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kipiarov [429]
2 years ago
6

A suspension of Bacillus subtilis cells is filtered under constant pressure for recovery of protease. A pilot-scale filter is us

ed to measure filtration properties. The filter area is 0.25 m2, the pressure drop is 360 mmHg, and the filtrate viscosity is 4.0 cP. The cell suspension deposits 22 g cake per liter of filtrate. The following data are measured: (Hints: Change the unit to standard units)
Time (min)​​ 2 ​ 3​ 6 ​ 10 ​15 ​20

Filtrate volume

(L)

10.8 12.1 18.0 21.8 28.4 32

a) Determine the specific cake resistance and filter medium resistance.
b) What size filter is required to process 4000 L cell suspension in 30 min at a pressure drop of 360 mm Hg?

Engineering
1 answer:
sattari [20]2 years ago
7 0

The answer & explanation for this question is given in the attachment below.

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As the porosity of a refractory ceramic brick increases:
ivolga24 [154]

Answer:

A) strength decreases, chemical resistance decreases, and thermal insulation increases

Explanation:

Strength always decreases, chemical resistence decreases, and thermal condictivity must be reduced therefore themal insulation must increase.

7 0
2 years ago
The 8-mm-thick bottom of a 220-mm-diameter pan may be made from aluminum (k = 240 W/m ⋅ K) or copper (k = 390 W/m ⋅ K). When use
Artemon [7]

Answer:

For aluminum 110.53 C

For copper 110.32 C

Explanation:

Heat transmission through a plate (considering it as an infinite plate, as in omitting the effects at the borders) follows this equation:

q = \frac{k * A * (th - tc)}{d}

Where

q: heat transferred

k: conduction coeficient

A: surface area

th: hot temperature

tc: cold temperature

d: thickness of the plate

Rearranging the terms:

d * q = k * A * (th - tc)

\frac{d * q}{k * A} = th - tc

th = \frac{d * q}{k * A} + tc

The surface area is:

A = \frac{\pi * d^2}{4}

A = \frac{\pi * 0.22^2}{4} = 0.038 m^2

If the pan is aluminum:

th = \frac{0.008 * 600}{240 * 0.038} + 110 = 110.53 C

If the pan is copper:

th = \frac{0.008 * 600}{390 * 0.038} + 110 = 110.32 C

7 0
2 years ago
Instructions given by traffic police or construction flaggers _____. A. Are sometimes important to follow B. Are usually not imp
Anestetic [448]

Answer:

D. Overrule any other laws and traffic control devices.

Explanation:

Laws and traffic control devices are undoubtedly compulsory to be followed at every point in time to control traffic and other related situations. However, there are cases when certain instructions overrule these laws and traffic control devices. For example, when a traffic police is giving instructions, and though the traffic control devices too (such as traffic lights) are displaying their own preset lights to control some traffic, the instructions from the traffic police take more priority. This is because at that point in time, the instructions from the traffic control devices might not be just applicable or sufficient.

Also, in the case of instructions given by construction flaggers, these instructions have priority over those from controlling devices. This is because during construction traffic controls are redirected from the norms. Therefore, the flaggers such be given more importance.

6 0
2 years ago
4. Two technicians are discussing the evaporative emission monitor. Technician A says that serious monitor faults cause a blinki
snow_lady [41]

Answer:

The correct option is;

Neither Technician A nor B

Explanation:

The evaporative emission monitor or Evaporaive Emission Control System EVAP System monitors enables the Power Control Module of the car to check fuel system leak integrity and the vapor consumption efficiency during engine combustion

It is a requirement of EPA on cars to check the emission of smug forming evaporates from cars

Serious monitor faults can cause the turning on of the check engine lights and the vehicle will not pass OBD II test, but it will not lead to engine shutdown

It runs when the engine is 15 to 85% full and the TP sensor is between 9% and 35%.

Therefore, the correct option is that neither Technician A nor B are correct.

3 0
2 years ago
2.5 kg of air at 150 kPa and 12°C is contained in a gas-tight, frictionless piston-cylinder device. The air is now compressed to
Korolek [52]

Answer:

Work input =283.47 KJ

Explanation:

Given that

P_1=150\ KPa

P_2=600\ KPa

T=12°C=285 K

m= 2.5 kg

Given that this is the constant temperature process.

e know that work for isothermal process  

W=P_1V_1\ln \dfrac{P_1}{P_2}

W=mRT\ln \dfrac{P_1}{P_2}

So now putting the values

W=mRT\ln \dfrac{P_1}{P_2}

W=2.5\times 0.287\times 285\ln \dfrac{150}{600}

W=-283.47 KJ

Negative sign indicates that work is done on the system.

So work input =283.47 KJ

8 0
2 years ago
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