Calcular el balance inicial (dividir el % y multiplicar por los Kgs.
)Proteínas (20% / 100) 300 Kg x (0.
2) = 60 KgGrasas (10% / 100) 300 Kg x (0.
1) = 30 KgCarbohidratos (1 % / 100) 300 Kg x (0.
01) = 3 KgHumedad (68 % / 100) 300 Kg x (0.
68) = 204 KgCeniza (1 % / 100) 300 Kg x (0.
01) = 3 Kg ```````````````````````` `````````````````````````````````````````````````````````````` Total 100 % 300 KgCalcular el contenido final de humedad (X) : (X / total) x 100 = 10 ( X / total ) = 0, 1Calcularel balance finalProteínas 60 KgGrasas 30 KgCarbohidratos 3 Kg Humedad XCeniza 3 Kg ``````````````````````Total X + 96 Kg Calcular el contenido final de la humedadX / (X + 96) = 0, 1 despejar XX = 0, 1 (X + 96)X = 0, 1 X + 96 (agrupar X)X - 0.
1 X = 96 0, 9 X = 9.
6X = 9.
6 / 0, 9X = 10.
67 Kg Calcular masa de agua eliminada m de agua eliminada = (contenido inicial - contenido final ) humedadmasa de agua eliminada = 204 Kg - 10.
67 Kg = 193.
33KgCalcular masa total final = 10.
67 kg + 96 Kg = 106.
67 KgCalcular la composición finalProteínas (60 Kg / 106.
67) x 100 = 56, 24%Grasas (30 Kg / 106.
67) x 100 = 28.
12 %Carbohidratos (3 Kg / 106.
67) x 100 = 2.
81 %Humedad (10.
67 Kg / 106.
67 Kg) x 100 = 10 %Ceniza (3 Kg / 106.
67) x 100 = 2.
81 % ````````````````````````````````````````````````````````````````````````````````` total = 100 %.