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labwork [276]
2 years ago
13

Water flows through a converging pipe at a mass flow rate of 25 kg/s. If the inside diameter of the pipes sections are 7.0 cm an

d 5.0 cm, find the volume flow rate and the average velocity in each pipe section.
Engineering
1 answer:
ser-zykov [4K]2 years ago
7 0

Answer:

volumetric flow rate = 0.0251 m^3/s

Velocity in pipe section 1 = 6.513m/s

velocity in pipe section 2 = 12.79 m/s

Explanation:

We can obtain the volume flow rate from the mass flow rate by utilizing the fact that the fluid has the same density when measuring the mass flow rate and the volumetric flow rates.

The density of water is = 997 kg/m³

density = mass/ volume

since we are given the mass, therefore, the  volume will be mass/density

25/997 = 0.0251 m^3/s

volumetric flow rate = 0.0251 m^3/s

Average velocity calculations:

<em>Pipe section A:</em>

cross-sectional area =

\pi \times d^2\\=\pi \times 0.07^2 = 3.85\times10^{-3}m^2

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

velocity = 25/(997 \times 3.85\times10^{-3}) = 6.513m/s

<em>Pipe section B:</em>

cross-sectional area =

\pi \times d^2\\=\pi \times 0.05^2= 1.96\times10^{-3}m^2

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

velocity = 25/(997 \times 1.96\times10^{-3}) = 12.79m/s

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The voltage and current at the terminals of the circuit element in Fig. 1.5 are zero fort &lt; 0. Fort 2 0 they areV =75 ~75e-10
masya89 [10]

Answer:

maximum value of the power delivered to the circuit =3.75W

energy delivered to the element = 3750e^{ -IOOOt} - 7000e ^{-2OOOt} -3750

Explanation:

V =75 - 75e-1000t V

l = 50e -IOOOt mA

power = IV = 50 * 10^-3 e -IOOOt * (75 - 75e-1000t)

=50 * 10^-3 e -IOOOt *75 (1 - e-1000t)

=

maximum value of the power delivered to the circuit =3.75W

the total energy delivered to the element = \int\limits^t_0  {3.75(e^{ -IOOOt} - e ^{-2OOOt} )} , dx \\\\

3750e^{ -IOOOt} - 7000e ^{-2OOOt} -3750

5 0
3 years ago
Transmission cleaners are used: A) Only in conjunction with fuel system cleanersB) Only in the colder monthsC) By themselvesD) I
valentina_108 [34]

Answer: D

Explanation: When you change your transmission fluid you change your filter. When you do that you clean where the filter was. Then you can put the new filter on and the new fluid.

5 0
3 years ago
Read 2 more answers
An inventor claims that he wants to build a dam to produce hydroelectric power. He correctly realizes that civilization uses a l
Mekhanik [1.2K]

Answer:

The problem is that the pumps would consume more energy than the generators would produce.

Explanation:

Water has a potential energy associated with the height it is at. The higher it is, the higher the potential energy. When water flows down into the turbines that energy is converted to kinetic energy and then into electricity.

A pump uses electricity to add energy to the water to send it to a higher potential energy state.

Ideally no net energy woul be hgenerate or lost, because the generators would release the potential energy and pumps would store it again in the water. However the systems are not ideal, everything has an efficiency and losses. The losses would accumulate and the generator would be generating less energy than the pumps consume, so that system wastes energy.

What should be done is closing the floodgates to keep the water up in the dam at night producing only the power that is needed and releasing more water during the day.

6 0
3 years ago
A hypothetical metal alloy has a grain diameter of 1.7 102 mm. After a heat treatment at 450C for 250 min, the grain diameter ha
Romashka-Z-Leto [24]

Answer:

the required time for the specimen is  1109.4 min

Explanation:

Given that;

diameter of metal alloy d₀ = 1.7 × 10² mm

Temperature of heat treatment T = 450°C = 450 + 273 = 723 K

Time period of heat treatment t = 250 min

Increased grain diameter 4.5 × 10² mm

grain diameter exponent n = 2.1

First we calculate the time independent constant K

dⁿ - d₀ⁿ = Kt

K = (dⁿ - d₀ⁿ) / t

we substitute

K = (( 4.5 × 10² )²'¹ - ( 1.7 × 10² )²'¹) / 250

K = (373032.163378 - 48299.511117) / 250

K = 1298.9306 mm²/min

Now, we calculate the time required for the specimen to achieve the given grain diameter ( 8.7 × 10² mm )

dⁿ - d₀ⁿ = Kt

t = (dⁿ - d₀ⁿ) / K

t = (( 8.7 × 10² )²'¹ - ( 1.7 × 10² )²'¹) / 1298.9306

t = ( 1489328.26061158 - 48299.511117) / 1298.9306

t = 1441028.74949458 / 1298.9306

t = 1109.4 min

Therefore, the required time for the specimen is  1109.4 min

4 0
3 years ago
1- The preexponential and activation energy for the diffusion of iron in cobalt are 1.1×10-5 m 2 /s and 253,300 J/mol, respectiv
n200080 [17]

The temperature at which the diffusion coefficient have a value of 2.1×10-5 m 2 /s is  -47078 K.

Using the relation;

logD = logDo - Ea/2.303RT

D = diffusion coefficient

Do =  preexponential

Ea = activation energy

R = gas constant

T = temperature

Substituting values;

log(2.1×10-5)= log (1.1×10-5 ) - 253,300/2.303 × 8.314 × T

log(2.1×10-5) -  log (1.1×10-5 ) =  - 253,300/2.303 × 8.314 × T

log[2.1×10-5/1.1×10-5] = - 253,300/2.303 × 8.314 × T

0.281 × (2.303 × 8.314 × T) = - 253,300

T =  - 253,300/2.303 × 0.281 × 8.314

T = -47078 K

Learn more: brainly.com/question/14283892

4 0
2 years ago
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