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Art [367]
3 years ago
6

he Weather Channel reports that it is a hot, muggy day with an air temperature of 90????F, a 10 mph breeze out of the southwest,

and bright sunshine with a solar insolation of 400 W/m2. Consider the wall of a metal building over which the prevailing wind blows. The length of the wall in the wind direction is 10 m, and the emissivity is 0.93. Assume that all the solar irradiation is absorbed, that irradiation from the sky is negligible, and that flow is fully turbulent over the wall. Estimate the average wall temperature.
Engineering
1 answer:
Lilit [14]3 years ago
7 0

Answer:

29\°c

Explanation:

We need to define a couple of properties

T_f = 305KP= 1atm\\v=16.27*10^{-6}m^2/s\\k=0.02658W/mK\\Pr=0.707

For this problem I took Steady-state conditions with isothermal temperature in T_s

As well as a flow turbulent over the wall and negligible heat transfer into the building.

We start making a energy balance on the wall,

E'_{in}-E'_{out}=0

-q'_{cv}+(\alpha_S G_S-E_S)L=0

-\bar{h}_L L(T_s-T_{\infty})+(\alpha_S G_S - \epsilon \sigma T_s^4)L = 0

Again, assuming fully turbulent flow over the leng of the wall,

\bar{Nu}_L = \frac{\bar{h}_L}{k} = 0.037 Re_L^{4/4} Pr^{1/3}

Re_L = u_{\infty}\frac{L}{v} = 4.47*\frac{10}{16.27*10^{-6}}= 2.748*10^6

\bar{h}_L = (0.02658/10)0.037(2.748*10^6)^{4/5}(0.707)^{1/3}=12-4W/m^K

Substituting to fin T_s,

-12.4W/m^2*10m[T_s-(32.2+273)]K+[1*400W/m^2-0.93*5.67*10^{-8}W/m^2.K^4T_s^4]*10m=0

T_s=302.2K=29\°c

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Heater power = 425 watts

Explanation:

Detailed explanation and calculation is shown in the image below

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3 years ago
A biotechnology company produced 225 doses of somatropin, including 11 which were defective. Quality control test 15 samples at
Radda [10]

Answer:

  • <u>0.59</u>

Explanation:

The <em>batch</em> is <em>rejected</em> if any of the <em>random samples are found defective</em>, or, what is the same, it will be accepted only if all 15 samples are good.

The probability that none be defective is the same probability that all the samples are good. Thus, start to calculate the probability that the batch is accepted.

The probability that the first sample is good is 214 /225, because there are 225 - 11 = 214 good samples in 225 doses.

The probability that the second samples is good too is 213/224, because there is 1 less good sample, in the 224 remaining samples.

By the same process, you conclude that the consecutive probabilities of selecting a good sample are: 212/223, 211/222, 210/221, . . . up to 199/211.

The joint probability of all the samples are good is the product of each probability:

\frac{214}{225}\cdot\frac{213}{224}\cdot\frac{212}{223}\cdot\frac{211}{222}\cdot\frac{210}{221}\cdot\frac{209}{220}\cdot\frac{208}{219}\cdot\frac{207}{218}\cdot\frac{206}{217}\cdot\frac{205}{216}\cdot\frac{204}{215}\cdot\frac{203}{214}\cdot\frac{202}{213}\cdot\frac{201}{212}\cdot\frac{200}{211}\cdot\frac{199}{210}

The result is: 0.41278 ≈ 0.41

The conclusion is that the probability that all the samples are good and the batch is accepted is 0.41.

Therefore, <em>the probability that the batch is rejected</em> is 1 - 0.41 = 0.59.

4 0
3 years ago
¿Qué áreas del conocimiento me pueden<br> aportar a la ejecución del proyecto?
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Answer:

la escuela,en casa y listo...............

8 0
3 years ago
A growing trend in urban design is the concept of a rooftop garden. If every building in a city were to install a rooftop garden
vlabodo [156]

Answer:The Urban heat island temperature will be REDUCED.

Two Impacts of Rooftop gardens

1) provision of shade against Sunlight.

2) It helps to purify the air around the building.

Explanation: Rooftop gardens are gardens made on top of the roofs of buildings, it is a Green initiative aimed at helping to improve the overall Environment.

Rooftop gardens have several significant benefits which includes

Reduction of the surrounding temperatures and the Urban heat Island temperatures.

Rooftop gardens helps to shade the roof from the direct impacts of harsh weather conditions.

Generally, plants are known as air purifiers as they remove the excess Carbondioxide around the environment through photosynthesis, and they also help to release water vapor which will help to improve the humidity of the environment.

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3 years ago
It is appropriate to use the following yield or failure criterion for ductile materials (a) Maximum shear stress or Tresca crite
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Answer:

(b)Distortion energy theory.

Explanation:

The best suitable theory for ductile material:

       (1)Maximum shear stress theory (Guest and Tresca theory)

It theory state that applied maximum shear stress should be less or equal to its maximum shear strength.

      (2)Maximum distortion energy theory(Von Mises henkey's        theory)

It states that maximum shear train energy per unit volume at any point  is equal to strain energy per unit volume under the state of uni axial stress condition.

But from these two Best theories ,suitable theory is distortion energy theory ,because it gives best suitable result for ductile material.

6 0
3 years ago
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