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damaskus [11]
3 years ago
8

Volume of sale (i.e., the number of parts sold) is a factorwhen determining which

Engineering
1 answer:
nadya68 [22]3 years ago
3 0

Answer: N has to be lesser than or equal to 1666.

Explanation:

Cost of parts N in FPGA = $15N

Cost of parts N in gate array = $3N + $20000

Cost of parts N in standard cell = $1N + $100000

So,

15N < 3N + 20000 lets say this is equation 1

(cost of FPGA lesser than that of gate array)

 Also. 15N < 1N + 100000  lets say this is equation 2  

(cost of FPGA lesser than that of standardcell)

Now

From equation 1

12N < 20000

N < 1666.67

From equation 2

14N < 100000

N < 7142.85

AT the same time, Both conditions must hold true

So N <= 1666 (Since N has to be an integer)

N has to be lesser than or equal to 1666.

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A column carries 5400 pounds of load and is supported on a spread footing. The footing rests on coarse sand. Design the smallest
Citrus2011 [14]

Answer:

Following are the responses to the given question:

Explanation:

7 0
2 years ago
B1) 20 pts. The thickness of each of the two sheets to be resistance spot welded is 3.5 mm. It is desired to form a weld nugget
kap26 [50]

Answer:

minimum current level required =  8975.95 amperes

Explanation:

Given data:

diameter = 5.5 mm

length = 5.0 mm

T = 0.3

unit melting energy = 9.5 j/mm^3

electrical resistance = 140 micro ohms

thickness of each of the two sheets = 3.5mm

Determine the minimum current level required

first we calculate the volume of the weld nugget

v = \frac{\pi }{4} * D^2 * l = \frac{\pi }{4} * 5.5^2 * 5 = 118.73 mm^3

next calculate the required melting energy

= volume of weld nugget * unit melting energy

= 118.73 * 9.5 = 1127.94 joules

next find the actual required electric energy

= required melting energy / efficiency

= 1127 .94 / ( 1/3 )  = 3383.84 J

TO DETERMINE THE CURRENT LEVEL REQUIRED  use the relation below

electrical energy =  I^2 * R * T

3383.84 / R*T = I^2

3383.84 / (( 140 * 10^-6 ) * 0.3 ) = I^2

therefore  8975.95 = I ( current )

4 0
3 years ago
A 5-mm-thick stainless steel strip (k = 21 W/m•K, rho = 8000 kg/m3, and cp = 570 J/kg•K) is being heat treated as it moves throu
Drupady [299]

Answer:

The temperature of the strip as it exits the furnace is 819.15 °C

Explanation:

The characteristic length of the strip is given by;

L_c = \frac{V}{A} = \frac{LA}{2A} = \frac{5*10^{-3}}{2} = 0.0025 \ m

The Biot number is given as;

B_i = \frac{h L_c}{k}\\\\B_i = \frac{80*0.0025}{21} \\\\B_i = 0.00952

B_i < 0.1,  thus apply lumped system approximation to determine the constant time for the process;

\tau = \frac{\rho C_p V}{hA_s} = \frac{\rho C_p L_c}{h}\\\\\tau = \frac{8000* 570* 0.0025}{80}\\\\\tau = 142.5 s

The time for the heating process is given as;

t = \frac{d}{V} \\\\t = \frac{3 \ m}{0.01 \ m/s} = 300 s

Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;

T(t) = T_{ \infty} + (T_i -T_{\infty})e^{-t/ \tau}\\\\T(t) = 930 + (20 -930)e^{-300/ 142.5}\\\\T(t) = 930 + (-110.85)\\\\T_{(t)} = 819.15 \ ^0 C

Therefore, the temperature of the strip as it exits the furnace is 819.15 °C

5 0
3 years ago
A natural-draft cooling tower receives 250,000 ft3/min of air at standard atmospheric pressure, 70oF, and 45 percent relative hu
notsponge [240]

Find the attachment for complete solution

5 0
3 years ago
Given below are the measured streamflows in cfs from a storm of 6-hour duration on a stream having a drainage area of 185 mi^2.
sertanlavr [38]

Answer:

33.56 ft^3/sec.in

Explanation:

Duration = 6 hours

drainage area = 185 mi^2

constant baseflow = 550 cfs

<u>Derive the unit hydrograph using the inverse procedure </u>

first step : calculate for the volume of direct runoff hydrograph using the details in table 2 attached below

Vdrh = sum of drh *  duration

        = 29700 * 6 hours ( 216000 secs )

        = 641,520,000 ft^3.

next step : Calculate the volume of runoff in equivalent depth

Vdrh / Area = 641,520,000  / 185 mi^2

                    = 1.49 in

Finally derive the unit hydrograph

Unit of hydrograph = drh /  volume of runoff in equivalent depth

                                = 50 ft^3 / 1.49 in  =  33.56 ft^3/sec.in

5 0
3 years ago
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