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Rashid [163]
3 years ago
12

Consider a fan located in a 3 ft by 3 ft square duct. Velocities at various points at the outlet are measured, and the average f

low velocity is determined to be 22 ft/s. Taking the air density to 0.075 lbm/ft3, estimate the minimum electric power consumption of the fan motor.
Engineering
1 answer:
natulia [17]3 years ago
5 0

Answer:

minimum electric power consumption of the fan motor is 0.1437 Btu/s

Explanation:

given data

area = 3 ft by 3 ft

air density = 0.075 lbm/ft³

to find out

minimum electric power consumption of the fan motor

solution

we know that energy balance equation that is express as

E in - E out  = \frac{dE \ system}{dt}    ......................1

and at steady state  \frac{dE \ system}{dt} = 0

so we can say from equation 1

E in = E out

so

minimum power required is

E in = W = m \frac{V^2}{2} = \rho A V \frac{V^2}{2}  

put here value

E in =  \rho A V \frac{V^2}{2}  

E in =  0.075 *3*3* 22 \frac{22^2}{2}  

E in = 0.1437 Btu/s

minimum electric power consumption of the fan motor is 0.1437 Btu/s

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posledela

Answer:

In the center and directed away from it.

Explanation:

The direction along the radius and directed away from the center is known as radial direction.

The velocity is highest in the radial direction pointing away from the center, this is because of the reason that  when the particle executes its motion in the direction that is radial, then it is not acted upon by any force that opposes the motion of the particle and thus there is no obstruction to the velocity of the particle and it is therefore, the highest in the radial direction.

8 0
3 years ago
The human body gets its energy via the combustion of blood sugar (glucose). if all of the chemical bond energy in 10 g of glucos
Gnesinka [82]

Answer:

speed by mass attain is 55.86 m/s

Explanation:

given data

glucose = 10 g

mass = 100 kg

to find out

speed by mass attain

solution

we know glucose have 180 g molecular weight and

that 1 g glucose produce energy = 2816/180 × 10³ J

so here 10 g of glucose produce energy =  1.56 × 10^{5} J

so here energy release = 1/2 × mv²

1.56 × 10^{5}  = 1/2 × (100)v²

v² = 3.12 × 10³

and v = 55.86 m/s

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4 0
3 years ago
A compound machine contains three simple machines with IMAs of 2, 4 and 5, respectively. What is the overall ideal mechanical ad
anygoal [31]

Answer:

Overall ideal mechanical advantage of the machine = 40

Explanation:

Given:

Ideal mechanical advantage of three machine = 2, 4, 5

Find:

Overall ideal mechanical advantage of the machine

Computation:

Overall ideal mechanical advantage of the machine = 2 × 4× 5

Overall ideal mechanical advantage of the machine = 40

3 0
3 years ago
PLEASE HELP ME!!!!!! 100 POINTS FOR HELPFUL ANSWERS + BRAINLIEST!!!!!
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6 0
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A steel bar is 150 mm square and has a hot-rolled finish. It will be used in a fully reversed bending application. Sut for the s
Xelga [282]

Answer:

See explanation

Explanation:

Given The bar is square and has a hot-rolled finish. The loading is fully reversed bending.

Tensile Strength

Sut: 600 MPa

Maximum temperature

Tmax: 500 °C

Bar side dimension

b: 150 mm

Alternating stress

σa: 100 MPa

Reliability

R: 0.999 Note 1.

Assumptions Infinite life is required and is obtainable since this ductile steel will have an endurance limit. A reliability factor of 99.9% will be used.

Solution See Excel file Ex06-01.xls.

1 Since no endurance-limit or fatigue strength information is given, we will estimate S'e based on the ultimate tensile strength using equation 6.5a.

S'e: 300 MPa = 0.5 * Sut

2 The loading is bending so the load factor from equation 6.7a is

Cload: 1

3 The part size is greater than the test specimen and the part is not round, so an equivalent diameter based on its 95% stressed area must be determined and used to find the size factor. For a rectangular section in nonrotating bending, the A95 area is defined in Figure 6-25c and the equivalent diameter is found from equation 6.7d

A95: 1125 mm2 = 0.05 * b * b Note 2.

dequiv: 121.2 mm = SQRT(A95val / 0.0766)

and the size factor is found for this equivalent diameter from equation 6.7b, to be

Csize: 0.747 = 1.189 * dequiv^-0.097

4 The surface factor is found from equation 6.7e and the data in Table 6-3 for the specified hot-rolled finish.

Table 6-3 constants

A: 57.7

b: -0.718 Note 3.

Csurf: 0.584 = Acoeff * Sut^bCoeff

5 The temperature factor is found from equation 6.7f :

Ctemp: 0.710 = 1 - 0.0058 * (Tmax - 450)

6 The reliability factor is taken from Table 6-4 for R = 0.999 and is

Creliab: 0.753

7 The corrected endurance limit Se can now be calculated from equation 6.6:

Se: 69.94 MPa = Cload * Csize * Csurf * Ctemp *

Creliab * Sprme

Let

Se: 70 MPa

8 To create the S-N diagram, we also need a value for the estimated strength Sm at 103 cycles based on equation 6.9 for bending loading.

Sm: 540 MPa = 0.9 * Sut

9 The estimated S-N diagram is shown in Figure 6-34 with the above values of Sm and Se. The expressions of the two lines are found from equations 6.10a through 6.10c assuming that Se begins at 106 cycles.

b: -0.2958 Note 4.

a: 4165.7

Plotting Sn as a function of N from equation 6.10a

N Sn (MPa)

1000 540 =aa*B73^bb

2000 440

4000 358

8000 292

16000 238

32000 194

64000 158

128000 129

256000 105

512000 85

1000000 70

FIGURE 6-34. S-N Diagram and Alternating Stress Line Showing Failure Point

10 The number of cycles of life for any alternating stress level can now be found from equation 6.10a by replacing σa for Sn.

At N = 103 cycles,

Sn3: 540 MPa = aa * 1000^bb

At N = 106 cycles,

Sn6: 70 MPa = aa * 1000000^bb

The figure above shows the intersection of the alternating stress line (σa = 100 MPa) with the failure line at N = 3.0 x 105 cycles.

8 0
3 years ago
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