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seraphim [82]
3 years ago
6

The hot-wire anemometer.' A hot-wire anemome ter is essentially a fine wire, usually made of platinum,which is heated electrical

ly and inserted into a flowingfluid. The wire temperature, which is a function of the fluidtemperature, fluid velocity, and the rate of heating, may bedetermined by measuring its electrical resistance. It is used for measuring velocities and velocity fluctuations in flow systems. In this problem we analyze the temperature distribution in the wire element. We consider a wire of diameter D and length 2L supported at its ends (z = -L and z =+L) and mounted perpendicular to an air stream. An electric current of density I amp/cm^2 flows through the wire, and the heat thus generated is partially lost by convection to the air stream and partially by conduction toward the ends of the wire. Because of their size and their high electrical and thermal conductivity, the supports are not appreciably heated by the current, but remain at the temperature TL, which is the same as that of the approaching air stream. Heat loss by radiation is to be neglected.
Required:
Derive an equation for the steady-state temperature distribution in the wire, assuming that T depends on z alone.
Engineering
1 answer:
Delicious77 [7]3 years ago
7 0

Answer:

attached below

Explanation:

<u>To derive the equation we have to make some assumptions </u>

assumptions:

Assume T depends on Z alone

Assume uniform thermal and electrical conductivities (k and ke) in the wire, Assume  a uniform heat

transfer coefficient h from the wire to the air stream.

attached below is the required derivation of the equation

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A team member who has been a good worker for many years has recently been doing poor work. You suspect that he may be tired of h
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Answer:

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Explanation:

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8 0
4 years ago
Type the correct answer in the box. Spell all words correctly.
sesenic [268]

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6 0
3 years ago
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This question allows you to practice proving a language is non-regular via the Pumping Lemma. Using the Pumping Lemma (Theorem 1
Ulleksa [173]

Answer:

<em>L is not a regular language with formal proofs  </em>

Explanation:

<em>(a) To prove that L is not a regular language, we will use a proof by contradiction. the assumption entails  that L is a regular language. Then by the Pumping Lemma for Regular Languages, </em>

<em>there exists a pumping length p for L such that for any string s ∈ L where |s| ≥ p, </em>

<em>s = xyz subject to the following conditions: </em>

<em>(a) |y| > 0 </em>

<em>(b) |xy| ≤ p, and </em>

<em>(c) ∀i > 0, xyi </em>

<em>z ∈ L</em>

<em />

<em>(b) To determine that L is not a regular language, we mke use of proof by contradiction.  lets assume, that L is regular. Then by the Pumping Lemma for Regular Languages, it states also,</em>

<em>The pumping length, p for L such that for any string s ∈ L where |s| ≥ p, s = xyz subject  to the condtions as follows : </em>

<em>(a) |y| > 0 </em>

<em>(b) |xy| ≤ p, and </em>

<em>(c) ∀i > 0, xyi </em>

<em>z ∈ L. </em>

<em>Choose s = 0p10p </em>

<em>. Clearly, |s| ≥ p and s ∈ L. By condition (b) above, it follows is shown. by the first condition x and y are zeros.</em>

<em>for some  k > 0. Per (c), we can take i = 0 and the resulting string will still be in L. Thus,  xy0 </em>

<em>z should be in L. xy0 </em>

<em>z = xz = 0(p−k)10p </em>

<em>It is shown that is is  not in L. This is a  contraption with the pumping lemma.  our assumption that L is regular is  incorrect, and L is not a regular language</em>

6 0
3 years ago
Coal fire burning at 1100 k delivers heat energy to a reservoir at 500 k. Find maximum efficiency.
Marizza181 [45]

Answer:

<em>55%</em>

Explanation:

hot reservoir = 1100 K

cold reservoir = 500 K

<em>This is a Carnot system</em>

For a Carnot system, maximum efficicency of the system is given as

Eff = 1 - \frac{Tc}{Th}

where Tc = temperature of cold reservoir = 500K

Th = temperature of hot reservoir = 1100 K

Eff = 1 - \frac{500}{1100}

Eff = 1 - 0.45 = 0.55 or<em> 55%</em>

7 0
3 years ago
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