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lord [1]
3 years ago
8

In the formula shown, k is a correction factor for nonideal mixing. In the worst case, k is usually estimated to be:_______. Cpp

m = KMRT/KQVPM * 10^6 (Select the best answer and then click 'Submit.) a. 0b. 0.1c. 0.5d. 1.0
Engineering
1 answer:
LuckyWell [14K]3 years ago
3 0

Answer:

d. 1.0

Explanation:

Correlation identifies the relationship between two variables. In the given scenario there is strong relation between non ideal mixing. The correction factor can be between -1 to 1 depending on the intensity of the relationship and dependency. The non ideal mixing efficiency is highly dependent on the factors that govern it this means there is high intensity relation so the k is estimated to be nearly 1.

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A resistance of 30 ohms is placed in a circuit with a 90 volt battery. What current flows in the circuit?
blagie [28]

Answer:

3A

Explanation:

Using Ohms law U=I×R solve for I by I=U/R

4 0
3 years ago
Air flows through a device such that the stagnation pressure is 0.4 MPa, the stagnation temperature is 400°C, and the velocity i
RoseWind [281]

To solve this problem it is necessary to apply the concepts related to temperature stagnation and adiabatic pressure in a system.

The stagnation temperature can be defined as

T_0 = T+\frac{V^2}{2c_p}

Where

T = Static temperature

V = Velocity of Fluid

c_p = Specific Heat

Re-arrange to find the static temperature we have that

T = T_0 - \frac{V^2}{2c_p}

T = 673.15-(\frac{528}{2*1.005})(\frac{1}{1000})

T = 672.88K

Now the pressure of helium by using the Adiabatic pressure temperature is

P = P_0 (\frac{T}{T_0})^{k/(k-1)}

Where,

P_0= Stagnation pressure of the fluid

k = Specific heat ratio

Replacing we have that

P = 0.4 (\frac{672.88}{673.15})^{1.4/(1.4-1)}

P = 0.399Mpa

Therefore the static temperature of air at given conditions is 72.88K and the static pressure is 0.399Mpa

<em>Note: I took the exactly temperature of 400 ° C the equivalent of 673.15K. The approach given in the 600K statement could be inaccurate.</em>

3 0
3 years ago
Steel bar A of 5 mm diameter is subject to a stress of 500 MPa. Aluminum bar B of 10 mm diameter is subject to a stress of 150 M
weqwewe [10]

Answer:

aluminum bar carrying a higher load than steel bar

Explanation:

Given data;

steel abr

diameter = 5 mm

stress = 500 MPa

aluminium bar

diameter = 10 mm

stress = 150 MPa

we know

stress = laod/area

for steel bar

500 = \frac{P}{\frac{\pi}{4} 5^2}

solving for P

P = 9817.47 N

for Aluminium bar

150 = \frac{P}{\frac{\pi}{4} 10^2}

solving for P

P = 11790 N

aluminum bar carrying a higher load than steel bar

6 0
3 years ago
Study the graphs below what type of trend and pattern in observed .
hichkok12 [17]

Answer:

negative and linear

Explanation:

its a straight line going down

8 0
3 years ago
Read 2 more answers
(2 points) A perfectly mixed aeration pond with no recycle serves as the biological reactor for a small community. The pond rece
FromTheMoon [43]

Answer:

a)  t = 165 days

b) 9.9 kg/day

Explanation:

Given data:

final lelvel of BOD is 20 mg/l

Ks = 100 mg BOD/L,

kd = 0.10 day-1 ,

μm = 1.6 day-1 ,

Y = 0.60 mg SS/mg BOD

a) we know that

c_{out} = \frac{c_{in}}{1 + kt}

20 = \frac{350}{1 + 0.1 t}

solving for t

t  =  165 days

b) mass of microbes = Q(mlD) × C(mg/l)

= 30\times 10^{-3} (350-20) = 9.9 kg/day

3 0
3 years ago
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