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mr Goodwill [35]
2 years ago
11

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 wel

l insulated, while the front surface is exposed to the flow of a coolant at To 15°C. From reliability considerations, the chip temperature must not exceed T 85°C. Coolant T, h Chip If the coolant is air and the corresponding convection 200 W/m2 K, what is the maximum coefficient is h allowable chip power? If the coolant is a dielectric liq- uid for which h 3000 W/m2 K, what is the maxi- mum allowable power?

Engineering
1 answer:
Lynna [10]2 years ago
6 0

Answer:

a) 0.35 W

b) 5.25 W

Explanation:

Given that

Width of the chip, W = 5 mm

The environment temperature, T(∞) = 15° C

Surface temperature, T(s) = 85° C

The initial convection heat coefficient, h1 = 200 W/m².K

The final convection heat coefficient, h2 = 3000 W/m².K

See attachment for solution

You might be interested in
If an airplane travels at a speed of 1120 km/hr at an altitude of 15 km, what is the required speed at an altitude of 8 km to sa
jek_recluse [69]

Answer:

1170 km/hr

Explanation:

Let's first state the formula to be used

c = √(KRT)

The temperature at an altitude of 15km is -56.5° C

Let's not convert this to °K, we have

-56.5° + 273.15 = 216.65° K

Also, the temperature at 8km is -36.9° C, on converting to °K we have

-36.9° + 273.15 = 236.25° K

Then again, we look for the speed at both 15 km and 8 km both of which are 295 m/s and 308 km

Finally, we use the mach similarity formula

(V/c) of 15km = (V/c) of 8km

V of 8km = c of 8km * (V/c) of 15km

V of 8km = 1170 km/hr

6 0
2 years ago
An incompressible fluid flows between two infinite stationary parallel plates. The velocity profile is given by u=umaxðAy2 + By+
nexus9112 [7]

Answer:

the volume flow rate per unit depth is:

\frac{Q}{b} = \frac{2}{3} u_{max} h

the ratio is : \frac{V}{u_{max}}=\frac{2}{3}

Explanation:

From the question; the  equations of the velocities profile in the system are:

u = u_{max}(Ay^2+By+C)   ----- equation (1)

The above boundary condition can now be written as :

At y= 0; u =0           ----- (a)

At y = h; u =0            -----(b)

At y = \frac{h}{2} ; u = u_{max}     ------(c)

where ;

A,B and C are constant

h = distance between two plates

u = velocity

u_{max} = maximum velocity

y = measured distance upward from the lower plate

Replacing the boundary condition in (a) into equation (1) ; we have:

u = u_{max}(Ay^2+By+C) \\ \\ 0 = u_{max}(A*0+B*0+C) \\ \\ 0=u_{max}C \\ \\ C= 0

Replacing the boundary condition (b) in equation (1); we have:

u = u_{max}(Ay^2+By+C) \\ \\ 0 = u_{max}(A*h^2+B*h+C) \\ \\ 0 = Ah^2 +Bh + C \\ \\ 0 = Ah^2 +Bh + 0 \\ \\ Bh = - Ah^2 \\ \\ B = - Ah   \ \ \ \ \   --- (d)

Replacing the boundary condition (c) in equation (1); we have:

u = u_{max}(Ay^2+By+C) \\ \\ u_{max}= u_{max}(A*(\frac{h^2}{2})+B*\frac{h}{2}+C) \\ \\ 1 = \frac{Ah^2}{4} +B \frac{h}{2} + 0 \\ \\ 1 =  \frac{Ah^2}{4} + \frac{h}{2}(-Ah)  \\ \\ 1=  \frac{Ah^2}{4}  - \frac{Ah^2}{2}  \\ \\ 1 = \frac{Ah^2 - Ah^2}{4}  \\ \\ A = -\frac{4}{h^2}

replacing A = -\frac{4}{h^2} for A in (d); we get:

B = - ( -\frac{4}{h^2})hB = \frac{4}{h}

replacing the values of A, B and C into the velocity profile expression; we have:

u = u_{max}(Ay^2+By+C) \\ \\ u = u_{max} (-\frac{4}{h^2}y^2+\frac{4}{h}y)

To determine the volume flow rate; we have:

Q = AV \\ \\ Q= \int\limits^h_0 (u.bdy)

