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irga5000 [103]
3 years ago
9

A liquid phase chemical reaction (A → B) takes place in a well-stirred tank. The concentration of compound A in the feed is CA0

(mol/m3 ), and that in the outlet stream is CA (mol/m3 ). Neither concentration varies with time. The volume of the tank contents is V (m3) and the volumetric flow rate of the inlet and outlet streams is Ẇ (m3 /sec). The reaction rate (the rate at which A is consumed by reaction in the tank) is given by the expression r (mol A consumed/s) = kVCA wherek is a constant.
(a) Is this process continuous, batch, or semi batch? Is ittransient or steady - state?(b) What would you expect the reactant concentrationCA to equal if k = 0 (no reaction) ? What should itapproach if k [infinity] ( infinitely rapid reaction) ?(c) Write a differential balance on A, starting which terms inthe general balance equation ( accumulation = input + generation -output - consumption) you discarded and why you discarded them. Usethe balance to derive the following relation between the inlet andoutlet reactant concentrations :CA = CA0 / (1 + kV/ ). Verify that this relation predicts the results in part(b).
Engineering
1 answer:
SCORPION-xisa [38]3 years ago
4 0

Answer:

EH buddy use a sparkplug use a drill through a hose im from da bronx

Explanation:

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Consider a system with two tasks, Task1 and Task2. Task1 has a period of 200 ms, and Task2 has a period of 300 ms. All tasks ini
Murrr4er [49]

<u>Explanation:</u>

Task 1 time period = 200ms, Task 2 time period = 300ms

Task ticked = \frac{1000ms}{200ms}= 5  →  5 times

Task 2 ticked =\frac{1000ms}{300ms} = 3.33 → 3 times

At 600 ms → 200ms 200ms 200ms

                     300ms → \frac{30ms}{60ms}

Largest time period = H.C.M of (200ms, 300ms)

                                 = 600ms

4 0
2 years ago
10. What does a profile of a river from its headwaters to its mouth typically show?
irina1246 [14]

Answer:

c. an abrupt increase followed by a gradual decrease

Explanation:

At the headwater, the flow gradient starts high but then slowly decreases as the river moves downstream to its mouth.

3 0
2 years ago
All brake lights are dimmer than normal. Technician A says that bad bulbs could be the cause. Technician B says that high resist
yarga [219]

Answer:

All Brake lights are dimmer than normal because high resistance in the brake switch could be the cause according to Technician B.

Explanation:

According to Technician A

When the bulb is faulty then no current will flow through bulb and it will be open circuit.So no light will produce in bulb .

According to Technician B

When a high resistance inserted in series  circuit the voltage across each resistance is reduced and this cause the light glow dimly.

Formula of resistance in series circuit

Rt=r1+r2+r3......

5 0
3 years ago
4.2 A vapor compression refrigeration machine uses 30kW of electric power to produce 50 tons of cooling. What is
stellarik [79]

Answer:

5.833

Explanation:

Coefficient of Perfomance (COP) is the ratio of refrigeration effect to power input.

COP=\frac {RE}{P} where RE is refrigeration effect and P is power input

Here, the power input is given as 30 kW

We also know that 1 ton cooling is equivalent to 3.5 kW hence for 50 tons, RE=50*3.5=175 kW

Now the COP=\frac {175}{30}=5.833

6 0
3 years ago
A medium-sized jet has a 3.8-mm-diameter fuselage and a loaded mass of 85,000 kg. The drag on an airplane is primarily due to th
SCORPION-xisa [38]

Answer:

F_{thrust} ≅ 111 KN

Explanation:

Given that;

A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8

mass = 85,000 kg

drag co-efficient (C) = 0.37

(velocity (v)= 230 m/s

density (ρ) = 1.0 kg/m³

To calculate the thrust; we need to determine the relation of the drag force; which is given as:

F_{drag} = \frac{1}{2} × CρAv²

where;

ρ = density of air wind.

C = drag co-efficient

A = Area of the jet

v = velocity of the jet

From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0

SO, F_{drag}-F_{thrust} = 0

We can as well say:

F_{drag}= F_{thrust}

We can now replace F_{thrust} with F_{drag} in the above equation.

Therefore, F_{thrust} = \frac{1}{2} × CρAv²

The A which stands as the area of the jet is given by the formula:

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

We can now have a new equation after substituting our A into the previous equation as:

F_{thrust} = \frac{1}{2} × Cρ (\frac{\pi d^2}{4})v^2

Substituting our data from above; we have:

F_{thrust} = \frac{1}{2} × (0.37)(1.0kg/m^3)(\frac{\pi(3.8m)^2 }{4})(230m/s)^2

F_{thrust} = \frac{1}{8}   (0.37)(1.0kg/m^3)({\pi(3.8m)^2 })(230m/s)^2

F_{thrust} = 110,990N

F_{thrust}  in N (newton) to KN (kilo-newton) will be:

F_{thrust} = (110,990N)*\frac{1KN}{1,000N}

F_{thrust} = 110.990 KN

F_{thrust} ≅ 111 KN

In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.

5 0
3 years ago
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