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2 years ago

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A thick-walled tube of stainless steel having a k = 21.63 W/m∙K with dimensions of 0.0254 m ID and 0.0508 m OD is covered with 0

.0254 m thick layer of an insulation (k = 0.2423 W/m∙K). The inside-wall temperature of the pipe is 811 K and the outside surface of the insulation is 310.8 K. For a 0.305 m length of pipe, calculate the heat loss and the temperature at the interface between the metal and the insulation.

1 answer:

Aneli [31]2 years ago

4 0

Answer:

Q=339.5W

T2=805.3K

Explanation:

Hi!

To solve this problem follow the steps below, the procedure is attached in an image

1. Draw the complete outline of the problem.

2.to find the heat Raise the heat transfer equation for cylinders from the inside of the metal tube, to the outside of the insulation.

3. Once the heat is found, Pose the heat transfer equation for cylinders from the inner part of the metal tube to the outside of the metal tube and solve to find the temperature

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Determine F12 and F21 for the following configurations: (a) A long semicircular duct with diameter of 0.1 meters: (b) A hemisphe

uysha [10]

Answer:

long duct: 1.0 and 0.424

Hemisphere 1.0 ; 0.125; 0.5

Explanation:

For a long duct:

By inspection, $F_{12} = 1.0$

Calculating by reciprocity, $F_{21} = \frac{A_{1} }{A_{2}F_{12} } = \frac{2RL}{\frac{3}{4}*2\pi RL }* 1.0\\ = 0.424$

Hemisphere:

By reciprocity gives = 0.125

using the summation rule: $F_{21} + F_{22} + F_{23} = 1$

However, because this is a hemisphere, the value will be= 0.5 * 1

= 0.5

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Which one of the following activities is not an example of incident coordination

Lady bird [3.3K]

Directing, ordering, or controlling

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Explain why failure of this garden hose occurred near its end and why the tear occurred along its length. Use numerical values t

alukav5142 [94]

Answer:

hoop stress
longitudinal stress
material used

all this could led to the failure of the garden hose and the tear along the length

Explanation:

For the flow of water to occur in any equipment, water has to flow from a high pressure to a low pressure. considering the pipe, water is flowing at a constant pressure of 30 psi inside the pipe which is assumed to be higher than the allowable operating pressure of the pipe. but the greatest change in pressure will occur at the end of the hose because at that point the water is trying to leave the hose into the atmosphere, therefore the great change in pressure along the length of the hose closest to the end of the hose will cause a tear there. also the other factors that might lead to the failure of the garden hose includes :

hoop stress ( which acts along the circumference of the pipe):

αh = $\frac{PD}{2T}$ EQUATION 1

and Longitudinal stress ( acting along the length of the pipe )

αl = $\frac{PD}{4T}$ EQUATION 2

where p = water pressure inside the hose

d = diameter of hose, T = thickness of hose

we can as well attribute the failure of the hose to the material used in making the hose .

assume for a thin cylindrical pipe material used to be

$\frac{D}{T}$ ≥ 20

insert this value into equation 1

αh = $\frac{20 *30}{2}$ = 60/2 = 30 psi

the allowable hoop stress was developed by the material which could have also led to the failure of the garden hose

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2 years ago

A group of statisticians at a local college has asked you to create a set of functions that compute the median and mode of a set

iVinArrow [24]

Answer:

Functions to create a median and mode of a set of numbers

Explanation:

def median(list):

if len(list) == 0:

return 0

list.sort()

midIndex = len(list) / 2

if len(list) % 2 == 1:

return list[midIndex]

else:

return (list[midIndex] + list[midIndex - 1]) / 2

def mean(list):

if len(list) == 0:

return 0

list.sort()

total = 0

for number in list:

total += number

return total / len(list)

def mode(list):

numberDictionary = {}

for digit in list:

number = numberDictionary.get(digit, None)

if number == None:

numberDictionary[digit] = 1

else:

numberDictionary[digit] = number + 1

maxValue = max(numberDictionary.values())

modeList = []

for key in numberDictionary:

if numberDictionary[key] == maxValue:

modeList.append(key)

return modeList

def main():

print "Mean of [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]: ", mean(range(1, 11))

print "Mode of [1, 1, 1, 1, 4, 4]:", mode([1, 1, 1, 1, 4, 4])

print "Median of [1, 2, 3, 4]:", median([1, 2, 3, 4])

main()

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Read 2 more answers

The extruder head in a fused- deposition modeling setup has a diameter of 1.25 mm (0.05 in) and produces layers that are 0.25mm

Angelina_Jolie [31]

Answer:

The time taken will be "1 hour 51 min". The further explanation is given below.

Explanation:

The given values are:

Number of required layers:

= $\frac{38}{0.25}$

= $152 \ layers$

Diameter (d):

= 1.25 mm

Velocity (v):

= 40 mm/s

Now,

The area of one layer will be:

= $38\times 38 \ mm^2$

= $1444 \ mm^2$

The area covered every \second will be:

= $d\times v$

= $1.25\times 40$

= $50 \ mm^2$

The time required to deposit one layer will be:

= $\frac{1444}{50}$

= $28.88 \ sec$

The time required for one layer will be:

= $15 \ sec$

∴ Total times required for one layer will be:

= $15+28.88$

= $43.88 \ sec$

So,

Number of layers = 152

Therefore,

Total time will be:

= $152\times 43.88$

= $6669.76 \ sec$

= $1 \ hour \ 51 \ min$

6 0

2 years ago