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OverLord2011 [107]

4 years ago

11

Consider a pipe with an inner radius of 5cm and an outer radius of 7cm.The inner surface is kept at 100C, and the outer surface

at 80C. Determine the heat loss per meter length of the pipe if the pipe is made of pure copper [k=387W/m.oC], pore aluminum [k=200W/m.C], and pure iron [k=62W/m.C].

1 answer:

Ivahew [28]4 years ago

7 0

Answer:

Explanation:

given data:

innner radius r_i = = 5cm

outer radius r_0 = 7 cm

temperature at outer surface = 80 degree celcius

temperature at inner surface = 100 degree celcius

thermal resistance per unit length is given as

$R _{th} = \frac{1}{ 2\pi kl} ln \frac{ ro}{ri}$

$= \frac{1}{ 2\pi k*1} ln \frac{ 7}{5}$

$= \frac{0.053}{k}$

heat loss rate per unit length $q = \frac{Ti-To}{R_{th}}$

$q = \frac{100-80}{\frac{0.053}{k}}$

q = 373.474 K

1) for pure COPPER

k = 387 W/m degree celcius

q = 373.474 * 387 = 144534.438 W

2) for pure ALUMINIUM

k = 200 W/m degree celcius

q = 373.474 * 200 = 74694.84 W

3)

1) for pure IRON

k = 62W/m degree celcius

q = 373.474 * 62 = 23155.416 W

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A composite shaft with length L = 46 in is made by fitting an aluminum sleeve (Ga = 5 x 10^3 ksi) over a

Xelga [282]

Answer:

Explanation:

Given the data in the question;

L = 46 in

Ga = 5 × 10³ ksi

Gs = 11 × 10³ ksi

Outside diameter da = 5 in

ds = 4 in

Tb = 3 kip.in

Now,

Ja = polar moment of Inertia of Aluminum;

Ja ⇒ π/32( 5⁴ - 4⁴ ) = π/32( 625 - 256 ) = π/32( 369 ) $in^u$

Js = polar moment of inertia of steel

Js ⇒ π/32 ds⁴ = π/32( 4⁴ ) = π/32( 256 )

Ta is torque transmitted by Aluminum

Ts is torque transmitted by steel

{composite member }

T = Ta + Ts ------ let this be equation m1

Now, we use the relation;

T/J = G∅/L

JG∅ = TL

∅ = TL/GJ

so, for aluminum rod ∅ $_{alu$ = TaLa/GaJa

for steel rod ∅ $_{steel$ = TsLs/GsJs

but we know that, ∅a = ∅s = ∅ $_B$

so

[TaLa/GaJa] = [TsLs/GsJs]

also, we know that, La = Ls = L

∴ [Ta/GaJa] = [Ts/GsJs]

we solve for Ta

TaGsJs = TsGaJa

Ta = TsGaJa / GsJs

we substitute

Ta = [Ts(5 × 10³)( π/32( 369) )] / [ (11 × 10³)( π/32( 256 ) ) ]

Ta = 0.66Ts

now, we substitute 0.66Ts for Ta and 3 for T in equation 1

T = Ta + Ts

3 = 0.66Ts + Ts

3 = 1.66Ts

Ts = 3 / 1.66

Ts = 1.8072 ≈ 1.81 kip-in

so

∅ $_{steel$ = TsLs / GsJs

we substitute

∅ $_{steel$ = (1.81 × 46 ) / ( 11 × 10³ × π/32( 256 ) )

∅ $_{steel$ = 83.26 / 276460.1535

∅ $_{steel$ = 0.000301

∅ $_{steel$ = 3.01 × 10⁻⁴ rad

so

∅ $_{steel$ = ∅ $_B$ = 3.01 × 10⁻⁴ rad

Therefore, the magnitude of the angle of twist at end B is 3.01 × 10⁻⁴ rad

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3 years ago

A rigid 10-L vessel initially contains a mixture of liquid and vapor water at 100 °C, with a quality factor of 0.123. The mixtur

masya89 [10]

Answer:

$Q_{in} = 46.454\,kJ$

Explanation:

The vessel is modelled after the First Law of Thermodynamics. Let suppose the inexistence of mass interaction at boundary between vessel and surroundings, changes in potential and kinectic energy are negligible and vessel is a rigid recipient.

$Q_{in} = U_{2} - U_{1}$

Properties of water at initial and final state are:

State 1 - (Liquid-Vapor Mixture)

$P = 101.42\,kPa$

$T = 100\,^{\textdegree}C$

$\nu = 0.2066\,\frac{m^{3}}{kg}$

$u = 675.761\,\frac{kJ}{kg}$

$x = 0.123$

State 2 - (Liquid-Vapor Mixture)

$P = 476.16\,kPa$

$T = 150\,^{\textdegree}C$

$\nu = 0.2066\,\frac{m^{3}}{kg}$

$u = 1643.545\,\frac{kJ}{kg}$

$x = 0.525$

The mass stored in the vessel is:

$m = \frac{V}{\nu}$

$m = \frac{10\times 10^{-3}\,m^{3}}{0.2066\,\frac{m^{3}}{kg} }$

$m = 0.048\,kg$

The heat transfer require to the process is:

$Q_{in} = m\cdot (u_{2}-u_{1})$

$Q_{in} = (0.048\,kg)\cdot (1643.545\,\frac{kJ}{kg} - 675.761\,\frac{kJ}{kg} )$

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3 years ago

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3 years ago

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A heat pump receives heat from a lake that has an average wintertime temperature of 6o C and supplies heat into a house having a

Dafna1 [17]

Answer:

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Explanation:

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$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

$COP_{HP} = \frac{298.15\,K}{298.15\,K-279.15\,K}$

$COP_{HP} = 15.692$

The minimum heat received by the house must be equal to the heat lost to keep the average temperature constant. Hence:

$\dot Q_{H} = 60000\,\frac{kJ}{h}$

The minimum power supplied to the heat pump is:

$\dot W = \frac{\dot Q_{H}}{COP}$

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3 years ago

A 11-cm-diameter horizontal jet of water with a velocity of 40 m/s relative to the ground strikes a flat plate that is moving in

Reika [66]

Answer:

F = 8552.7N

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3 years ago