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Margarita [4]
3 years ago
13

An equation used to evaluate vacuum filtration is Q = ΔpA2 α(VRw + ARf) , Where Q ≐ L3/T is the filtrate volume flow rate, Δp ≐

F/L2 the vacuum pressure differential, A ≐ L2 the filter area, α the filtrate "viscosity," V ≐ L3 the filtrate volume, R ≐ L/F the sludge specific resistance, w ≐ F/L3 the weight of dry sludge per unit volume of filtrate, and Rf the specific resistance of the filter medium. What are the dimensions of Rf and and α?

Engineering
1 answer:
larisa86 [58]3 years ago
6 0

Answer:

Explanation:

The explanations and answers are shown in the following attachments

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A fatigue test is performed on 69 rotating specimens made of 5160H steel. The measured number of cycles to failure (L in kcycles
tensa zangetsu [6.8K]

Answer:

(a) Mean = 122.9, σ = 30.071

(b) No. of failed specimens at less than 115k cycles are 27.

(c) μ = 39.07

Explanation:

We are given:

L  60  70  80  90  100  110  120  130  140  150  160  170  180  190  200  210

f    2     1    3     5     8     12     6     10     8     5     2      3      2      1       0      1

(a) First we need to calculate the mean and standard deviation. The formula for calculating mean is:

Mean = ∑fx/∑f

And for standard deviation we have:

S.D. = √Var

Var = ∑fx²/∑f - (Mean)²

∑fx = (2*60) + (1*70) + (3*80) + (5*90) + (8*100) + (12*110) + (6*120) + (10*130) + (8*140) + (5*150) + (2*160) + (3*170) + (2*180) + (1*190) + (0*200) + (1*210)

         = 120 + 70 + 240 + 450 + 800 + 1320 + 720 + 1300 + 1120 + 750 + 320 + 510 + 360 + 190 + 0 + 210

∑fx = 8480

Mean = ∑fx/∑f

          = 8480/69

Mean = 122.9  

∑fx² = (2*60²) + (1*70²) + (3*80²) + (5*90²) + (8*100²) + (12*110²) + (6*120²) + (10*130²) + (8*140²) + (5*150²) + (2*160²) + (3*170²) + (2*180²) + (1*190²) + (0*200²) + (1*210²)

   =7200+4900+19200+40500+80000+145200+86400+169000+156800+112500+51200+86700+64800+36100+0+44100

∑fx² = 1104600

Var = ∑fx²/∑f - (Mean)²

     = 1104600/69 - (122.9)²

     = 16008.69565 - 15104.41

Var = 904.2856

S.D = √Var

σ = √904.2856

σ = 30.071

(b) Let X be the number of failed specimen.

We will use the z-score to calculate the probability. The formula for z-score is:

z = (X-μ)/σ

P(X<115) = P(z<(115-122.9)/30.071)

              = P(z<-0.26)

Using the normal distribution probability table, we can compute the value of  P(z<-0.26).

P(X<115) = 0.3974

So, no. of failed specimens at less than 115k cycles are: 0.3974*69 = 27 specimens

(c) σ = 30.071

P(x<115) = 0.99

P(z<(115-μ)/30.071) = 0.99

From the normal distribution table we find that 0.99 lies between the z values 2.52 and 2.33. Hence, we get 2.525 as the z-value at which the probability is 0.99.

z = (x-μ)/σ

2.525 = (115 - μ)/30.071

75.93 = 115 - μ

μ = 115 - 75.93

μ = 39.07

4 0
3 years ago
Hi im ***ar and im doing sculptural but what should it be about star wars or Marvel
Anarel [89]

Answer:

Cool I think u should do Marvel first

4 0
3 years ago
A hub a signal that refreshes the signal strength.
vekshin1

Answer:You are a network engineer. While moving a handheld wireless LAN device, you notice that the signal strength increases when the device is moved from a ...

Explanation:

7 0
2 years ago
Six forces act on a beam that forms part of a building's
IrinaVladis [17]

Answer:

<h2> FA = 13 kN </h2><h2>FG = 15.3 kN</h2>

Explanation:

write each force in terms of magnitude and directions  

Fx = F sin Ф

Fy = F cos Ф

where Ф is to be measured from x axis.

∑F at y = o

FAy + FBy + FCy + FDy + FEy + FGy = 0

∑F at x = o

FAx + FBx + FCx + FDx + FEx + FGx = 0

Let  

FA = FA sin (110)   +   FA cos (110)

FB = 20 sin (270)  +  20 cos (270)

FC = 16 sin (140)    +  16 cos (140)

FD = 9 sin (40)       +  9 cos (40)

FE = 20 sin (270)    +  20 cos (270)

FG = FG sin (50)     +  FG cos (50)

add x and y forces:

FAx + FBx + FCx + FDx + FEx + FGx = 0

FAy + FBy + FCy + FDy + FEy + FGy = 0

FA sin (110)  + 0  + 16 sin (140)  + 9 sin (40)  + 0   + FG sin (50) = 0

FA cos (110) - 20 + 16 cos (140) + 9 cos (40) - 20 + FG cos (50 = 0

FA sin (110)  + 0  + 10.285  + 5.785  + 0   + FG sin (50) = 0

FA cos (110) - 20 - 12.257 + 6.894 - 20 + FG cos (50) = 0

FA sin (110)  + 16.070 + FG sin (50) = 0        

FA cos (110) - 45.363 + FG cos (50) = 0

solving for FA, and FG

FA = 13 kN

FG = 15.3 kN

7 0
3 years ago
A traffic flow has density 61 veh/km when the speed is 59 veh/hr. If a flow has a jam density of 122 veh/km, what is the maximum
antoniya [11.8K]

Since this traffic flow has a jam density of 122 veh/km, the maximum flow is equal to 3,599 veh/hr.

<u>Given the following data:</u>

  • Density = 61 veh/km.
  • Speed = 59 km/hr.
  • Jam density = 122 veh/km.

<h3>How to calculate the maximum flow.</h3>

According to Greenshield Model, maximum flow is given by this formula:

q_{max}=\frac{V_f \times K_i}{4}

<u>Where:</u>

  • V_f is the free flow speed.
  • K_i is the Jam density.

In order to calculate the free flow speed, we would use this formula:

V_f =2 V\\\\V_f =2\times 59\\\\V_f=118\;km/hr

Substituting the parameters into the model, we have:

q_{max}=\frac{118 \times 122}{4}\\\\q_{max}=\frac{14396}{4}

Max flow = 3,599 veh/hr.

Read more on traffic flow here: brainly.com/question/15236911

6 0
2 years ago
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