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gogolik [260]
3 years ago
7

Generally, final design results are rounded to or fixed to three digits because the given data cannot justify a greater display.

In addition, prefixes should be selected so as to limit number strings to no more than four digits to the left of the decimal point. Using these rules, as well as those for the choice of prefixes, solve the following relations:
Captionless Image

Engineering
1 answer:
creativ13 [48]3 years ago
8 0

Answer:

(a) 1.90 kpsi

(b) 0.40 kpsi

(c) 0.61 in.

(d) 0.009

(a) 8 MPa

(b) 1.30 cm⁴

(c) 2.04 cm⁴

(d) 62.2 MPa

Explanation:

(a) σ = M/Z, where M = 1770 lbf·in and Z = 0.943 in³.

1770/0.943 = 1876.988 lbf/in² = 1.90 kpsi

(b) σ = F/A, where F = 9440 lbf and A = 23.8 in².

9440 /23.8 = 396.639 lbf/in² = 0.4 kpsi

(c) y = Fl³/(3EI)

F = 270 lbf

l = 31.5 in.

E = 30 Mpsi

I = 0.154 in.⁴

y = 270×31.5³/(3×30×10⁶×0.154) = 0.61 in.

(d) θ = Tl/(GJ), where T = 9740 lbf·in, l = 9.85 in. G = 11.3 Mpsi, and d = 1.00 in.

J = π·d⁴/32 = π/32 in.⁴

∴ θ = 9740  × 9.85 /(11.3 × 10⁶× π/32) = 0.009

(a) σ = F/wt, where F = 1 kN, w = 25 mm, and t = 5 mm

∴ σ = 1000/(0.025 × 0.005) = 8 MPa

(b) I = bh³/12, where b = 10 mm and h = 25 mm.

10×25³/12 = 1.30 cm⁴

(c) I = π·d⁴/64 where d = 25.4 mm.

I = π × 25.4⁴/64 = 2.04 cm⁴

(d) τ = 16×T/(π×d³), where T = 25 N·m, and d = 12.7 mm.

16×25/(π×0.0127³) = 62.2 MPa.

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Answer:

\mathbf{C_{10} = 137.611 \ kN}

Explanation:

From the information given:

Life requirement = 40 kh = 40 40 \times 10^{3} \ h

Speed (N) = 520 rev/min

Reliability goal (R_D) = 0.9

Radial load (F_D) = 2600 lbf

To find C10 value by using the formula:

C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}

where;

x_D = \text{bearing life in million revolution} \\  \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}

\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}

The Weibull parameters include:

x_o = 0.02

(\theta - x_o) = 4.439

b= 1.483

∴

Using the above formula:

C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}

C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}

C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}

C_{10} = 30962.449 \ lbf

Recall that:

1 kN = 225 lbf

∴

C_{10} = \dfrac{30962.449}{225}

\mathbf{C_{10} = 137.611 \ kN}

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Bind hole, 38 diameter, .50 deep
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Answer:

59.69021

Explanation:

38/.5 x 3.14159

4 0
2 years ago
it creates parts from thin plastic sheets as opposed to plastic pellets. is it 1. Pickling 2. Thermoforming 3. Extrusion​
iVinArrow [24]

Answer:

thermoforming

Explanation:

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4 0
3 years ago
Read 2 more answers
A spherical hot air balloon is initially filled with air having 120 kPa pressure and 24 °C temperature. Initial diameter of the
tino4ka555 [31]

Answer:

v = 1.076 m /s

Explanation:

Initial volume of balloon = 4/3 x  3.14 x (9.905/2)³

=508.56 m³

Final volume of balloon = 4/3 x 3.14 x (16.502/2)³

= 2351.73 m³

Increase in volume = 1843.17 m³

Cross sectional area of inlet  A  = 3.14 x( 1.458/2)²

A = 1.6687 m²

Volume rate of flow of air = cross sectional area x velocity of inflow

= 1 .6687 V [ V is velocity of inflow ]

Total time taken = Increase in volume / rate of flow of air

17.108 X 60 = 1843.17 / 1.6687 V

V = \frac{1843.17}{1.6687\times17.108\times60}

v = 1.076 m /s

8 0
3 years ago
An air-conditioning system is used to maintain a house at 70^\circ{} ∘ F when the temperature outside is 100^\circ{} ∘ F. The ho
Vika [28.1K]

Answer:

the minimum power input required for this air-conditioning system is 1.20 hp

Explanation:

Given the data in the question;

Temperature inside ( Sink ) T_L = 70°F = ( 70 + 460 )R = 530 R

temperature outside ( source )T_H = 100°F = ( 100 + 460 )R = 560 R

Q_W = 800 Btu/min

Q_L = 100  Btu/min

Now, from the equation of coefficient of performance of refrigerator;

COP_{Ref = Desired output / Required input

COP_{Ref = Q_{out / W_{net --------- let this be equation 1

COP_{Ref = ( Q_L + Q_W ) / W_{net

where Q_W is the rate heat gained through the wall

Q_L is the heat generation from people and lights and appliances.

Now, lets consider the equation coefficient of performance of refrigerator in terms of temperatures;

COP_{Ref = T_L / ( T_H - T_L )

we substitute

COP_{Ref = 530 / ( 560 - 530 )

COP_{Ref = 530 / 30

COP_{Ref = 17.667

so we substitute into equation 1;

COP_{Ref = Q_{out / W_{net ---------

COP_{Ref = ( Q_L + Q_W ) / W_{net

17.667 = ( 100 + 800 ) / W_{net

17.667 = 900 / W_{net

W_{net = 900 / 17.667

W_{net = 50.94 Btu/min

W_{net = ( 50.94 / 42.53 ) hp

W_{net = 1.198 hp ≈ 1.20 hp

Therefore, the minimum power input required for this air-conditioning system is 1.20 hp

5 0
3 years ago
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