1. In a safe approach to electrical work,

For precipitation hardening, how many phases (in numeric form) exist (a) following the solution heat treatment, and (b) followin

Nitella [24]

The correct answer is B

4 0

2 years ago

The following are the results of a sieve analysis. U.S. sieve no. Mass of soil retained (g) 4 0 10 18.5 20 53.2 40 90.5 60 81.8

il63 [147K]

Answer:

a.)

US Sieve no. % finer (C₅ )

4 100

10 95.61

20 82.98

40 61.50

60 42.08

100 20.19

200 6.3

Pan 0

b.) D10 = 0.12, D30 = 0.22, and D60 = 0.4

c.) Cu = 3.33

d.) Cc = 1

Explanation:

As given ,

US Sieve no. Mass of soil retained (C₂ )

4 0

10 18.5

20 53.2

40 90.5

60 81.8

100 92.2

200 58.5

Pan 26.5

Now,

Total weight of the soil = w = 0 + 18.5 + 53.2 + 90.5 + 81.8 + 92.2 + 58.5 + 26.5 = 421.2 g

⇒ w = 421.2 g

As we know that ,

% Retained = C₃ = C₂× $\frac{100}{w}$

∴ we get

US Sieve no. % retained (C₃ ) Cummulative % retained (C₄)

4 0 0

10 4.39 4.39

20 12.63 17.02

40 21.48 38.50

60 19.42 57.92

100 21.89 79.81

200 13.89 93.70

Pan 6.30 100

Now,

% finer = C₅ = 100 - C₄

∴ we get

US Sieve no. Cummulative % retained (C₄) % finer (C₅ )

4 0 100

10 4.39 95.61

20 17.02 82.98

40 38.50 61.50

60 57.92 42.08

100 79.81 20.19

200 93.70 6.3

Pan 100 0

The grain-size distribution is :

b.)

From the diagram , we can see that

D10 = 0.12

D30 = 0.22

D60 = 0.12

c.)

Uniformity Coefficient = Cu = $\frac{D60}{D10}$

⇒ Cu = $\frac{0.4}{0.12} = 3.33$

d.)

Coefficient of Graduation = Cc = $\frac{D30^{2}}{D10 . D60}$

⇒ Cc = $\frac{0.22^{2}}{(0.4) . (0.12)}$ = 1

3 0

2 years ago

Which of the following is correct regarding the principal stresses and maximum in-plane shear stresses? a. Principal stresses ca

ziro4ka [17]

Answer:

option B.

Explanation:

The correct answer is option B.

Principal stress is the maximum normal stress a body can have. In principal stress, there is purely normal stress. On principal plane shear stress is zero.

In-plane shear stress are the shear stress which is acting on the plane.

The statement which is correct regarding principle plane and shear stress is that The shear stress over principal stress planes is always zero.

4 0

2 years ago

The connection is made using a bolt and nut and two washers. If the allowable bearing stress of the washers on the boards is (sb

Yuki888 [10]

Answer:

P = 0.490 kip

Explanation:

given data

allowable bearing stress = 2 ksi

allowable tensile stress = 18 ksi

diameter = 0.31 in

outer diameter = 0.75 in

inner diameter (hole) = 0.50 in

solution

we find here cross section area of shank that is express as

Area = $\frac{\pi }{4} \times d^2$ ..................1

area = $\frac{\pi }{4} \times 0.31^2$

area = 0.0754 in²

and

now we get here allowable load in bolt will be

$\sigma = \frac{P}{A}$ ...................2

P = $\sigma \times A$

P = 18 × 10³ × 0.0754

P = 1357.2 = 1.357 kip

and

now find here area of washer is

Area = $\frac{\pi }{4} \times (d^2-d1^2)$ .......................3

put here value

Area = $\frac{\pi }{4} \times (0.75^2-0.5^2)$

area = 0.2454 in²

so now we get here allowable load of washer will be

$\sigma = \frac{P}{Area}$ .....................4

P = 2 × 10³ × 0.245

P = 490 = 0.490 kip

6 0

2 years ago

Derive an expression for the specific heat difference of a substance whose equation of state is 1 2 ( ) RT a P b b T ν ν ν = − −

sergij07 [2.7K]

Answer:

Given data:

Equation of the state $p=\frac{RT}{v-b}-\frac{a}{v(v+b) T^{1/2} }$

Where p = pressure of fluid, pα

T = Temperature of fluid, k

V = Specific volume of fluid $m^{3} / k g$

R = gas constant , $j/k g k$

a, b = Constants

Solution:

Specific heat difference, $\begin{array}{c}c_{p}-c_{v}=-T\left(\frac{\partial v}{\partial T}\right)^{2} p \\\left(\frac{\partial P}{\partial v}\right)_{r}\end{array}$

According to cyclic reaction

$\left(\frac{\ dv}{\ dT}\right)_{p}=-\frac{\left(\frac{\ d P}{\ d T}\right)_{v}}{\left(\frac{\ d P}{\ d v}\right)_{v}}$

Hence specific heat difference is

$c_{p}-c_{v}=\frac{-T\left(\frac{\ d v}{\ d T}\right)_{p}^{2}}{\left(\frac{\ d p}{\ dv}\right)_{v}}$

Equation of state, $p=\frac{R T}{v-b}-\frac{a}{v(v+b)^{\ 1/2}}$

Differentiating the equation of state with respect to temperature at constant volume,

$\left(\frac{\ d P}{\ d T}\right)_{v}=\frac{R}{v-b}-\frac{1}{2}- \frac{a}{v(v+b)^} T^{\frac{-1}{2}}$

$\begin{aligned}&\left(\frac{\ dP}{\ dT}\right)_{V}=\frac{R}{v-b}+\frac{a}{2 v(v+b) T^{3 / 2}}\end{aligned}$

Differentiating the equation of the state with respect to volume at constant temperature.

$\left(\frac{\ dP}{\ dv}\right)_{\gamma}=+(-1) \times R T(v-b)^{-1-1}+\frac{a}{b T^{1 / 2}}\left(\frac{1}{v^{2}}-\frac{1}{(v+b)^{2}}\right)\)\\\(\left(\frac{\ dP}{\ dv}\right)_{r}=-\frac{R T}{(v-b)^{2}}+\frac{a}{T^{1 / 2}}\left(\frac{2 v+b}{v^{2}(v+b)^{2}}\right)$

Substituting both eq (3) and eq (4) in eq (2)

We get,

${cp{} - } c_{v}=\frac{T\left(\frac{R}{v-b}+\frac{a}{2 v(v+b) T^{3 / 2}}\right)^{2}}{\left(\frac{R T}{(v-b)^{2}}-\frac{a(2 v+b)}{T^{1 / 2} v^{2}(v+b)^{2}}\right)}$

Specific heat difference equation,

$c_{p} -c_{v}}=\frac{T\left(\frac{R}{v-b}+\frac{a}{2 v(v+b)^{T}^{3 / 2}}\right)^{2}}{\left(\frac{R T}{(v-b)^{2}}-\frac{a(2 v+b)}{T^{1 / 2} v^{2}(v+b)^{2}}\right)}$

7 0

2 years ago