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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 0.25" diameter A36 steel rivet connects two 1" wide by .25" thick 6061-T6 Al strips in a single lap shear joint. The shear str
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Option B

1025 psi

Explanation:

In a single shear, the shear area is \frac {\pi d^{2}}{4}=\frac {\pi 0.25^{2}}{4}

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6-What is the difference between the critical point and the triple point?
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2 years ago
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1. (5 pts) An adiabatic steam turbine operating reversibly in a powerplant receives 5 kg/s steam at 3000 kPa, 500 °C. Twenty per
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Answer:

temperature of first extraction 330.8°C

temperature of second extraction 140.8°C

power output=3168Kw

Explanation:

Hello!

To solve this problem we must use the following steps.

1. We will call 1 the water vapor inlet, 2 the first extraction at 100kPa and 3 the second extraction at 200kPa

2. We use the continuity equation that states that the mass flow that enters must equal the two mass flows that leave

m1=m2+m3

As the problem says, 20% of the flow represents the first extraction for which 5 * 20% = 1kg / s

solving

5=1+m3

m3=4kg/s

3.

we find the enthalpies and temeperatures in each of the states, using thermodynamic tables

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties

4.we find the enthalpy and entropy of state 1 using pressure and temperature

h1=Enthalpy(Water;T=T1;P=P1)

h1=3457KJ/kg

s1=Entropy(Water;T=T1;P=P1)

s1=7.234KJ/kg

4.

remembering that it is a reversible process we find the enthalpy and the temperature in the first extraction with the pressure 1000 kPa and the entropy of state 1

h2=Enthalpy(Water;s=s1;P=P2)

h2=3116KJ/kg

T2=Temperature(Water;P=P2;s=s1)

T2=330.8°C

5.we find the enthalpy and the temperature in the second extraction with the pressure 200 kPav y the entropy of state 1

h3=Enthalpy(Water;s=s1;P=P3)

h3=2750KJ/kg

T3=Temperature(Water;P=P3;s=s1)

T3=140.8°C

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Finally, to find the power of the turbine, we must use the first law of thermodynamics that states that the energy that enters is the same that must come out.

For this case, the turbine uses a mass flow of 5kg / s until the first extraction, and then uses a mass flow of 4kg / s for the second extraction, taking into account the above we infer the following equation

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