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Usimov [2.4K]
3 years ago
11

Para conseguir jugo de naranja concentrada, se parte de un extracto con 7% en peso de sólidos el cual se mete a un evaporador al

vacío. En el evaporador se elimina el agua necesaria para que el jugo salga con una concentración del 60% de peso de sólidos. Si se introducen al proceso 1000 kg/h de jugo diluido, calcule la cantidad de agua evaporada y de jugo concentrado saliente.
Engineering
1 answer:
padilas [110]3 years ago
4 0

Answer:

Se obtendrán 116.66 litros de jugo concentrado, y el agua evaporada será por un total de 883.33 litros.

Explanation:

Dado que para conseguir jugo de naranja concentrada, se parte de un extracto con 7% en peso de sólidos el cual se mete a un evaporador al vacío, y en el evaporador se elimina el agua necesaria para que el jugo salga con una concentración del 60% de peso de sólidos, si se introducen al proceso 1000 kg/h de jugo diluido, para calcular la cantidad de agua evaporada y de jugo concentrado saliente se debe realizar el siguiente cálculo;

1000 x 0.07 = 70

60 = 70

100 = X

100 x 70 / 60 = X

7000 / 60 = X

116.66 = X

Por lo tanto, se obtendrán 116.66 litros de jugo concentrado, y el agua evaporada será por un total de 883.33 litros.

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  150 W

Explanation:

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3 years ago
Find the phasor form of:
mixer [17]

Answer:

I=24.598\angle 50.377

Explanation:

A tension or current expressed in cosine form and with a positive sign can be converted directly into a phasor. This is done by indicating the tension and the offset angle:

Acos(10\omega t +\phi)=A\angle \phi

So:

i(t)=10cos(10t+63)+15cos(10t-42)=10\angle 63 + 15\angle42=I

You can sum the phasors simply using a calculator, however, let's do it manually:

Let's find the rectangular form of each phasor using the next formulas:

A=\sqrt{a^2+b^2} \\\phi=arctan(\frac{b}{a})

For 10\angle 63

63=arctan(\frac{b}{a} )\\\\tan(63)=\frac{b}{a} \\\\b=a*tan(63)

10=\sqrt{a^2+(a*tan(63))^2} \\\\10^2=a^2+a^2*(tan(63))^2\\\\Solving\hspace{3}for\hspace{3}a\\\\a=\sqrt{\frac{100}{1+tan(63)^2} } =4.539904997\\\\and\hspace{3}b\\b=\sqrt{100-a^2} =8.910065242

So:

Z_1=a+bj=4.539904997+8.910065242j

For 15\angle 42

42=arctan(\frac{b_2}{a_2} )\\\\tan(42)=\frac{b_2}{a_2} \\\\b_2=a_2*tan(42)

15=\sqrt{a_2^2+(a_2*tan(42))^2} \\\\15^2=a_2^2+a_2^2*(tan(42))^2\\\\Solving\hspace{3}for\hspace{3}a_2\\\\a_2=\sqrt{\frac{225}{1+tan(42)^2} } =11.14717238\\\\and\hspace{3}b_2\\\\b_2=\sqrt{225-a^2} =10.0369591

So:

Z_2=a_2+b_2j=11.14717238+10.0369591j

Hence:

Z_T=Z_1+Z_2=(4.539904997+11.14717238)+(8.910065242+10.0369591)j\\Z_T=15.68707738+18.94702434j

Finally:

I=\sqrt{15.68707738^2+18.94702434^2} \angle arctan (\frac{18.94702434}{15.68707738} )=24.598\angle 50.377

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