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morpeh [17]
2 years ago
9

The specifications of an electronic device are 24 /- 0.4 Amps. When the device exceeds specifications the average quality cost i

s $32.00. A sample of the devices had an average value of 23.9 Amps and a standard deviation of 0.211 Amps. What is the average loss for this device
Engineering
1 answer:
Misha Larkins [42]2 years ago
4 0

Answer:

The average loss for this device is $10.90

Explanation:

Given data

Specification = 24 +/- 0.4 Amps

average quality cost = $32.00

average value y = 23.9

standard deviation s = 0.211 Amps

32 = k(24.4 - 24)²

32 = k(0.4)² = k(0.16)

k = 32/0.16 = 200

To evaluate average loss, use

L = k{s² + (y - T)²}

T = 24A

L = 200{0.211² + (23.9 - 24)²}

L = $10.90

The average loss for this device is $10.90

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A piston–cylinder device contains a mixture of 0.5 kg of H2 and 1.2 kg of N2 at 100 kPa and 300 K. Heat is now transferred to th
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Answer:

(a) The heat transferred is 2552.64 kJ    

(b) The entropy change of the mixture is 1066.0279 J/K

Explanation:

Here we have

Molar mass of H₂ = 2.01588 g/mol

Molar mass of N₂ = 28.0134 g/mol

Number of moles of H₂ = 500/2.01588  = 248 moles

Number of moles of N₂ = 1200/28.0134 = 42.8 moles

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At constant pressure, the temperature is doubled, therefore

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If we assume constant specific heat at the average temperature, we have

Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂

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cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K

Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 =  2552.64 kJ    

b)  \Delta S = - R(n_A \times lnx_A + n_B \times ln x_B)

Where:

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Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks necessary data to solve. But I have found the similar question on the internet. So, I will be using the data from that question to solve this question for the sack of concept and understanding.

Data Given:

x = 27 , 44 , 32 , 47, 23 , 40, 34, 52

y = 30, 19,  24,  13 , 29,  19,  21,  14

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We are asked to verify the above values manually in this question.

So,

1. ∑x = 299

Let's verify it:

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∑x = 299

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2. ∑y = 167

Let's verify it:

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∑y = 169

No, it is not equal to the given value.

3. ∑x^{2} = 11887

Let's verify it:

For this to find,  first we need to square all the value of x individually and then add them together to verify.

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Yes, it is equal to the given value. Hence, verified.

4. ∑y^{2} = 3773

Let's verify it:

Again, for this we need to find the squares of all the y values and then add them together to verify it.

∑y^{2} = 30^{2} + 19^{2} +  24^{2} + 13^{2} + 29^{2} + 19^{2} +  21^{2} +  14^{2}

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Answer:

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