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A cylinder with a frictionless piston contains 0.05 m3 of air at 60kPa. The linear spring holding the piston is in tension. The

AleksAgata [21]

Answer:

18 kJ

Explanation:

Given:

Initial volume of air = 0.05 m³

Initial pressure = 60 kPa

Final volume = 0.2 m³

Final pressure = 180 kPa

Now,

the Work done by air will be calculated as:

Work Done = Average pressure × Change in volume

thus,

Average pressure = $\frac{60+180}{2}$ = 120 kPa

and,

Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³

Therefore,

the work done = 120 × 0.15 = 18 kJ

4 0

3 years ago

Given a 12-bit A/D converter operating over a voltage range from ????5 V to 5 V, how much does the input voltage have to change,

Doss [256]

Answer:

2.44 mV

Explanation:

This question has to be one of analog quantization size questions and as such, we use the formula

Q = (V₂ - V₁) / 2^n

Where

n = 12

V₂ = higher voltage, 5 V

V₁ = lower voltage, -5 V

Q = is the change in voltage were looking for

On applying the formula and substitutiting the values we have

Q = (5 - -5) / 2^12

Q = 10 / 4096

Q = 0.00244 V, or we say, 2.44 mV

6 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

15

AveGali [126]

Answer:

b

Explanation:

4 0

2 years ago

What is the total inductance of a circuit that contains two 10 uh inductors connected in a parallel?

kolbaska11 [484]

Answer:

5 microhenries

Explanation:

The effective value of inductors in parallel "add" in the same way that resistors in parallel do. The value is the reciprocal of the sum of the reciprocals of the inductances that are in parallel.

10 uH ║ 10 uH = 5 uH

The effective inductance is 5 uH.

6 0

2 years ago

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