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Stolb23 [73]
3 years ago
15

A gasoline engine takes in air at 290 K, 90 kPa and then compresses it. The combustion adds 1000 kJ/kg to the air after which th

e temperature is 2050 K.
Using a cold air-standard analysis, i.e., constant specific heats at 300 K, find the compression ratio, compression specific work, and the highest pressure in the cycle. Repeat your analysis assuming temperature-dependent specific heats and comment on your results.

Engineering
1 answer:
Inessa [10]3 years ago
4 0

Answer:

attached below

Explanation:

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Question 5 (20 pts) The rated current of a three-phase transmission line is 300 A. The currents flowing by the line are measured
prisoha [69]

Answer:

Check the explanation

Explanation:

Question 1.

The secondary current of 250/5 amps CT when 300 amps(rated current of transmission line ) flow in TL is

(5/250 ) X 300 = 6 amps

Question 2

The correct answer to this second question is yes, when Over current relay coil will operate and relay contacts gets close, if the pickup value( Ip) of relay is set as 6 amps in relay. ( because primary current of TL is 1.2 times of CT primary)

Question 3

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3 years ago
What have you learned about designing solutions? How does this apply to engineering? Think of some engineering solutions that st
Andrew [12]

Answer:

In engineering design, failure is expected. It helps you find the best solutions before implementing them in the “real world”. Having a prototype fail is a GOOD thing, because that means you have learned something new about the problem and potential solutions.

Explanation:

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3 years ago
An ideal gas mixture has a volume base composition of 40% Ar and 60% Ne (monatomic gases). The mixture is now heated at constant
baherus [9]

[Find the attachment]

6 0
3 years ago
Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series
Zina [86]

Answer:

0.245 m^3/s

Explanation:

Flow rate through pipe a is 0.4 m3/s Parallel pipes have a diameter D = 30 cm => r = 15 cm = 0.15 m Length of Pipe a = 1000m Length of Pipe b = 2650m Temperature = 15 degrees Va = V / A = (0.4m3/s) / (3.14 (0.15m)^2) = 5.66 m/s h = (f(LV^2)) / D2g (fa(LaVa^2)) / Da2g = (fb(LbVb^2)) / Da2g and Da = Db; fa = fb LaVa^2 = LbVb^2 => La/Lb = Vb^2/Va^2 Vd^2 = Va^2(La/Lb) => Vb = Va(La/Lb)^(1/2) Vb = 5.66 (1000/2650)^(1/2) => 5.66 x 0.6143 = 3.4769 m/s Vb = 3.4769 m/s V = AVb = 3.14(0.15)^2 x 3.4769 m/s = 0.245 m^3/s

5 0
3 years ago
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