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dimaraw [331]
4 years ago
15

(a) Determine the specific volume of nitrogen gas at 8.5 MPa and 139 K based on the generalized compressibility chart and compar

e this result with the experimental value of 0.002002 m³/kg to (b).
(b) Determine the error involved.
(R = 0.2968 kPa.m³/kg.K, Tcr=126.2 K, Pcr=3.39 MPa)
Engineering
1 answer:
Jet001 [13]4 years ago
8 0

Answer:

V = 0.002524m³/kg

%Error = 0.2851

Explanation:

Using the ideal gas equation, we can determine the specific volume of nitrogen as

PV = RT

V = RT/P

Given the following data:

R = 296.8j/kg.k

T = 139k

P = 8.5Mpa

Therefore, V = (296.8j/kg.k)×(139k)/8500000

V = 0.004853m³/kg

Using the compressibility chart, we can see that the compressibility factor of Nitrogen at this state is equal to 0.52. Using this, we can find out the specific volume of nitrogen to be

Z = PV/RT

0.52 = (8500000Pa)V/(296.8J/kg⋅K)(139K)

V = 0.52×296.87×139/8500000

V = 0.002524m³/kg

%Error = V(ideal)-V(actual)/V(actual)

%Error = 0.002524-0.002002/0.002002

%Error = 0.2609

Error for this compared with the actual value is

%Error = V(ideal)-V(actual)/V(actual)

%Error = (0.004853-0.002002)/0.004853

%Error = 0.2851

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Answer:

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Suppose that the weights for newborn kittens are normally distributed with a mean of 125 grams and a standard deviation of 15 gr
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(a) If a kitten weighs 99 grams at birth, it is at 5.72 percentile of the weight distribution.

(b) For a kitten to be at 90th percentile, the minimum weight is 146.45 g.

<h3>Weight distribution of the kitten</h3>

In a normal distribution curve;

  • 2 standard deviation (2d) below the mean (M), (M - 2d) is at 2%
  • 1 standard deviation (d) below the mean (M), (M - d) is at 16 %
  • 1 standard deviation (d) above the mean (M), (M + d) is at 84%
  • 2 standard deviation (2d) above the mean (M), (M + 2d) is at 98%

M - 2d = 125 g - 2(15g) = 95 g

M - d = 125 g - 15 g = 110 g

95 g is at 2% and 110 g is at 16%

(16% - 2%) = 14%

(110 - 95) = 15 g

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From 95 g to 99 g:

99 g - 95 g  = 4 g

4g x 0.93%/g = 3.72%

99 g will be at:

(2% + 3.72%) = 5.72%

Thus, if a kitten weighs 99 grams at birth, it is at 5.72 percentile of the weight distribution.

<h3>Weight of the kitten in the 90th percentile</h3>

M + d = 125 + 15 = 140 g      (at 84%)

M + 2d = 125 + 2(15) = 155 g   ( at 98%)

155 g - 140 g = 15 g

14% / 15g = 0.93%/g

84% + x(0.93%/g) = 90%

84 + 0.93x = 90

0.93x = 6

x = 6.45 g

weight of a kitten in 90th percentile = 140 g + 6.45 g  = 146.45 g

Thus, for a kitten to be at 90th percentile, the approximate weight is 146.45 g

Learn more about standard deviation here: brainly.com/question/475676

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