HELP PLEASE!!!!!!!!!!!

The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

8 0

3 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()

3 0

4 years ago

Read 2 more answers

Suppose you want to buy a new car. The car will be used mainly for going to work, shopping, running errands, and visiting friend

Sergeu [11.5K]

1. So you can complete the things you NEED to do faster. (I.e work, groceries)

2. So it’s easier for you to do leisure activities. (I.e visiting friends, shopping)

5 0

3 years ago

An insulated mixing chamber receives 0.5 kg/s of steam at 3 MPa and 300°C through one inlet, and saturated liquid water at 3 MPa

maxonik [38]

Answer:

$\dot m_{2} = 0.199\,\frac{kg}{s}$

Explanation:

The mixing chamber can be modelled by applying the First Law of Thermodynamics:

$\dot W_{in}+\dot m_{1}\cdot h_{1} +\dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0$

Since that mass flow rate of water at inlet 1 is the only known variable, the expression has to be simplified like this:

$\frac{\dot W_{in}}{\dot m_{1}} + h_{1}+ y\cdot h_{2} - z\cdot h_{3} = 0$

Besides, the following expression derived from the Principle of Mass Conservation is presented below:

$1 + y = z$

Then, the expression is simplified afterwards:

$\frac{\dot W_{in}}{\dot m_{1}} + h_{1}+ y\cdot h_{2} - (1+y)\cdot h_{3} = 0$

$\frac{\dot W_{in}}{\dot m_{1}} +h_{1} - h_{3} + y\cdot (h_{2}-h_{3}) = 0$

Specific enthalpies are obtained from steam tables and described as follows:

State 1 (Superheated vapor)

$h = 2994.3\,\frac{kJ}{kg}$

State 2 (Saturated liquid)

$h = 1008.3\,\frac{kJ}{kg}$

State 3 (Liquid-Vapor mixture)

$h = 2444.22\,\frac{kJ}{kg}$

The ratio of the stream at state 2 to the stream at state 1 is:

$y = \frac{\frac{\dot W_{in}}{\dot m_{1}}+h_{1}-h_{3}}{h_{3}-h_{2}}$

$y = \frac{\frac{10\,kW}{0.5\,\frac{kg}{s} }+2994.3\,\frac{kJ}{kg}-2444.22\,\frac{kJ}{kg} }{2444.22\,\frac{kJ}{kg}-1008.3\,\frac{kJ}{kg} }$

$y = 0.397$

The mass flow rate of the saturated liquid is:

$\dot m_{2} = y\cdot \dot m_{1}$

$\dot m_{2} = 0.397\cdot (0.5\,\frac{kg}{s} )$

$\dot m_{2} = 0.199\,\frac{kg}{s}$

5 0

3 years ago

Which of the following ranges depicts the 2% tolerance range to the full 9 digits provided?

Lyrx [107]

Answer:

the only one that meets the requirements is option C .

Explanation:

The tolerance of a quantity is the maximum limit of variation allowed for that quantity.

To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula

x_average = ∑ $x_{i}$ / n

The tolerance or error is the current value over the mean value per 100

Δx₁ = x₁ / x_average

tolerance = | 100 -Δx₁ 100 |

bars indicate absolute value

let's look for these values for each case

a)

x_average = (2.1700000+ 2.258571429) / 2

x_average = 2.2142857145

fluctuation for x₁

Δx₁ = 2.17000 / 2.2142857145

Tolerance = 100 - 97.999999991

Tolerance = 2.000000001%

fluctuation x₂

Δx₂ = 2.258571429 / 2.2142857145

Δx2 = 1.02

tolerance = 100 - 102.000000009

tolerance 2.000000001%

b)

x_average = (2.2 + 2.29) / 2

x_average = 2,245

fluctuation x₁

Δx₁ = 2.2 / 2.245

Δx₁ = 0.9799554

tolerance = 100 - 97,999

Tolerance = 2.00446%

fluctuation x₂

Δx₂ = 2.29 / 2.245

Δx₂ = 1.0200445

Tolerance = 2.00445%

c)

x_average = (2.211445 +2.3) / 2

x_average = 2.2557225

Δx₁ = 2.211445 / 2.2557225 = 0.9803710

tolerance = 100 - 98.0371

tolerance = 1.96%

Δx₂ = 2.3 / 2.2557225 = 1.024624

tolerance = 100 -101.962896

tolerance = 1.96%

d)

x_average = (2.20144927 + 2.29130435) / 2

x_average = 2.24637681

Δx₁ = 2.20144927 / 2.24637681 = 0.98000043

tolerance = 100 - 98.000043

tolerance = 2.000002%

Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017

tolerance = 2.0000002%

e)

x_average = (2 +2,3) / 2

x_average = 2.15

Δx₁ = 2 / 2.15 = 0.93023

tolerance = 100 -93.023

tolerance = 6.98%

Δx₂ = 2.3 / 2.15 = 1.0698

tolerance = 6.97%

Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .

4 0

4 years ago