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Alekssandra [29.7K]
4 years ago
5

A heated long cylindrical rod is placed in a cross flow of air at 20°C (1 atm) with velocity of 10 m/s. The rod has a diameter o

f 5 mm and its surface has an emissivity of 0.95. If the surrounding temperature is 20°C and the heat flux dissipated from the rod is 17000 W/m2, determine the surface temperature of the rod. Evaluate the air properties at 70°C. The properties of air (1 atm) at 70°C are k
Engineering
1 answer:
postnew [5]4 years ago
5 0

Answer:

Ts = 413.66 K

Explanation:

given data

temperature = 20°C

velocity = 10 m/s

diameter = 5 mm

surface emissivity = 0.95

surrounding temperature = 20°C

heat flux dissipated = 17000 W/m²

to find out

surface temperature

solution

we know that here properties of air at 70°C

k = 0.02881 W/m.K

v = 1.995 ×10^{-5} m²/s

Pr = 0.7177

we find here reynolds no for air flow that is

Re = \frac{\rho V D }{\mu } = \frac{VD}{v}    

Re = \frac{10*0.005}{1.99*10^{-5}}

Re = 2506

now we use churchill and bernstein relation for nusselt no

Nu = \frac{hD}{k} = 0.3 + \frac{0.62 Re6{0.5}Pr^{0.33}}{[1+(0.4/Pr)^{2/3}]^{1/4}} [1+ (\frac{2506}{282000})^{5/8}]^{4/5}

h = \frac{0.02881}{0.005}0.3 + \frac{0.62*2506{0.5}0.7177^{0.33}}{[1+(0.4/0.7177)^{2/3}]^{1/4}} [1+ (\frac{2506}{282000})^{5/8}]^{4/5}

h = 148.3 W/m².K

so

q conv = h∈(Ts- T∞ )

17000 = 148.3 ( 0.95) ( Ts - (20 + 273 ))

Ts = 413.66 K

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The ABC Corporation manufactures and sells two products: T1 and T2. 20XX budget for the company is given below:
Sliva [168]

Answer:

The ABC Corporation

a) Total Expected Revenue (in dollars) for 20XX:

Revenue from T1 = 60,000 x $165 = $26,400,000

Revenue from T2 = 40,000 x $250 = $10,000,000

Total Revenue from T1 and T2 = $36,400,000

b) Production Level (in units) for T1 and T2

                                           T1                       T2

Total Units sold             160,000           40,000

Add Closing Inventory   25,000             9,000

Units Available for sale 185,000           49,000

less opening inventory  20,000             8,000

Production Level          165,000 units 41,000 units

c) Total Direct Material Purchases (in dollars):

Cost of direct materials used    T1                T2

A:       (165,000 x 4 x $12)   $7,920,000   $2,460,000 (41,000 x 5 x $12)

B:       (165,000 x 2 x $5)       1,650,000          615,000 (41,000 x 3 x $5)

C:                                                           0          123,000 (41,000 x 1 x$3)

Total cost                            $9,570,000     $3,198,000 Total = $12,768,000

Cost of direct per unit = $58 ($9,570,000/165,000) for T1 and $78 ($3,198,000/41,000) for T2

Cost of direct materials used for production $12,768,000

Cost of closing direct materials:

                 A  (36,000 x $12)  $432,000

                 B (32,000 x $5)        160,000

                 C (7,000 x $3)            21,000             $613,000

Cost of direct materials available for prodn   $13,381,000

Less cost of beginning direct materials:

                 A  (32,000 x $12)        $384,000

                 B  (29,000 x $5)            145,000

                 C  (6,000 x $3)                18,000        $547,000

Cost of direct materials purchases               $12,834,000

d) The Total Direct Manufacturing Labor Cost (in dollars):

                                             T1                         T2

Direct labor per unit              2 hours                  3 hours

Direct labor rate per hour    $12                        $16

Direct labor cost per unit   $24                          $48

Production level              165,000 units        41,000 units

Labor Cost ($)                $3,960,000        $1,968,000

Total labor cost  $5,928,000 ($3,960,000 + $1,968,000)

e) Total Overhead cost (in dollars):

Overhead rate  = $20 per labor hour

Overhead cost per unit: T1 = $40 ($20 x 2) and T2 = $60 ($20 x 3)

T1 overhead = $20 x 2  x 165,000) = $6,600,000

T2 overhead = $20 x 3 x 41,000) =    $2,460,000

Total Overhead cost =                        $9,060,000

Cost of goods produced:

Cost of opening inventory of materials  = $547,000

Purchases of directials materials             12,834,000

less closing inventory of materials     =      $613,000

Cost of materials used for production    12,768,000

add Labor cost                                           5,928,000

add Overhead cost                                    9,060,000

Total production cost                            $27,756,000

f) Total cost of goods sold (in dollars):

