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SCORPION-xisa [38]
3 years ago
12

Thermal energy generated by the electrical resistance of a 5-mm-diameter and 4-m-long bare cable is dissipated to the surroundin

g air at 20°C. The voltage drop and the electric current across the cable in steady operation are measured to be 60 V and 1.5 A, respectively. Disregarding radiation, estimate the surface temperature of the cable. Evaluate air properties at a film temperature of 60°C and 1 atm pressure. Is this a good assumption?

Engineering
1 answer:
ella [17]3 years ago
6 0

Answer:

surface temperature = 128.74⁰c

Explanation:

Given data

diameter of cable = 5 mm = 0.005 m

length of cable = 4 m

T∞ ( surrounding temperature ) = 20⁰c

voltage drop across cable ( dv )= 60 V

current across cable = 1.5 A

attached to this answer is the comprehensive analysis and solution to the problem.

The assumption made is not a good one since the calculated Ts ( surface temperature ) is very much different from the assumed Ts

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Air (ideal gas) is contained in a cylinder/piston assembly at a pressure of 150 kPa and a temperature of 127°C. Assume that the
Stels [109]

Answer:

The process is not possible.

Explanation:

We know for ideal condition, the work done for isothermal process is

W_{ideal} = P_{1}.V_{1} ln\frac{V_{2}}{V_{1}}

and for ideal gas, we know  PV = mRT

Therefore, W_{ideal} = mRTln\frac{V_{2}}{V_{1}}

                                                  = mRTln\frac{P_{1}}{P_{2}}

                                                  =  0.287 x 400ln\frac{150}{450}

                                                  = -126.12 kJ/kg (negative sign indicates that the process is compressive, so work input to the compressor is 126.12 kJ/kg )

Now we know for adiabatic compression process

                    PV^{\gamma } = C

We know \frac{T_{2}}{T_{1}}=(\frac{P_{2}}{P_{1}})^{\frac{\gamma -1}{\gamma }}

T_{2} = 556 K

For adiabatic work done, W_{adiabatic} = \frac{P_{1}\times V_{1}-P_{2}\times V_{2}}{\gamma -1}

                                                                       = \frac{mR(T_{1}-T_{2})}{\gamma -1}

                                                                       = \frac{0.287(400-556)}{1.45 -1}

                                                                       = -99.49 kJ/kg (negative sign indicates that the process is compressive, so work input to the compressor is 99.49 kJ/kg )

We know that in isothermal process, work input to the compressor is minimum. But in the above adiabatic polytropic process, work input to the compressor is less than the work done in the isothermal process.

Thus the process is not possible.

                                                             

7 0
3 years ago
Although many countries have issues with soil erosion due to deforestation, some of the most serious effects are seen
Tju [1.3M]
Nigeria
According to the FAO, Nigeria has the world's highest deforestation rate of primary forests. It has lost more than half of its primary forest in the last five years.
Please give brainliest
4 0
2 years ago
6.3.3 Marks on an exam in a statistics course are assumed to be normally distributed
bekas [8.4K]

Answer:

- The calculated p-value (0.392452) is higher than the significance level at which the test was performed, hence, the null hypothesis is true and μ = 60

- 95% Confidence interval for the population mean score = (47.4, 84.1)

Explanation:

The sample of 4 students had scores of 52, 63, 64, 84.

First of, we need to compute the sample mean, we do not need the sample standard deviation as the population variance is given as 5

Mean = (Σx)/N

x = each variable

N = number of variables = 4

Mean = (52 + 63 + 64 + 84)/4

Mean = 65.75

Sample Standard deviation = σ

= √[Σ(x - xbar)²/N]

xbar = mean = 65.75

Σ(x - xbar)² = 532.75

σ = √[532.75/4] = 11.54

in hypothesis testing, the first thing is usually to state the null and alternative hypothesis.

From the question, the null hypothesis has already been stated as

H₀: μ = 60

The alternative hypothesis would then be that the population mean score isn't equal to 60

Hₐ: μ ≠ 60

Since the population distribution is normal and the sample standard deviation is to be used, we use the t-test statistic

t = (x - μ₀)/σₓ

x = sample mean = 65.75

μ₀ = Standard to be compared against = 60

σₓ = standard error = (σ/√n) = (11.54/√4) = 5.77

t = (65.75 - 60)/5.77 = 0.9965 = 1.00

checking the tables for the p-value of this t-statistic

Degree of freedom = df = n - 1 = 4 - 1 = 3

Significance level = 0.05 (95% confidence level)

The hypothesis test uses a two-tailed condition because we're testing in two directions.

p-value (for t = 1.00, at 0.05 significance level, df = 3, with a two tailed condition) = 0.392452

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.05

p-value = 0.392452

0.392452 > 0.05

Hence,

p-value > significance level

This means that we fail to reject the null hypothesis & say that there is enough evidence to conclude that the populatiom mean score is equal to 60.

b) Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Sample Mean = 65.75

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the t-distribution.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 4 - 1 = 3

Significance level for 95% confidence interval

(100% - 95%)/2 = 2.5% = 0.025

t (0.025, 3) = 3.18 (from the t-tables)

Standard error of the mean = 5.77

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 65.75 ± (3.18 × 5.77)

CI = 65.75 ± 18.3486

95% CI = (47.4014, 84.0986)

95% Confidence interval = (47.4, 84.1)

Hope this Helps!!!

7 0
3 years ago
Do heavier cars really use more gasoline? Suppose a car is chosen at random. Let x be the weight of the car (in hundreds of poun
RoseWind [281]

Answer:

Do heavier cars really use more gasoline? Suppose a car is chosen at random. Let x be the weight of the car (in hundreds of pounds), and let y be the miles per gallon (mpg)

Explanation:

6 0
3 years ago
What is the definition of insert view and why do we use it
maw [93]

Answer:

its a view point for auto cad

Explanation:

from my knowlege in IED we learned about it as a way of sing how an object would look in inventor or auto CAD

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3 years ago
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