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DedPeter [7]
3 years ago
10

Is it possible to interface an IC with a different technology such as TTL to HCS12 ports? What are the conditions in terms of el

ectrical parameters that need to be satisfied for this purpose?
Engineering
1 answer:
damaskus [11]3 years ago
6 0

Answer:

The condition is true  when their voltage and current specifications with their impedance are matched or complementary to each other.

Explanation:

Solution

Yes it is possible or true to interface an IC with a different technology like the TTL to HCS12 ports. but the condition is that their current and voltage specifications should be matched and their impedance and power also should be matched.

What this implies is that both their voltage and current requirements should be complementary to each other so as their impedance.

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Air at 1 atm enters a thin-walled ( 5-mm diameter) long tube ( 2 m) at an inlet temperature of 100°C. A constant heat flux is ap
alexdok [17]

Answer:

heat rate   = 7.38 W

Explanation:

Given Data:

Pressure = 1atm

diameter (D) = 5mm = 0.005m

length = 2

mass flow rate (m) = 140*10^-6 kg/s

Exit temperature = 160°C,

At 400K,

Dynamic viscosity (μ) = 22.87 *10^-6

Prandtl number (pr) = 0.688

Thermal conductivity (k) = 33.65 *10^-3 W/m-k

Specific heat (Cp) = 1.013kj/kg.K

Step 1: Calculating Reynolds number using the formula;

Re = 4m/πDμ

     = (4*140*10^-6)/(π* 0.005*22.87 *10^-6)

     = 5.6*10^-4/3.59*10^-7

     = 1559.

Step 2: Calculating the thermal entry length using the formula

Le = 0.05*Re*Pr*D

Substituting, we have

Le = 0.05 * 1559 * 0.688 *0.005

Le = 0.268

Step 3: Calculate the heat transfer coefficient  using the formula;

Nu = hD/k

h = Nu*k/D

Since Le is less than given length, Nusselt number (Nu) for fully developed flow and uniform surface heat flux is 4.36.

h = 4.36 * 33.65 *10^-3/0.005

h = 0.1467/0.005

h = 29.34 W/m²-k

Step 4: Calculating the surface area using the formula;

A = πDl

   =π * 0.005 * 2

    =0.0314 m²

Step 5: Calculating the temperature Tm

For energy balance,

Qc = Qh

Therefore,

H*A(Te-Tm) = MCp(Tm - Ti)

29.34* 0.0314(160-Tm) =  140 × 10-6* 1.013*10^3 (Tm-100)

0.921(160-Tm) = 0.14182(Tm-100)

     147.36 -0.921Tm = 0.14182Tm - 14.182

1.06282Tm = 161.542

Tm = 161.542/1.06282

Tm = 151.99 K

Step 6: Calculate the rate of heat transferred using the formula

Q = H*A(Te-Tm)

   = 29.34* 0.0314(160-151.99)

  = 7.38 W

the Prandtl number using the formula

5 0
3 years ago
In a lab, scientists grew several generations of offspring of a plant using the method shown. What conclusion can you make about
exis [7]
Answer: c) they have low genetic variability among them.


When a plant is grown for several generations of offspring of a plant, then there are some common things which are to be noted which are found similar in the offspring and in the parent of the offspring. The flowers and fruits and the time or season they come in are absolutely the same.
6 0
1 year ago
Which career related to architecture deals with the planning of entire cities and focuses on designing and arranging buildings,
snow_lady [41]

Answer:

City planner would be a good answer. Unless there are any options, this seems like the best bet. Hope this helps!

8 0
3 years ago
Read 2 more answers
Suppose that the cost of electrical energy is $0.15 per kilowatt hour and that your electrical bill for 30 days is $80. Assume t
mr_godi [17]

Answer:

a) 740 W b) 6.2 A c) 8.1%

Explanation:

We need first to get the total energy spent during the 30 days, that can be calculated as follows:

1 month = 30 days = 720 hr

If the total cost amounts $80 (for 720 hr), and the cost per kwh is 0.15, we have:

80 $/mo  =  0.15 $/Kwh*x Kwh/mo

Solving for the total energy spent in the month:

x (kwh/mo) = \frac{80}{0.15} = 533.3 kwh

Assuming that the power delivered is constant over the entire 30 days, as power is the rate of change of energy, we can find the power as follows:

P = \frac{E}{t} = \frac{533.3 kWh}{720 h}  = 0.74 kW = 740 W

b) If the power is supplied by a voltage of 120 V, we can find the current I as follows:

I =\frac{P}{V} =\frac{740W}{120V} = 6.2A

c) If part of the electrical load is a 60-W light, we can substract this power from the one we have just found, as follows:

P = 740 W - 60 W = 680 W

The new value of the energy spent during the entire month will be as follows:

E = 0.68 kW*24(hr/day)*30(days/mo) = 490 kWh/mo

The reduction in percentage regarding the total energy spent can be calculated as follows:

ΔE = \frac{533.3-490}{533.3} = 0.081*100= 8.1%

⇒ ΔE(%) = 8.1%

%(E) =\frac{533.3-490}{533.3}  = 0.081 * 100 = 8.1%

4 0
4 years ago
Air enters the compressor of a regenerative gas-turbine engine at 300 K and 100 kPa, where it is compressed to 800 kPa and 580 K
Bad White [126]

Answer:

a) The amount of heat transfer in the regenerator, q = 114.12 kJ/kg

b) Thermal efficiency = 35.9%

Explanation:

The calculations are neatly handwritten and attached as files to this solution for easiness of expression and clarity. The cycle is also drawn on the T - S diagram and included in the attached files. Check the files below for the complete calculation.

I hope this helps!

3 0
3 years ago
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