Cost of opening inventory =          $3,928,000

Total Production cost             =    $27,756,000

Cost of goods available for sale  $31,684,000

Less cost of closing inventory       $4,724,000

Total cost of goods sold            $26,960,000

g) Total expected operating income (in dollars)

Sales Revenue:  T1 and T2  $36,400,000

Cost of goods sold                 26,960,000

Gross profit                             $9,440,000

less marketing & distribution      400,000

Total Expected Operating Income = $9,040,000

Explanation:

a) Cost of beginning inventory of finished goods:

T1, (Direct materials + Labor + Overhead) X inventory units =

T1 = 20,000 x ($58 + 24 + 40) = $2,440,000

T2 = 8,000 ($78 + 48 + 60) = $1,488,000

Total cost of beginning inventory = $3,928,000

b) Cost of closing Inventory of finished goods:

T1 = 25,000 x ($58 + 24 + 40) = $3,050,000

T2 = 9,000 ($78 + 48 + 60) = $1,674,000

Total cost of closing inventory = $4,724,000

5 0
3 years ago
How do I find v0 using Kirchhoff's laws and Ohm's law? The answer is 25V but I'm confused about how to get that.
telo118 [61]

Answer:

  25 V

Explanation:

It is convenient to use Kirchoff's current law (KCL), which tells you the sum of currents into a node is zero. The node of interest is the top left node.

The currents into it are ...

  20 mA + (-5 -Vo)/(2kΩ) -(Vo/(5kΩ)) = 0

  20 mA -2.5 mA = Vo(1/(2kΩ) +1/(5kΩ)) . . . . add the opposite of Vo terms

  (17.5 mA)(10/7 kΩ) = Vo = 25 . . . volts . . . . divide by the coefficient of Vo

_____

You will notice that the equation resolves to what you would get if you drew the Norton equivalent of the voltage source with its 2k impedance. You have two current sources, one of +20 mA, and one of -2.5 mA supplying current to a load of 2k║5k = (10/7)kΩ. KCL tells you the total current into the node is equal to the current through that load (out of the node).

5 0
3 years ago
Consider a cylindrical nickel wire 1.8 mm in diameter and 2.6 × 104 mm long. Calculate its elongation when a load of 290 N is ap
telo118 [61]

Answer:

e = 3.97*10^-4

Explanation:

1.8 mm = 0.0018 m

2.6*10^4 mm = 26 m

Elongation is The ratio between the stretched length and the original length.

e = L/L0

This is calculated with Hooke's law:

e = σ/E

Where

σ: normal stress

E: elastic constant

σ = P/A

Where

P: normal load

A: cross section

A = π/4 * d^2

Therefore:

e = P / (A * E)

e = 4 * P / (π * d^2 * E)

e = 4 * 290 / (π * 0.0018^2 * 207*10^9) = 3.97*10^-4

8 0
3 years ago
Hellpppppppppp will give brainliest just help!!!!!!
aliya0001 [1]
Turbooooooooooi


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3 0
3 years ago
Atmospheric air at 25 °C and 8 m/s flows over both surfaces of an isothermal (179C) flat plate that is 2.75m long. Determine the
vekshin1

Answer:

Re=100,000⇒Q=275.25 \frac{W}{m^2}

Re=500,000⇒Q=1,757.77\frac{W}{m^2}

Re=1,000,000⇒Q=3060.36 \frac{W}{m^2}

Explanation:

Given:

For air      T_∞=25°C  ,V=8 m/s

  For surface T_s=179°C

     L=2.75 m    ,b=3 m

We know that for flat plate

Re⇒Laminar flow

Re>30\times10^5⇒Turbulent flow

<u> Take Re=100,000:</u>

 So this is case of laminar flow

  Nu=0.664Re^{\frac{1}{2}}Pr^{\frac{1}{3}}

From standard air property table at 25°C

  Pr= is 0.71  ,K=26.24\times 10^{-3}

So    Nu=0.664\times 100,000^{\frac{1}{2}}\times 0.71^{\frac{1}{3}}

Nu=187.32   (\dfrac{hL}{K_{air}})

187.32=\dfrac{h\times2.75}{26.24\times 10^{-3}}

     ⇒h=1.78\frac{W}{m^2-K}

heat transfer rate =h(T_∞-T_s)

                           =275.25 \frac{W}{m^2}

<u> Take Re=500,000:</u>

So this is case of turbulent flow

  Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}

Nu=0.037\times 500,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}

Nu=1196.18  ⇒h=11.14 \frac{W}{m^2-K}

heat transfer rate =h(T_∞-T_s)

                             =11.14(179-25)

                           = 1,757.77\frac{W}{m^2}

<u> Take Re=1,000,000:</u>

So this is case of turbulent flow

  Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}

Nu=0.037\times 1,000,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}

Nu=2082.6  ⇒h=19.87 \frac{W}{m^2-K}

heat transfer rate =h(T_∞-T_s)

                             =19.87(179-25)

                           = 3060.36 \frac{W}{m^2}

7 0
3 years ago
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