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Artist 52 [7]
3 years ago
14

The irreversible losses in the penstock of a hydroelectric dam are estimated to be 7 m. The elevation difference between the res

ervoir surface upstream of the dam and the surface of the water exiting the dam is 140 m. If the flow rate through the turbine is 4000 L/min, determine (a) the power loss due to irreversible head loss, (b) the efficiency of the piping, and (c) the electric power output if the turbine-generator efficiency is 84 percent.
Engineering
1 answer:
geniusboy [140]3 years ago
8 0

Answer:

a) the power loss due to irreversible head loss is 4.57 kW

b) the efficiency of the piping is 95%

c) the electric power output is 72.9918 kW

Explanation:

Given the data in the question below;

Irreversible loses h_L = 7m

Total head, H = 140 m

flow rate Q = 4000 L/min = 0.0666 m³/s

Generator efficiency n₀ = 84% = 0.84

we know that density of water is 1000 kg/m³

g = 9.81 m/s²

a) power loss due to irreversible head loss P_L is;    

P_L = p × Q × g × h_L

we substitute

P_L = 1000 × 0.0666 × 9.81 × 7

P_L = 4573.422 W

P_L = 4573.422 / 1000

P_L = 4.57 kW

Therefore, the power loss due to irreversible head loss is 4.57 kW

b) the efficiency of the piping n is;

n = (Actual head / maximum head) × 100

n = (( H - h_L ) / H) × 100

so we substitute

n = (( 140 - 7 ) / 140) × 100

n = (133/140) × 100

n = 0.95 × 100

n = 95%

Therefore, the efficiency of the piping is 95%

c) the electric power output if the turbine-generator efficiency is 84 percent;

n₀ = Power_{outpu / power_{inpu

Power_{outpu = n₀ × power_{inpu

Power_{outpu = n₀ × ( pQg( H - h_L ))

so we substitute

Power_{outpu = 0.84 × ( 1000 × 0.0666 × 9.81( 140 - 7 ))

Power_{outpu = 0.84 × 653.346( 133)

Power_{outpu = 0.84 × 86895.018

Power_{outpu = 72991.815 W

Power_{outpu = 72991.815 / 1000

Power_{outpu = 72.9918 kW

Therefore, the electric power output is 72.9918 kW

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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

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