2.

A square isothermal chip is of width w = 5 mm on a side and is mounted in a substrate such that its side and back surfaces are w

adelina 88 [10]

Answer:

Maximum allowable chip power is 0.35 W

Explanation:

This question is incomplete. The complete question is

A square isothermal chip is of width w = 5 mm on a side and is mounted in a substrate such that its side and back surfaces are well insulated; the front surface is exposed to the flow of a coolant at t[infinity] = 15°c. from reliability considerations, the chip temperature must not exceed t = 85°c. f the coolant is air and the corresponding convection 200 w/m2 k, what is the maximum allowable chip power?

ANSWER:

The heat transfer through convection, we have the equation:

q = hA(T - T∞)

where,

q = power transfer through convection = ?

h = convection coefficient = 200 W/m²K

A = Area of convection surface = (0.005 m)² = 0.000025 m²

T = Chip surface temperature = 85° C

T∞ = Fluid temperature = 15° C

Therefore,

q = (200 W/m².K)(0.000025 m²)(85° C - 15° C)

q = 0.35 W

Since, difference in temperature is same on both Celsius and kelvin scale. Therefore, Celsius is written as kelvin for difference and they shall be cancelled.

3 0

4 years ago

Air initially at 15 psla and 60 F is compressed to 75 psia and 400 F. The power input to air under steady state condition is 5 h

Triss [41]

Answer: $\dot{m}=3.46lbm/min$

Explanation:

Initial conditions

$P_1=15 psia$

$T_1=60 F^{\circ}$

Final conditions

$P_2=75 psia$

$T_2=400F^{\circ}$

Steady flow energy equation

$\dot{m}\left [ h_1+\frac{v_1^2}{2}+gz_1\right ]+\dot{Q}=\dot{m}\left [ h_2+[tex]\frac{v_2^2}{2}+gz_2\right ]+\dot{W}$

$\dot{m}\left [ c_pT_1+\frac{0^2}{2}+g0\right ]+\dot{Q}=\dot{m}\left [ c_pT_2+\frac{0^2}{2}+g0\right ]+\dot{W}$

$\dot{m}c_p\left [ T_1-T_2\right ]+\left [ -5hp\right ]=\dot{W} -5\times 746\times 3.4121$

$-4\dot{m}-\dot{m}\times 0.24\times \left [ 400-60\right ]$

$-81.6\dot{m}-4\dot{m}=-4.949 BTU/sec$

$\dot{m}=0.057821lbm/sec$

$\dot{m}=3.46lbm/min$

3 0

3 years ago

A journeyman electrician with 16 years experience on-the-job was removing metal fish tape from a hole at the base of a metal lig

Romashka-Z-Leto [24]

Answer: He should have ensured the following before beginning work;

1) All circuits must be de-energized before beginning work

2) All controls must be deactivated during work. LOTO(Log out tag out) must be practiced irrespective of the experience of the electrian or worker.

3) Technicians and Electrical workers must be instructed to know the unsafe conditions associated with their work.

Explanation:

We cannot assume that because he is experienced he cannot be prone to all this wrong practices. Humans tend to think that with experience people become less prone to errors and study has shown that this is not true. Safety must be a priority.

5 0

3 years ago

Given: A graphite-moderated nuclear reactor. Heat is generated uniformly in uranium rods of 0.05m diameter at the rate of 7.5 x

sineoko [7]

Answer:

The maximum temperature at the center of the rod is found to be 517.24 °C

Explanation:

Assumptions:

1- Heat transfer is steady.

2- Heat transfer is in one dimension, due to axial symmetry.

3- Heat generation is uniform.

Now, we consider an inner imaginary cylinder of radius R inside the actual uranium rod of radius Ro. So, from steady state conditions, we know that, heat generated within the rod will be equal to the heat conducted at any point of the rod. So, from Fourier's Law, we write:

Heat Conduction Through Rod = Heat Generation

-kAdT/dr = qV

where,

k = thermal conductivity = 29.5 W/m.K

q = heat generation per unit volume = 7.5 x 10^7 W/m³

V = volume of rod = π r² l

A = area of rod = 2π r l

using these values, we get:

dT = - (q/2k)(r dr)

integrating from r = 0, where T(0) = To = Maximum center temperature, to r = Ro, where, T(Ro) = Ts = surface temperature = 120°C.

To -Ts = qr²/4k

To = Ts + qr²/4k

To = 120°C + (7.5 x 10^7 W/m³)(0.025 m)²/(4)(29.5 W/m.°C)

To = 120° C + 397.24° C

To = 517.24° C

5 0

4 years ago

Yfchggfcvkuhgvjbjhbsdfhjkibgihdfgb

Molodets [167]

Answer:

dysurygfyubiugyunhuhfnbgjdsbnm

Explanation: Because ngubduykrtvytiudhurthgvniuonssdf

8 0

3 years ago