Replacing u_{max} (-\frac{4}{h^2}y^2+\frac{4}{h}y) \ for \ u

\frac{Q}{b} = \int\limits^h_0 u_{max}(-\frac{4}{h^2} y^2+\frac{4}{h}y)dy \\ \\  \frac{Q}{b} = u_{max}  \int\limits^h_0 (-\frac{4}{h^2} y^2+\frac{4}{h}y)dy \\ \\ \frac{Q}{b} = u_{max} (-\frac{-4}{h^2}\frac{y^3}{3} +\frac{4}{h}\frac{y^2}{y})^ ^ h}}__0  }} \\ \\ \frac{Q}{b} =u_{max} (-\frac{-4}{h^2}\frac{h^3}{3} +\frac{4}{h}\frac{h^2}{y})^ ^ h}}__0  }} \\ \\ \frac{Q}{b} = u_{max}(\frac{-4h}{3}+\frac{4h}2} ) \\ \\ \frac{Q}{b} = u_{max}(\frac{-8h+12h}{6}) \\ \\ \frac{Q}{b} =u_{max}(\frac{4h}{6})

\frac{Q}{b} = u_{max}(\frac{2h}{3}) \\ \\ \frac{Q}{b} = \frac{2}{3} u_{max} h

Thus; the volume flow rate per unit depth is:

\frac{Q}{b} = \frac{2}{3} u_{max} h

Consider the discharge ;

Q = VA

where :

A = bh

Q = Vbh

\frac{Q}{b}= Vh

Also;  \frac{Q}{b} = \frac{2}{3} u_{max} h

Then;

\frac{2}{3} u_{max} h = Vh \\ \\ \frac{V}{u_{max}}=\frac{2}{3}

Thus; the ratio is : \frac{V}{u_{max}}=\frac{2}{3}

5 0
3 years ago
Estimate the chronic daily intake of toluene from exposure to a city water supply that contains a toluene concentration equal to
vredina [299]

Answer: the total chronic daily intake is  0.03296 mg/ kg.d  

Explanation:

First of  we have to determine the CDI due to ingestion of drinking water

so

CDI = (C x IR x EF x ED) / (BW x AT)     let leave this as equation 1

where C is the concentration of the life time risk of drinking water(1.0mg/L)

ED is the risk dying per year((70),

EF is the number of days per year(365 day/year)

BW is the weight of body(65.4)kg

AT is the average tenure time of life(75years)

IR = 2.3 L/days

so we substitute

CDI₁ = (1(mg/L) x 2.3(L/day) x 365 x 70) / (65.4 x 365 x 75)

= 3.29 x 10⁻² mg/kg.d  

next we determine the CDI due to dermal contact with water.  

find the ingestion rate which is equal to the exposure time in hour/day.

IR = ET

(20 min/day) /  (60 min/hour)

= 0.333 h/day

now lets substitute for CD1₂

1.0 mg/L for C , 0.333 h/day for m , 365 day/year for EF, 70 year for ED, 65.4 kg for Bw , and 75 year for AT

0.8 for submergence and 1.69m² for area of skin of adult female in our equation 1

CDI₂ = [(1 mg/1) × (1.69 m²) × (9.0 x 10⁻⁶(m/h)) × 0.333(h/day) x 365 x 70)) / (65.4 x 365 x 75)]  × ( 0.8 x 10³ L/m³)

CDI₂ = 5.41 x 10⁻⁵ mg/kg.d  

now we find the CDI due to inhalation during bath

we substitute 1.0 mg/L for C, 11.3m³/day for IR , 365 day/year for EF , 70 year for ED. 65.4 kg for BW , and 75 year for AT in our equation 1

CDI₃ = [( 1ug/m³ × 1mg/10³ug) x (11.3m³/day x 1day/24 hr) x 365 x 70)] / (65.4 kg x 365 x 75 )

= 6.71 x 10⁻⁶ mg/kg.d

finally we Calculate the total chronic daily intake value

CDI_total = CDI₁ + CDI₂ + CDI₃  

we substitute

CDI_total = 3.29 x 10⁻² + 5.41 x 10⁻⁵ + 6.71 x 10⁻⁶

= 0.03296 mg/kg.d  

so the total chronic daily intake is  0.03296 mg/ kg.d  

4 0
3 years ago
A wet electrode can cause arc blow ?
irga5000 [103]

Answer:

yes it can

Explanation:

6 0
2 years ago
Design a synchronous 3-bit binary UP/DOWN counter uses the following counting pattern: (0,1,3,7,6,4,0,1,3….) the counter will co
telo118 [61]

Complete Question:

Find attached below file 1 and 2 contains the complete questions

Explanation:

image 3 shows the following

state diagram

state transition table and

answer to question 3

other added solutions are comprehensive and additional

4 0
2 years ago
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