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seraphim [82]
4 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]4 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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The pressure in a water line is 1500 kPa. What is the line pressure in (a) lb/ft2units and (b) lbf/in2(psi) units?
n200080 [17]

Answer:

Part A:

1500\ KPa= 31328.145 \frac{lb}{ft^2}

Part B:

1500 KPa=217.55656 \frac{lb}{in^2}(Psi)

Explanation:

Part A:

Line Pressure is 1500 KPa

We need a conversion factor which converts KPa to lb/ft^2.

20.88543 \frac{lb}{ft^2}= 1\ KPa

In order to convert 1500 KPa to lb/ft^2, we proceed as:

1\ KPa=20.88543 \frac{lb}{ft^2} \\1500\ KPa= 1500 KPa*20.88543 \frac{lb}{ft^2.KPa}\\1500\ KPa= 31328.145 \frac{lb}{ft^2}

1500 KPa is 31328.145 lb/ft^2

Part B:

We will use the same procedure we did in Part A:

1 ft= 12 in

(1\ ft)^2=(12\ in)^2\\1 ft^2=144 in^2

Converting 1500 KPa\  into\  \frac{lb}{in^2}

1500\ KPa= 1500 KPa*20.88543 \frac{lb}{ft^2.KPa} * \frac{ft^2}{144\ in^2}

1500 KPa=217.55656 \frac{lb}{in^2}(Psi)

1500 KPa is 217.55656 lb/in^2 (psi)

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3 years ago
How did dodge become a company
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6 0
3 years ago
A mass weighing 22 lb stretches a spring 4.5 in. The mass is also attached to a damper with Y coefficient . Determine the value
Dominik [7]

Answer:

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Explanation:

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x= 0.375 ft

Spring constant ,K

K=\dfrac{m}{x}=\dfrac{22}{0.375}

K= 58.66  lb/ft

The damper coefficient for critically damped system

C_c=2\sqrt{mK}

C_c=2\sqrt{0.6875\times 58.66}

Cc= 12.7 lb.sec/ft

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Afina-wow [57]

Option C (making lessons learned a regular part of meetings) is the correct approach.

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Aside from this, none of the choices are viable methods to learning lessons or gaining knowledge from the past. As a result, the methodology outlined above is the appropriate one.

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