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diamong [38]
2 years ago
6

An energy system can be approximated to simply show the interactions with its environment including cold air in and warm air out

, heat transfer inflow and power input. Write a first law balance around the system and solve for the rate of work input to the system.
Engineering
1 answer:
Elenna [48]2 years ago
5 0

Answer: The energy system related to your question is missing attached below is the energy system.

answer:

a) Work done = Net heat transfer

  Q1 - Q2 + Q + W = 0

b)  rate of work input ( W ) = 6.88 kW

Explanation:

Assuming CPair = 1.005 KJ/Kg/K

<u>Write the First law balance around the system and rate of work input to the system</u>

First law balance ( thermodynamics ) :

Work done = Net heat transfer

Q1 - Q2 + Q + W = 0 ---- ( 1 )

rate of work input into the system

W = Q2 - Q1 - Q -------- ( 2 )

where : Q2 = mCp T  = 1.65 * 1.005 * 293 = 485.86 Kw

             Q2 = mCp T = 1.65 * 1.005 * 308 = 510.74 Kw

              Q = 18 Kw

Insert values into equation 2 above

W = 6.88 Kw

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GREYUIT [131]

Answer:

i think is B

Explanation:

6 0
3 years ago
Nec ________ covers selection of time-delay fuses for motor- overload protection.
Murljashka [212]

Nec Article 430 covers selection of time-delay fuses for motor- overload protection.

<h3>What article in the NEC covers motor overloads?</h3>

Article 430 that is found in  National Electrical Code (NEC) is known to be state as “Motors, Motor Circuits and Controllers.” .

Note that the article tells that it covers areas such as motors, motor branch-circuit as well as feeder conductors, motor branch-circuit and others.

Therefore, Nec Article 430 covers selection of time-delay fuses for motor- overload protection.

Learn more about motor- overload from

brainly.com/question/20738481

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6 0
1 year ago
You are working in a lab where RC circuits are used to delay the initiation of a process. One particular experiment involves an
Ymorist [56]

Answer:

t'_{1\2} = 6.6 sec

Explanation:

the half life of the given circuit is given by

t_{1\2} =\tau ln2

where [/tex]\tau = RC[/tex]

t_{1\2} = RCln2

Given t_{1\2} = 3 sec

resistance in the circuit is 40 ohm and to extend the half cycle we added new resister of 48 ohm. the net resitance is 40+48 = 88 ohms

now the new half life is

t'_{1\2} =R'Cln2

Divide equation 2 by 1

\frac{t'_{1\2}}{t_{1\2}} = \frac{R'Cln2}{RCln2} = \frac{R'}{R}

t'_{1\2} = t'_{1\2}\frac{R'}{R}

putting all value we get new half life

t'_{1\2} = 3 * \frac{88}{40}  = 6.6 sec

t'_{1\2} = 6.6 sec

7 0
3 years ago
At what times should you use your headlights?
SVETLANKA909090 [29]

Answer:

Headlights are required to be used 1/2 hour after sunset to 1/2 hour before sunrise, when windshield wipers are being used, when visibility is less than 1000 feet, or when there is insufficient light or adverse weather.

Explanation:

hope this helps

8 0
3 years ago
Read 2 more answers
1. Examine the following circuit. Find RT, I3, R1, R2, R3, V1, V2 and V3. Show all of your work clearly below.
Mkey [24]

Explanation:

Ohm's law is used here. V = IR, and variations. The voltage across all elements is the same in this parallel circuit. (V1 =V2 =V3)

The total supply current is the sum of the currents in each of the branches. (It = I1 +I2 +I3)

Rt = (8 V)/(8 A) = 1 Ω . . . . supply voltage divided by supply current

I3 = 8A -3A -4A = 1 A . . . . supply current not flowing through other branches

R1 = (8 V)/(3 A) = 8/3 Ω

R2 = (8 V)/(4 A) = 2 Ω

R3 = (8 V)/(I3) = (8 V)/(1 A) = 8 Ω

V1 = V2 = V3 = 8 V

6 0
2 years ago
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