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2 years ago

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Lab scale tests performed on a cell broth with a viscosity of 5cP gave a specific cake resistance of 1 x1011 cm/g and a negligib

le medium resistance. The cake solids (dry basis) per volume of filtrate was 20 g/liter. It is desired to operate a larger rotary vacuum filter (diameter 8 m and length 12 m) at a vacuum pressure of 80 kPA with a cake formation time of 20 s and a cycle time of 60 s. Determine the filtration rate in volumes/hr expected for the rotary vacuum filter.

1 answer:

insens350 [35]2 years ago

4 0

Answer:

5.118 m^3/hr

Explanation:

Given data:

viscosity of cell broth = 5cP

cake resistance = 1*1011 cm/g

dry basis per volume of filtrate = 20 g/liter

Diameter = 8m , Length = 12m

vacuum pressure = 80 kpa

cake formation time = 20 s

cycle time = 60 s

<u>Determine the filtration rate in volumes/hr expected fir the rotary vacuum filter</u>

attached below is a detailed solution of the question

Hence The filtration rate in volumes/hr expected for the rotary vacuum filter

V' = ( $\frac{60}{20}$ ) * 1706.0670

= 5118.201 liters ≈ 5.118 m^3/hr

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A reversible refrigeration cycle operates between cold and hot reservoirs at temperatures TC and TH, respectively. (a) If the co

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Answer:

a) $T_{H} = 1.967\,^{\circ}F$ , b) $COP_{R} = 9.105$ , c) $T_{H} = 115.934\,^{\circ}F$ , d) $COP_{R} = 6.995$ , e) $T_{H} = 25.129\,^{\circ}C$

Explanation:

a) The coefficient of performance of the reversible refrigeration cycle is:

$COP_{R} = \frac{T_{C}}{T_{H}-T_{C}}$

$10 = \frac{419.67\,R}{T_{H}-419.67\,R}$

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$10\cdot T_{H} - 4196.7 = 419.67$

$T_{H} = 461.637\,R$

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b) The coefficient of performance is:

$COP_{R} = \frac{273.15\,K}{303.15\,K-273.15\,K}$

$COP_{R} = 9.105$

c) The temperature of the hot reservoir can be determined with the help of the following relation:

$\frac{Q_{C}}{Q_{H}-Q_{C}} = \frac{T_{C}}{T_{H}-T_{C}}$

$\frac{500\,BTU}{600\,BTU-500\,BTU} = \frac{479.67\,R}{T_{H}-479.67\,R}$

$5 = \frac{479.67\,R}{T_{H}-479.67\,R}$

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$T_{H} = 115.934\,^{\circ}F$

d) The coefficient of performance is:

$COP_{R} = \frac{489.67\,R}{559.67\,R-489.67\,R}$

$COP_{R} = 6.995$

e) The temperature of the cold reservoir is:

$8.9 = \frac{268.15\,K}{T_{H}-268.15\,K}$

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$T_{H} = 298.279\,K$

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3 years ago

The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is

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Answer:

a) 0.697*10³ lb.in

b) 6.352 ksi

Explanation:

a)

For cylinder AB:

Let Length of AB = 12 in

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$\phi_B=\frac{T_{AB}L}{GJ}=\frac{T_{AB}*12}{3.3*10^6*0.1437} =2.53*10^{-5}T_{AB}$

For cylinder BC:

Let Length of BC = 18 in

$c=\frac{1}{2}d=\frac{1}{2} *2.2=1.1in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}1.1^4=2.2998\ in^4\\$

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$T_{AB}+T_{BC}-T=0\\T_{AB}+T_{BC}=T\\T_{AB}+T_{BC}=14*10^3\ lb.in\\but\ T_{BC}=19.0717T_{AB}\\T_{AB}+19.0717T_{AB}=14*10^3\\20.0717T_{AB}=14*10^3\\T_{AB}=0.697*10^3\ lb.in\\T_{BC}=13.302*10^3\ lb.in$

b) Maximum shear stress in BC

$\tau_{BC}=\frac{T_{BC}}{J}c=13.302*10^3*1.1/2.2998=6.352\ ksi$

Maximum shear stress in AB

$\tau_{AB}=\frac{T_{AB}}{J}c=0.697*10^3*0.55/0.1437=2.667\ ksi$

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Flip-flops are normally used for all of the following applications, except logic gates.

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Flip flops are known to be tools that are used for counting. They come in different ranges.

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Hello, I have a question, I would be glad if you can help.

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Answer:

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