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KatRina [158]
3 years ago
10

A 1200-W iron is left on the iron board with its base exposed to ambient air at 26°C. The base plate of the iron has a thickness

of L = 0.5 cm, base area of A = 150 cm2 , and thermal conductivity of k = 18 W/m·K. The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. The outer surface of the base plate whose emissivity is ε = 0.7, loses heat by convection to ambient air with an average heat transfer coefficient of h = 30 W/m2 ·K as well as by radiation to the surrounding surfaces at an average temperature of Tsurr = 295 K. Disregarding any heat loss through the upper part of the iron,
(a) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the plate,


(b) obtain a relation for the temperature of the outer surface of the plate by solving the differential equation, and


(c) evaluate the outer surface temperature.

Engineering
1 answer:
faust18 [17]3 years ago
4 0

Answer:

See attachments for steo by step procedures into getting answers.

Explanation:

Given that;

A 1200-W iron is left on the iron board with its base exposed to ambient air at 26°C. The base plate of the iron has a thickness of L = 0.5 cm, base area of A = 150 cm2 , and thermal conductivity of k = 18 W/m·K. The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. The outer surface of the base plate whose emissivity is ε = 0.7, loses heat by convection to ambient air with an average heat transfer coefficient of h = 30 W/m2 ·K as well as by radiation to the surrounding surfaces at an average temperature of Tsurr = 295 K. Disregarding any heat loss through the upper part of the iron,

(a) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the plate,

(b) obtain a relation for the temperature of the outer surface of the plate by solving the differential equation, and

(c) evaluate the outer surface temperature.

See attachments for comolete solving

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Explain and show work:
belka [17]

Answer:

  5

Explanation:

The sum of the digits of the number is ...

  (4+1+3)+(4+6+5)+(7+8+9) = 8+15+24 = 47

The sum of those digits is 4+7=11, and those digits sum to 1+1 = 2.

That is, the value of the number mod 9 (or 3) is 2.

The ones digit is odd, so the value of the number mod 2 is 1.

This combination of modulo values tells you the mod 6 result is 5.

_____

<em>Additional comment</em>

We can look at the (mod2, mod3) values of the numbers 0 to 5:

  0 ⇒ (0, 0)

  1 ⇒ (1, 1)

  2 ⇒ (0, 2)

  3 ⇒ (1, 0)

  4 ⇒ (0, 1)

  5 ⇒ (1, 2) . . . . the mod {2, 3} results we have for the number of interest.

This process of adding up the digits repeatedly is referred to as "casting out 9s." The result of it is the modulo 9 value of the number (with 0 mapped to 9). Checking the mod 9 result of arithmetic operations is one quick way to spot certain kinds of errors. It can also be used as part of a divisibility test for 3 or 9.

3 0
3 years ago
What is the function of filters in communication system ?​
mihalych1998 [28]

In signal processing, a filter is a device or process that removes some unwanted components or features from a signal. Filtering is a class of signal processing, the defining feature of filters being the complete or partial suppression of some aspect of the signal. Most often, this means removing some frequencies or frequency bands. However, filters do not exclusively act in the frequency domain; especially in the field of image processing many other targets for filtering exist.

Explanation:

8 0
3 years ago
If coolant is mixed 50-50 with water, what is the freezing point?
sashaice [31]

Answer:

-35 degrees F

When mixed in equal parts with water (50/50), antifreeze lowers the freezing point to -35 degrees F and raises the boiling temperature to 223 degrees F. Antifreeze also includes corrosion inhibitors to protect the engine and cooling system against rust and corrosion.

8 0
3 years ago
A three-phase voltage source with a terminal voltage of 22kV is connected to a three-phase transformer rated 5MVA 22kV/220V. The
amid [387]

Answer:

A) See Attachment B) See Attachment C) 186.153

Explanation:

C) Per phase Voltage= 220/∛3

                                  = 6.037

S=V²/Z

  = (6.037)²/(0.01+i0.05)

  = 62051.282-i310256.41

Active power losses per phase= 62.051kW

total Active power losses= 62.051×3

                                         =186.153kW

6 0
4 years ago
Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a)
vekshin1

Answer:

a) Using Ideal gas Equation, T = 434.98°R = 435°R

b) Using Van Der Waal's Equation, T = 637.32°R = 637°R

c) T obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is T = 559.67°R = 560°R

Explanation:

a) Ideal gas Equation

PV = mRT

T = PV/mR

P = pressure = 400 psia

V/m = specific volume = 0.1144 ft³/lbm

R = gas constant = 0.1052 psia.ft³/lbm.°R

T = 400 × 0.1144/0.1052 = 434.98 °R

b) Van Der Waal's Equation

T = (1/R) (P + (a/v²)) (v - b)

a = Van Der Waal's constant = (27R²(T꜀ᵣ)²)/(64P꜀ᵣ)

R = 0.1052 psia.ft³/lbm.°R

T꜀ᵣ = critical temperature for refrigerant-134a (from the refrigerant tables) = 673.6°R

P꜀ᵣ = critical pressure for refrigerant-134a (from the refrigerant tables) = 588.7 psia

a = (27 × 0.1052² × 673.6²)/(64 × 588.7)

a = 3.596 ft⁶.psia/lbm²

b = (RT꜀ᵣ)/8P꜀ᵣ

b = (0.1052 × 673.6)/(8 × 588.7) = 0.01504 ft³/lbm

T = (1/0.1052) (400 + (3.596/0.1144²) (0.1144 - 0.01504) = 637.32°R

c) The temperature for the refrigerant-134a as obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is

T = 100°F = 559.67°R

7 0
4 years ago
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