Answer to Question #238718 in Chemical Engineering for grace

Question

a crusher and a crusher are connected to the same electric drive power 2700 kg / h of the material pass successively through the crusher and then the crusher. Sieve particle size analysis of feed, crusher product and mill product respectively indicated specific areas of 2.9, 103 and 865 m2 / kg. Calculate the horsepower required to operate the crusher-crusher assembly, if the crusher's energy efficiency is 20% and the crusher's energy efficiency is 25%. The Rittinger's number for material is 77.4 m2 / kJ.

Answer 1

$K_R = 77.4 m² / kJ\\ f_c =2700 kg / h \\ A_1 = 2.9m^2 / kg ,\ L_1 = 1.9m/kg\\ A_2 =103m^2 / kg, \ L_2 = 11.5m/kg \\ A_3 = 865 m^2 / kg,\ L_3 = 33.2m/kg$

$E = K_Rf_c(\dfrac1{L_2}– \dfrac{1}{L_1})$

$E_1 = 77.4 × 2700(\dfrac1{1.9}-\dfrac1{11.5})= 91.8kJ$

$E_2 = 77.4× 2700(\dfrac1{11.5}-\dfrac1{33.2})= 11.89kJ$

for Crusher 1, with an efficiency of 20%,

$100×91.8×\dfrac1{20 } =459\textsf{ kW/(kg/hr)}= 0.171\textsf{ hp/(kg/s)}$

for Crusher 2, with an efficiency of 25%,

$100×11.89×\dfrac1{25}=47.56\textsf{ kW/(kg/hr)}= 0.018\textsf{ hp/(kg/s)}$