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ss7ja [257]
3 years ago
12

The temperature controller for a clothes dryer consists of a bimetallic switch mounted on an electrical heater attached to a wal

l-mounted insulation pad. The switch is set to open at 70 °C, the maximum dryer air temperature. In order to operate the dryer at a lower air temperature, sufficient power is supplied to the heater such that the switch reaches 70 °C (Tset) when the air temperature T is less than Tset. If the convection heat transfer coefficient between the air and the exposed switch surface of 30 mm2 is 25 W/m2 ꞏ K.
Required:

How much heater power Pe is required when the desired dryer air temperature is T[infinity]=50°C?
Engineering
1 answer:
Naddika [18.5K]3 years ago
6 0

Answer:

0.015\ \text{W}

Explanation:

T_{set} = Set temperature = 70^{\circ}\text{C}

T_\infty = Air temperature = 50^{\circ}\text{C}

A = Surface area = 30\ \text{mm}^2=30\times 10^{-6}\ \text{m}^2

h = Convection heat transfer coefficient = 25\ \text{W/m}^2\text{K}

Heater power is given by

P_e=hA(T_{set}-T_\infty)\\\Rightarrow P_e=25\times 30\times 10^{-6}(70-50)\\\Rightarrow P_e=0.015\ \text{W}

The required heater power is 0.015\ \text{W}

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Calculate the wire pressure for a round copper bar with an original cross-sectional area of 12.56 mm2 to a 30% reduction of area
dybincka [34]

Answer:153.76 MPa

Explanation:

Initial Area\left ( A_0\right )=12.56 mm^2

Final Area\left ( A_f\right )=0.7\times 12.56 mm^2=8.792 mm^2

Die angle=30^{\circ}

\alpha =\frac{30}{2}=15^{\circ}

\mu =0.08

Yield stress\left ( \sigma _y \right )=350 MPa

B=\mu cot\left ( \aplha\right )=0.2985

\sigma _{pressure}=\sigma _y\left [\frac{1+B}{B}\right ]\left [ 1-\frac{A_f}{A_0}\right ]^B

\sigma _{pressure}=350\left [\frac{1+0.2985}{0.2985}\right ]\left [ 1-\frac{8.792}{12.56}\right ]^{0.2985}

\sigma _{pressure}=153.76 MPa

8 0
3 years ago
A concentrated load P is applied to the upper end of a 1.47-m-long pipe. The outside diameter of the pipe is D = 112 mm and the
myrzilka [38]

Answer:

Pmax = 38251.73 N

Explanation:

Given info

L = 1.47 m

D = 112 mm ⇒ R = D/2 = 112/2 mm = 56 mm

d = 101 mm  ⇒ r = D/2 = 101/2 mm = 50.5 mm

a) We can apply the following equation in order to get Q (First Moment of Area):

Q = 2*(A₁*y₁-A₂*y₂)

where

A₁ = π*R² = π*(56 mm)² = 3136 π mm²  

y₁ = 4*R/(3*π) = 4*56/(3*π) mm = 224/(3*π) mm

A₂ = π*r² = π*(50.5 mm)² = 2550.25 π mm²

y₂ = 4*r/(3*π) = 4*50.5/(3*π) mm = 202/(3*π) mm

then

Q = 2*(3136 π mm²*224/(3*π) mm-2550.25 π mm²*202/(3*π) mm)

⇒ Q = 62437.833 mm³

b) If  τallow = 83 MPa = 83 N/mm²

P = ?

We can use the equation

τ = V*Q / (t*I)   ⇒  V = τ*t*I / Q

where

t = D - d = 112 mm - 101 mm = 11 mm

I = (π/64)*(D⁴-d⁴) = (π/64)*((112 mm)⁴-(101 mm)⁴) = 2615942.11 mm⁴

Q = 62437.833 mm³

we could also use this equation in order to get Q:

Q = (4/3)*(R³-r³)

⇒  Q = (4/3)*((56 mm)³-(50.5 mm)³) = 62437.833 mm³

then we have

V = (83 N/mm²)*(11 mm)*(2615942.11 mm⁴) / (62437.833 mm³)

⇒ V = 2942.255 N

Finally Pmax = V = 38251.73 N

6 0
3 years ago
Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr
son4ous [18]

Answer:

The entropy change of the air is 0.240kJ/kgK

Explanation:

T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa

T_{2}  is unknown

we can apply the following expression to find T_{2}

-w_{out} =mc_{v} (T_{2} -T_{1} )

T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }

now substitute

T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K

To find entropy change of the air we can apply the ideal gas relationship

Δs_{air}=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }

Δs_{air} =1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )

Δs_{air} =0.240kJ/kgK

4 0
3 years ago
Which of the following types of protective equipment protects workers who are passing by from stray sparks or metal while anothe
lawyer [7]

A protective equipment which protects workers who are passing by from stray sparks or metal while another worker is welding is: E. Welding Screens.

A wielder refers to an individual who is saddled with responsibility of joining two or more metals together by wielding.

During the process of wielding, sparks and minute metallic objects are produced, which are usually hazardous to both the wielder and other workers within the vicinity.

Hence, the following protective equipment are meant to be worn or used directly by a wielder (worker) who is wielding:

  • Visors.
  • Goggles.
  • Protective clothing.
  • Dark walls.

However, a protective equipment which protects other workers who are passing by from stray sparks or metallic objects while wielder (worker) is welding is referred to as welding screens.

Find more information: brainly.com/question/15442363

4 0
2 years ago
How much does It cost to make a tracking chip
Basile [38]

Answer:

about 4 grand

Explanation:

6 0
2 years ago
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