Engineering College

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?

Answers

Answer 1

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

An Electronic Device, Such As A Power Transistor Mounted On A Finned Heat Sink, Can Be Modeled As A Spatially

Related Questions

A resistor and a capacitor are connected in series across a 150 Vac supply. When
the frequency is 40 Hz the current is 5 A, and when the frequency is 50 Hz the
current is 6 A. Find the resistance and capacitance for the of the resistor and
capacitor respectively.

Answers

Explanation:

for 40hz

since they are in series they will have the same amper

so R=30ohm

C=30F

for 50Hz

R=25

C=25

The entropy of the working fluid of the ideal Carnot cycle (increases, decreases, remains the same) during the isothermal heat rejection process. (Explain the reasoning behind your answer selection!)

Answers

Answer:

Decreases

Explanation:

The Entropy change for a reversible process is;

ΔS = nRIn(V_b/V_a)

Now, applying this formula to an isothermal expansion in an irreversible process, it will have same initial and final states as we got for the reversible expansion process. Now, due to the fact that entropy is a state function, the change in entropy of the system will be the same as we got for the equation in reversible process and is given by the same formula.

In both the reversible and irreversible process, the surroundings of which they differ by the entropy there are usually at a constant temperature, T in a way that we have the change jn entropy relationship as;

ΔS_sur = −Q/T

Where;

- the negative sign is used since the heat transferred to the surroundings is equal in magnitude and opposite in sign to the heat, Q, transferred to the system.

Thus, it is clear that the change in the entropy decreases due to the negative sign.

Example 5 10, P 236 9th edition Refrigerant 134 a at 700 kPa , 70 deg C and 8 kg/min is cooled by water in a condenser until it exists as a saturated liquid at the same pressure. The cooling water enters the condenser at 300 kPa and 15 deg C and leaves at 25 deg C at the same pressure. Determine the mass flow rate of the cooling water required to cool the refrigerant.

Answers

Solution :

The mass flow rate of the cooling water is determined from the energy balance equation. The enthalpies of the refrigerant at the initial and the final states were obtained from A-13 and A-12 for the given parameters.

Therefore by energy balance equation,

[tex]$\dot m_r h_{r1}=\dot m_w \Delta h_w +\dot m_r h_{r2} $[/tex]

[tex]$\dot m_w = \frac{\dot m(h_1-h_2)_r}{c \Delta T}$[/tex]

[tex]$\dot m_w = \frac{8(308.34-88.82)}{4.18 \times 10}$[/tex]

= 42 kg / min

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