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2 years ago

Write down about the water source selection criteria

Engineering

1 answer:

ololo11 [35]2 years ago

5 0

Answer:

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WARNING:<br><br> when people put links in the answer it is a virus DO NOT DOWNLOAD IT

kobusy [5.1K]

Answer:

Thank you for this!

Explanation:

I was about to click it on a question I saw.

6 0

2 years ago

Read 2 more answers

One kilogram of air, initially at 5 bar, 350 K, and 3 kg of carbon dioxide (CO2), initially at 2 bar, 450 K, are confined to opp

pentagon [3]

Answer:

Check the explanation

Explanation:

Energy alance of 2 closed systems: Heat from CO2 equals the heat that is added to air in

$m_{a} c_{v,a}(T_{eq} -T_{a,i)} =m_{co2} c_{v,co2} (T_{co2,i} -T_{eq)}$

1x0.723x $(T_{eq} -350)$ =3x0.780x $(450-T_{eq} )$ ⇒ $T_{eq}$ = 426.4 °K

The initail volumes of the gases can be determined by the ideal gas equation of state,

$V_{a,i} = \frac{mRT_{a,i} }{P_{a,i} }$ = $\frac{1x (8.314 28.97 kJ kg • °K)x 350°K}{5 bar x 100KPa bar}$ = 0.201 $m^{3}$

The equilibrium pressure of the gases can also be obtained by the ideal gas equation

$P_{eq=\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,eq}+V_{CO2,eq)} } =\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,i}+V_{CO2,i)} }$

$P_{eq}$ = 1x(8.314 28.97)x426.4+3x(8.314 44)x426.4

(0.201+1.275)

= 246.67 KPa = 2.47 bar

6 0

3 years ago

A program is seeded with 30 faults. During testing, 21 faults are detected, 15 of which are seeded faults and 6 of which are ind

Vesna [10]

Answer:

Estimated number of indigenous faults remaining undetected is 6

Explanation:

The maximum likelihood estimate of indigenous faults is given by,

$N_F=n_F\times \frac{N_S}{n_S}$ here,

$n_F$ = the number of unseeded faults = 6

$N_S$ = number of seeded faults = 30

$n_s$ = number of seeded faults found = 15

So NF will be calculated as,

$N_F=6\times \frac{30}{15}=12$

And the estimate of faults remaining is $N_F-n_F$ = 12 - 6 = 6

8 0

3 years ago

4. The instant the ignition switch is turned to the start position,

geniusboy [140]

Answer:

D. Both pull-in and hold-in windings are energized.

Explanation:

The instant the ignition switch is turned to the start position, "Both pull-in and hold-in windings are energized." This is because the moment the ignition switch is turned to the start position, voltage passes through to the S terminal of the solenoid.

The hold-in winding is attached to the case of the solenoid. Similarly, the pull-in winding is also attached to the starter motor. Thereby, the current will move across both windings by getting energized to generate a strong magnetic field.

4 0

3 years ago

Can you prove that the two bleu areas are the same without numbers please?

Svet_ta [14]

Answer:

$\small{\boxed{\tt{\colorbox{green}{✓Verified\:answer}}}}\:$

Just draw a line from point D join to point E

The triangle formed DME will be congruent to AMC

6 0

3 years ago

Write down about the water source selection criteria​

Write down about the water source selection criteria