Basic concepts surrounding electrical circuitry?

Ai r is compressed by a 30-kW compressor from P1 to P2. The air t emperature i s maintained constant at 25oC during thi s proces

RUDIKE [14]

Answer:

The rate of entropy change of the air is -0.10067kW/K

Explanation:

We'll assume the following

1. It is a steady-flow process;

2. The changes in the kinetic energy and the potential energy are negligible;

3. Lastly, the air is an ideal gas

Energy balance will be required to calculate heat loss;

mh1 + W = mh2 + Q where W = Q.

Also note that the rate of entropy change of the air is calculated by calculating the rate of heat transfer and temperature of the air, as follows;

Rate of Entropy Change = -Q/T

Where Q = 30Kw

T = Temperature of air = 25°C = 298K

Rate = -30/298

Rate = -0.100671140939597 KW/K

Rate = -0.10067kW/K

Hence, the rate of entropy change of the air is -0.10067kW/K

3 0

3 years ago

. Analyze the following scenarios and mention in which normal form it is with reason? a) Book- bookID bookTitle PublisherName Pu

svet-max [94.6K]

Answer:

You so good is ué KKKKKK a gente não não é possível que o seu ponto de vista da minha irmã me ligou

Explanation:

Irieiwy Wwowuwowywbrbbdnd

8 0

3 years ago

Evaluate (204 mm)(0.004 57 kg) / (34.6 N) to three

vesna_86 [32]

Answer:

the evaluation in SI unit will be $2.69\times 10^{-5}sec^{2}$

Explanation:

We have evaluate $\frac{(204mm\times 0.00457kg)}{34.6N}$

We know that 1 mm $=10^{-3}m$

So 240 mm $=204\times 10^{-3}m$

Newton can be written as $kgm/sec^2$

So $\frac{(204\times 10^{-3}m)\times 0.00457kg}{34.6kgm/sec^2}=2.69\times 10^{-5}sec^{2}$

So the evaluation in SI unit will be $2.69\times 10^{-5}sec^{2}$

4 0

3 years ago

Which of the following is an example of a composite material?

poizon [28]

Answer:

B

Explanation:

Did the assignment

4 0

3 years ago

4. A 25 km2 watershed has a time of concentration of 1.6 hr. Calculate the NRCS triangular UH for a 10-minute rainfall event and

Alika [10]

Answer:

The NRCS triangular UH for a 10-minute rainfall is Qp = 49.84 m³/s

Peak flow of the aggregated runoff hydrograph is 420.58 m³/s

The total volume of runoff is 2125000 m³/s

Explanation:

We have

A = 25 km²

tr = 10 min = 1/6 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/12 + 0.96 = 1.043 hr

Qp = 2.08×25/1.043 = 49.84 m³/s

Tb = 8/3×Tp = 8/3×1.043 = 2.782 hr

Since the area is

Time (min) Runoff (cm) Volume of runoff m³

0 0 0

10 4 1000000 m³

20 2.5 625000 m³

30 2 500000 m³

Total volume of runoff = 1000000 + 625000 + 500000 = 2125000 m³/s

For the 1st 10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr

Qp = 2.08×25×4/1.043 = 197.92 m³/s

Tb = 8/3×Tp = 8/3×1.21 = 3.227 hr

For the 2nd 10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr

Qp = 2.08×25×2.5/1.043 = 123.7 m³/s

Tb = 8/3×Tp = 8/3×1.21 = 3.227 hr

For the 3rd 10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr

Qp = 2.08×25×2.5/1.043 = 98.96 m³/s

Tb = 8/3×Tp = 8/3×1.21 = 3.227 hr

Peak flow of aggregate runoff is given by

Qp (total) = 98.96 + 123.7 +197.92 = 420.58 m³/s

Total volume of runoff is given by

Total volume of runoff = 1000000 + 625000 + 500000 = 2125000 m³/s

6 0

3 years ago

Basic concepts surrounding electrical circuitry?​

Basic concepts surrounding electrical circuitry?