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Ira Lisetskai [31]
3 years ago
12

A 1000 W iron utilizes a resistance wire which is 20 inches long and has a diameter of 0.08 inches. Determine the rate of heat g

eneration in the wire per unit volume, in Btu/hrft3 , and the heat flux on the outer surface of the wire, in Btu/hrft2 , as a result of this generation.
Engineering
1 answer:
SSSSS [86.1K]3 years ago
3 0

Answer:

The rate of heat generation in the wire per unit volume is 5.79×10^7 Btu/hrft^3

Heat flux is 9.67×10^7 Btu/hrft^2

Explanation:

Rate of heat generation = 1000 W = 1000/0.29307 = 3412.15 Btu/hr

Area (A) = πD^2/4

Diameter (D) = 0.08 inches = 0.08 in × 3.2808 ft/39.37 in = 0.0067 ft

A = 3.142×0.0067^2/4 = 3.53×10^-5 ft^2

Volume (V) = A × Length

L = 20 inches = 20 in × 3.2808 ft/39.37 in = 1.67 ft

V = 3.53×10^-5 × 1.67 = 5.8951×10^-5 ft^3

Rate of heat generation in the wire per unit volume = 3412.15 Btu/hr ÷ 5.8951×10^-5 ft^3 = 5.79×10^7 Btu/hrft^3

Heat flux = 3412.15 Btu/hr ÷ 3.53×10^-5 ft^2 = 9.67×10^7 Btu/hrft^2

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Sati [7]

Answer:

The rate of cell metabolism is limited by mass transfer since the value of maximum cell concentration obtained (38 g/l) is lower than 50 g l-1, the value planed.

Explanation:

                                                     Data

<u>kLa</u> = 0.17/s

<u>Solubility of oxygen</u> =  8 × 10^-3 kg / m^3

<u>The maximum specific oxygen uptake rate </u>= 4 mmol O2 / g h.

<u>Concentration of oxygen</u> =  0.5 × 10^-3 kg/ m^3

<u>**The maximum cell density</u> = 50 g/l

___________________

The calculated maximum cell concentration:

xmax=  kLa · CAL*/ qo

CAL* is the solubility of oxygen in the broth and qo is the specific oxygen uptake rate

Replacing the data given

xmax= ( 0.17/s ) ·   (8 × 10^-3 kg / m^3)  /  4 mmol O2 / g h

4 mmol O2 / g h  to kg O2/ g s

4 \frac{mmol}{gh} \frac{1 gmol}{1000mmol}\frac{1h}{3600s}\frac{32 g}{gmol} \frac{1Kg}{1000g}

= 3.56 x 10^-3 kg O2/ g s

So then,

xmax= ( 0.17/s ) ·   (8 × 10^-3 kg / m^3)  / 3.56 x 10^-3 o kg O2/ g s

xmax= 3. 8 x 10^4 g/ m^3   = 38 g/l

_____________________

5 0
3 years ago
Solve the problem with conditions if a wall has inner and outer surface temperatures of 16 and 6 C, respectively. The interior a
Angelina_Jolie [31]

Answer:

Heat flux is 20 W/m^2

Explanation:

Heat flux (Q) is computed as

Q = h \, \Delta T

where h is heat transfer coefficient and ΔT is the difference between body's temperature

From the interior air to the inner wall

Q = 5 \frac{W}{m^2 K} \, 4 K

Q = 20 \frac{W}{m^2}

From the the outer wall to the exterior air

Q = 20 \frac{W}{m^2 K} \, 1 K

Q = 20 \frac{W}{m^2}

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Cutting and abrasive machining are the two major material processes. List the differences between Cutting tool and Abrasive mach
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Answer:

Explained

Explanation:

Cutting tools:

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2. Cutting tools can have variety of material depending on use like ceramics, diamonds, metals, CBN, etc.

3.Cutting tools have definite shapes and geometry.

Abrasive machining tools

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2. Abrasive tools composed of abrasives bounded in medium of resin or metal.

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7 0
4 years ago
What separates the work of technology transfer research from implementation of the products of such research?
tatuchka [14]

Answer:

The thing that separates the work of technology transfer research from implementation of the products of such research is:

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Answer:

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