10%7D%7B%5Csqrt%7B3175%5Cmu%20m%7D%7D%20-%5Cfrac%7B10%7D%7B%5Csqrt%7B76200mm%7D%7D%20%29%5C%5C%5C%5C%5C%5CW%3D9.45%2A%280.1774%2B0.0362%29%3D2.019%20kWh%2Ft)
The power needed to process 50 ton/hor is
![P=2.0194\frac{kWh}{Ton}*\frac{50Ton}{h}*\frac{1.341HP}{1kW}= 135.4 \, HP](https://tex.z-dn.net/?f=P%3D2.0194%5Cfrac%7BkWh%7D%7BTon%7D%2A%5Cfrac%7B50Ton%7D%7Bh%7D%2A%5Cfrac%7B1.341HP%7D%7B1kW%7D%3D%20%20%20135.4%20%5C%2C%20HP)
2) The density of the packed bed can be expressed as
![\rho=f_v*\rho_v+f_s*\rho_s](https://tex.z-dn.net/?f=%5Crho%3Df_v%2A%5Crho_v%2Bf_s%2A%5Crho_s)
being f the fraction and ρ the density of every fraction. We know that the density of the void is 0 (ρv=0) and that fv=1-fs (the sum of the fractions ois equal to the total space).
Then we can rearrange
![\rho=f_v*\rho_v+f_s*\rho_s\\\\\rho=f_v*0+(1-f_v)*\rho_s\\\\\rho/\rho_s=1-f_v\\\\f_v=1-\rho/\rho_s=1-990/1570=1-0.63=0.37](https://tex.z-dn.net/?f=%5Crho%3Df_v%2A%5Crho_v%2Bf_s%2A%5Crho_s%5C%5C%5C%5C%5Crho%3Df_v%2A0%2B%281-f_v%29%2A%5Crho_s%5C%5C%5C%5C%5Crho%2F%5Crho_s%3D1-f_v%5C%5C%5C%5Cf_v%3D1-%5Crho%2F%5Crho_s%3D1-990%2F1570%3D1-0.63%3D0.37)
The void fraction of the bed is 0.37.
Answer:
divide the volume value by 1000
So 3828/1000=3.828
True.
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