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Yuliya22 [10]
3 years ago
9

An electronic device, such as a power transistor mounted on a finned heat sink, can be modeled as a spatially isothermal object

with internal heat generation and an external convection resistance. (a) Consider such system of mas M, specific heat c, and surface area As, which is initially in equilibrium with the environment at To. Suddenly, the electronic device is energized such that a constant heat generation Eg(W) occurs. Show that the temperature response of the device is: θ = exp RC where θ Ξ T-T(w): T(oo) is the steady-state temperature corresponding to t → oo; 0Ti T(; T: initial temperature of the device; R: thermal resistance 1/hAs; and C: thermal capacitance Mc. (b) An electronic device, which generates 60W of heat, is mounted on an aluminum heat sink weighing 0.31 kg and reaches a temperature of 100°C in ambient air at 20°C under steady- after the power is switched on?
Engineering
1 answer:
Kobotan [32]3 years ago
7 0

Answer:

hello your question is incomplete attached below is the complete question

answer :

a) attached below

b) 63.71°c

Explanation:

A) Given data

The system has ; mass ( M ) , specific heat capacity , and surface area

attached below is a detailed solution

B) Given data

power generated = 60w

weight of heat sink = 0.31 kg

temperature given = 100°c

temp in ambient air = 20°c

lets take

specific heat capacity = 918.18 J /kg-k

density = 2702 kg/m^2

attached below is the detailed solution

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If E = 94.2 mJ of energy is transferred when Q = 1.66 C of charge flows through a circuit element, what is the voltage across th
Anni [7]

Answer:

V = 56.8 mV

Explanation:

When a current I flows across a circuit element, if we assume that the dimensions of the circuit are much less than the wavelength of the power source creating this current, there exists a fixed relationship between the power dissipated in the circuit element, the current I and the voltage V across it, as follows:

P = V*I

By definition, power is the rate of change of energy, and current, the rate of change of the charge Q, so we can replace P and I, as follows:

E/t = V*q/t ⇒ E = V*Q

Solving for V:

V = E/Q = 94.2 mJ /1.66 C = 56.8 mV

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3 years ago
The temperature in a pressure cooker is 130 degree C while the water is boiling. Determine the pressure inside the cooker.
emmasim [6.3K]

Answer:

The pressure inside the cooker is 1.0804 atm.

Explanation:

Boiling occurs when the vapor pressure becomes equal to atmospheric pressure.

<u>For water, At standard conditions (Pressure = 1 atm) boiling occurs at 373.15 K.</u>

So, Standard conditions:

T₁ = 373.15 K

P₁ = 1 atm

Given ,

The water boils at Temperature =  130 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So, the temperature, T₂ = (130 + 273.15) K = 403.15 K

To find pressure inside the cooker (P₂) :

<u>Applying Amontons's Law as:</u>

\frac {P_1}{T_1}=\frac {P_2}{T_2}

So,

P_2=\frac {P_1}{T_1} \times {T_2}

P_2=\frac {1 atm}{373.15 K} \times {403.15 K}

P_2=1.0804 atm

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Answer:

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3 years ago
I am making composites of silicone rubber and copper particles; by mixing thermally conductive particles into the thermally insu
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Answer:

The composition of Composite of Volume basis is 2.154% Copper and 97.83% Rubber.

Explanation:

Let us assume that the total mass of composite is 100 lbm So as per the given conditions

  • 15 lbm is copper and 85 lbm is rubber.
  • Density of rubber is 70 lbm/ft3
  • Specific gravity of Copper is 9

So As per the formula of specific gravity

                                         S_{cu}=\frac{\rho_{cu}}{\rho_w}

here density of water is 62.4 lbm/ft3

Solving for Density of Copper gives

                                        S_{cu}=\frac{\rho_{cu}}{\rho_w}\\9=\frac{\rho_{cu}}{62.4}\\\rho_{cu}=9 \times 62.4\\\rho_{cu}=561.5 lbm/ft3

For composition on volume basis, volume of individual components and composite are calculated as

                                          V_{cu}=\frac{m_{cu}}{\rho_{cu}}\\V_{cu}=\frac{15}{561.5}\\V_{cu}=0.0267 ft^3\\\\V_{r}=\frac{m_{r}}{\rho_{r}}\\V_{r}=\frac{85}{70}\\V_{r}=1.214 ft^3\\\\V_{c}=V_{r}+V_{cu}\\V_{c}=1.214+0.0267 \\V_{c}=1.2409 ft^3

The composition is given as

c_{cu}=\frac{V_{cu}}{V_{c}}\\c_{cu}=\frac{0.0267}{1.2409} \times 100 \%\\c_{cu}=2.154 \%\\\\c_{r}=\frac{V_{r}}{V_{c}}\\c_{r}=\frac{1.214}{1.2409} \times 100 \%\\c_{r}=97.83 \%

So the composition of Composite of Volume basis is 2.154% Copper and 97.83% Rubber.

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