6. Question

A reversible process and an irreversible process both have the same________ between the same two states. a. Internal energy b. W

Vlad1618 [11]

Answer:

a) Internal energy

Explanation:

As we know that internal energy is a point function so it did not depends on the path ,it depends at the initial and final states of process.All point function property did not depends on the path.Internal energy is a exact function.

Work and heat is a path function so these depend on the path.They have different values for different path between two states.Work and heat are in exact function.

We know that in ir-reversible process entropy will increase so entropy will be different for reversible and ir-reversible processes.

5 0

3 years ago

A pipeline (NPS = 14 in; schedule = 80) has a length of 200 m. Water (15℃) is flowing at 0.16 m3/s. What is the pipe head loss f

dangina [55]

Answer:

Head loss is 1.64

Explanation:

Given data:

Length (L) = 200 m

Discharge (Q) = 0.16 m3/s

According to table of nominal pipe size , for schedule 80 , NPS 14, pipe has diameter (D)= 12.5 in or 31.8 cm 0.318 m

We know, $head\ loss = \frac{f L V^2}{( 2 g D)}$

where, f = Darcy friction factor

V = flow velocity

g = acceleration due to gravity

We know, flow rate Q = A x V

solving for V

$V = \frac{Q}{A}$

$= \frac{0.16}{\frac{\pi}{4} (0.318)^2} = 2.015 m/s$

obtained Darcy friction factor

calculate Reynold number (Re) ,

$Re = \frac{\rho V D}{\mu}$

where, $\rho$ = density of water

$\mu$ = Dynamic viscosity of water at 15 degree C = 0.001 Ns/m2

so reynold number is

$Re = \frac{1000\times 2.015\times 0.318}{0.001}$

= 6.4 x 10^5

For Schedule 80 PVC pipes , roughness (e) is 0.0015 mm

Relative roughness (e/D) = 0.0015 / 318 = 0.00005

from Moody diagram, for Re = 640000 and e/D = 0.00005 , Darcy friction factor , f = 0.0126

Therefore head loss is

$HL = \frac{0.0126 (200)(2.015)^2}{( 2 \times 9.81 \times 0.318)}$

HL = 1.64 m

7 0

3 years ago

If the slotted arm rotates counterclockwise with a constant angular velocity of thetadot = 2rad/s, determine the magnitudes of t

astraxan [27]

Answer:

Magnitude of velocity=10.67 m/s

Magnitude of acceleration=24.62 ft/ $s^{2}$

Explanation:

The solution of the problem is given in the attachments

3 0

3 years ago

You are an engineer in an electric-generation station. You know that the flames in the boiler reach a temperature of 1200 K and

andre [41]

Answer:

...wat

Explanation:

6 0

3 years ago

A compressible clay layer has a thickness of 3.8 m. After 1.5 yr, when the clay is 50% consolidated, 7.3 cm of settlement has oc

grigory [225]

The amount of settlement that would occur at the end of 1.5 year and 5 year are 7.3 cm and 13.14 cm respectively.

<h3>How to determine the amount of settlement?</h3>

For a layer of 3.8 m thickness, we were given the following parameters:

U = 50% = 0.5.

Sc = 7.3 cm.

For Sf, we have:

Sf = Sc/U

Sf = 7.3/0.5

Sf = 14.6

Therefore, Sf for a layer of 38 m thickness is given by:

Sf = 14.6 × 38/3.8

Sf = 146 cm.

At 50%, the time for a layer of 3.8 m thickness is: $t_{50}$ = 1.5 year.

At 50%, the time for a layer of 38 m thickness is:

$t_{50}$ = 1.5 × (38/3.8)²

$t_{50}$ = 150 years.

For the thickness of 38 m, U₂ is given by:

$\frac{U_1^2}{U_2^2} =\frac{(T_v)_1}{(T_v)_2} = \frac{t_1}{t_2} \\\\U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{1.5}{150} ]\\\\U_2^2 = 0.25 \times 0.01\\\\U_2=\sqrt{0.0025} \\\\U_2=0.05$

The new settlement after 1.5 year is:

Sc = U₂Sf

Sc = 0.05 × 146

Sc = 7.3 cm.

For time, t₂ = 5 year:

$U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{5}{150} ]\\\\U_2^2 = 0.25 \times 0.03\\\\U_2=\sqrt{0.0075} \\\\U_2=0.09$

The new settlement after 5 year is:

Sc = U₂Sf

Sc = 0.09 × 146

Sc = 13.14 cm.

Read more on clay layer here: brainly.com/question/22238205

8 0

2 years ago