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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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AnswerWhat Are the Classifications of Burns? Burns are classified as first-, second-, or third-degree, depending on how deep and severe they penetrate the skin's surface. First-degree burns affect only the epidermis, or outer layer of skin. The burn site is red, painful, dry, and with no blisters.

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Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th
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Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine =1 - \frac{Tl}{Th} \\

where Th =temperature at which the heat enters the engine

Tl is the  temperature of the environment

since both engines have the same thermal capacities <em>n_{th} </em> therefore n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\

We have now that

\frac{-1400}{T}+\frac{T}{300}=0\\

multiplying through by T

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multiplying through by 300

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5 0
3 years ago
I need a thesis statement about Engineers as Leaders.
algol [13]

Answer:

Engineers are a very beneficial contribution in which offers great solutions to national problems.

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3 years ago
What mass of LP gas is necessary to heat 1.4 L of water from room temperature (25.0 ∘C) to boiling (100.0 ∘C)? Assume that durin
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m_{LP}=0.45\,kg

Explanation:

Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:

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Q_{water} = 3599.435\,kJ

The heat liberated by the LP gas is:

Q_{LP} = \frac{3599.435\,kJ}{0.16}

Q_{LP} = 22496.469\,kJ

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Answer:

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