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Ugo [173]
2 years ago
8

A long rod of 60-mm diameter and thermophysical properties rho=8000 kg/m^3, c=500J/kgK, and k=50 W/mK is initally at a uniform t

emperature in a forced convection furnace maintained at 750K. The convection coeffiecent is estimated to be 1000 W/m^2K.
What is the centerline temperature of the rod when the surface temperature is 550K?
Engineering
1 answer:
Monica [59]2 years ago
3 0

Answer:

Tc = 424.85 K

Explanation:

Given that,

D = 60 mm = 0.06 m

\rho = 8000 kg/m^3

k = 50 w/m . kc = 500 j/kg.k

h_{\infty} = 1000 w/m^2t_{\infity} = 750 kt_w = 500 K

surface area = As = \pi dL \\\frac{As}{L} = \pi D = \pi \timeS 0.06

HEAT FLOW Q  is

Q = h_{\infty} As (T_[\infty} - Tw)  = 1000 \pi\times 0.06 (750-500)

 = 47123.88 w per unit length of rod

volumetric heat rate

q = \frac{Q}{LAs}

= \frac{47123.88}{\frac{\pi}{4} D^2 \times 1}

q = 1.66\times 10^{7} w/m^3

Tc = \frac{- qR^2}{4K} + Tw

= \frac{ - 1.67\times 10^7 \times (\frac{0.06}{2})^2}{4\times 50} +  500

  = 424.85 K

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Two Carnot engines operate in series such that the heat rejected from one is the heat input to the other. The heat transfer from
kykrilka [37]

Answer:

Given:

high temperature reservoir T_{H} =1000k

low temperature reservoir T_{L} =400k

thermal efficiency n_{1}= n_{2}

The engines are said to  operate on Carnot cycle which is totally reversible.

To find the intermediate temperature between the two engines, The thermal efficiency of the first heat engine can be defined as

n_{1} =1-\frac{T}{T_{H} }

The thermal efficiency of second heat engine can be written as

n_{2} =1-\frac{T_{L} }{T}

The temperature of intermediate reservoir can be defined as  

1-\frac{T}{T_{H} } =1-\frac{T_{L} }{T} \\T^2=T_{L} T_{H} \\T=\sqrt{T_{L} T_{H} }\\T=\sqrt{400*1000} =632k

8 0
3 years ago
The 40-ft-long A-36 steel rails on a train track are laid with a small gap between them to allow for thermal expansion. Determin
vfiekz [6]

Answer:

Ф = 0.02838 ft

F  = 1,032 N

Explanation:

To find out gap delta,

As it is case of free thermal expansion,

First we start with, some assumptions we have to made to solve this problem.

1. Thermal Expansion Coefficient of Steel is ∝= 6.45 ×10^(-6)

2. Modulas of elasticity for A-36 steel is E= 200 GPa

3. Area of rail is assumed to be unit area.

The gape required can be given by,

Ф = ∝  × ΔT  × L  ... where Ф= Gap Delta in ft

                                          ΔT= Temperature rise in F

                                               = 90- (-20)

                                               =  110 F

Ф =  6.45 ×10^(-6) × 110 × 40

Ф =  28,380 × 10^(-6) ft

Ф = 0.02838 ft     .... total gape required for expansion of steel rails

Stress induced in rails is given by,

   σ     =  ∝  × ΔT  × E

          =  6.45 ×10^(-6)   × 110  × 200

  σ      =  1,41,900 Pa

Now, let's find axial force in rails,

Here,we have to consider  ΔT= 20 F.

As due to temperature change, axial force generated in rails can be find by,

F = A × ∝ × ΔT× E × L

F = 1 × 6.45 × 10^(-6) × 20 × 200 × 10^(-9) × 40

F = 25,800 × 40 × 10^(-3)

F = 10,32,000 × 10^(-3)

F= 1,032 N

Finally, due to temperature change, rail is subjected to axial force, axial stress.

8 0
3 years ago
Affordability is most concerned with:
leva [86]

Answer:

feasibility study

Explanation:

7 0
3 years ago
Where loads are likely to be on continuously, the calculated load for branch circuits and feeders must be figured at (100%) (125
Anna [14]

Where loads are likely to be on continuously, the calculated load for branch circuits and feeders must be figured at 125%.

Section 210.19(A)(1) permits the bigger of the two values listed below to be utilized as the connectors 's ultimate size for sizing an ungrounded branch circuit conductor:

Without any extra adjustments or corrections, either 125% of the continuous load, OR

When adjustment and corrective factors are applied, the load is 100% (not 125% as stated previously).

This will be the same in the 2020 NEC. The introduction of new exception 2 is what has changed. To comprehend this new exception, one must study it very carefully. A part of a branch circuit connected to pressure connectors (such as power distribution blocks) that complies with 110.14(C)(2) may now be sized using the continuous load plus the noncontiguous load instead of 125% of the continuous load thanks to the new exception.

To know more about connectors click here:

brainly.com/question/16987039

#SPJ4

4 0
1 year ago
The velocity of a particle which moves along the s-axis is given by = 40 − 3 2/ , ℎ t is in seconds. Calculate the displacement
scoundrel [369]

The displacement ∆S of the particle during the interval from t = 2sec to 4sec is; 210 sec

<h3>How to find the displacement?</h3>

We are given the velocity equation as;

s' = 40 - 3t²

Thus, the speed equation will be gotten by integration of the velocity equation to get;

s = ∫40 - 3t²

s = 40t - ¹/₂t³

Thus, the displacement between times of t = 2 sec and t = 4 sec is;

∆S = [40(4) - ¹/₂(4)³] - [40(2) - ¹/₂(2)³]

∆S = 210 m

Read more about Displacement at; brainly.com/question/4931057

#SPJ1

8 0
2 years ago
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