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Amanda [17]
3 years ago
8

Consider a very long, slender rod. One end of the rod is attached to a base surface maintained at Tb, while the surface of the r

od is exposed to an air temperature of 400°C. Thermocouples imbedded in the rod at locations 25 mm and 120 mm from the base surface register temperatures of 335°C and 375°C, respectively.
(a) Calculate the rod base temperature (°C).
(b) Determine the rod length (mm) for the case where the ratio of the heat transfer from a finite length fin to the heat transfer from a very long fin under the same conditions is 99 percent.

Engineering
1 answer:
9966 [12]3 years ago
3 0

Answer:

(a) Calculate the rod base temperature (°C). = 299.86°C

(b) Determine the rod length (mm) for the case where the ratio of the heat transfer from a finite length fin to the heat transfer from a very long fin under the same conditions is 99 percent.  = 0.4325m

Explanation:

see attached file below

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Air is compressed by a 40-kW compressor from P1 to P2. The air temperature is maintained constant at 25°C during this process a
AlexFokin [52]

Answer:

the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

Explanation:

Given the data in the question;

From the first law of thermodynamics;

dQ = dU + dW ------ let this be equation 1

where dQ is the heat transfer, dU is internal energy and dW is the work done.

from the question, the process is isothermal ( internally reversible process )

Thus, the change in internal energy is 0

dU = 0

given that; Air is compressed by a 40-kW compressor from P1 to P2

since it is compressed, dW = -40 kW

we substitute into equation 1

dQ = 0 + ( -40 kW )

dQ = -40 kW

Now, change in entropy of air is;

ΔS_{air = dQ / T

given that T = 25 °C = ( 25 + 273.15 ) K = 298.15 K

so we substitute

ΔS_{air =  -40 kW / 298.15 K

ΔS_{air =  -0.13416 ≈ -0.1342 kW/K

Therefore, the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

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2 years ago
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kow [346]

Answer:

i wish i knew what you meant by that

7 0
3 years ago
Natural-draft furnaces are no longer available for new installations because they are _____.
MAXImum [283]

Answer:

They are made of an material that is no longer available.

Explanation:

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3 years ago
An aluminum alloy [E = 73 GPa] cylinder (2) is held snugly between a rigid plate and a foundation by two steel bolts (1) [E = 18
V125BC [204]

Answer:

\sigma_A = 58.43 N/mm^2

Explanation:

Given data:

length of Steel bolt L_1 = 335 mm

Length of aluminium cylinder L_2 = 275 mm

Pitch of bolt p = 1mm

Modulus of elasticity of steel E = 215 GPa

Modulus of elasticity of aluminium =  74 GP

Area of bolt = \frac{\pi}{4} 14^2 =  153.93 mm^2

Area of cylinder  = 2300 mm^2

n =1

By equilibrium

\sum F_y = 0

P_A -2P_S = 0

P_A =2P_S

By the compatibility

\delta _s + \delta_A = nP

Displacement in steel is \delta_s = \frac{P_sL_s}{E_sA_s}

Displacement in Aluminium is \delta_A = \frac{P_AL_A}{E_AA_A}

from compatibility equation we have

\frac{P_sL_s}{E_sA_s} +  \frac{P_AL_A}{E_AA_A} = nP

\frac{P_s\times 335}{180\times 10^3\times 153.93} +  \frac{P_A\times 275}{73\times 10^3\times 2300} = 1\times 1

1.20\times 10^[-5} P_s  + 1.44\times 10^{-6}P_A = 1

substituteP_A =2P_S

1.20\times 10^[-5} P_s  + 1.44\times 10^{-6} (2\times P_s) = 1

1.488\times 10^{-5} P_s = 1

P_s = 67204.30 N

P_A = 134,408.60 N

Stress in Aluminium \sigma  = \frac{P_A}{A_A}

                                               = \frac{134,408.60}{2300}

                             \sigma_A = 58.43 N/mm^2

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3 years ago
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