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Andreas93 [3]
4 years ago
15

A large spherical tank contains gas at a pressure of 400 psi. The tank is constructed of high-strength steel having a yield stre

ss in tension of 80 ksi. For volume requirements, the inner diameter of the tank is 45-ft. Determine the required thickness (to the nearest ¼ inch) of the wall of the tank if a factor of safety of 3.0 with respect to yielding is required.
Engineering
1 answer:
jok3333 [9.3K]4 years ago
5 0

Answer:

t=2.025 inches

Explanation:

Given that

P = 400 Psi

Yield stress ,σ = 80 ksi

Diameter ,d= 45 ft

We know that

1 ft = 12 inches

d= 540 inches

Factor of safety ,K= 3

The required thickness given as

\dfrac {Pd}{4t}=\dfrac{\sigma}{K}

t=thickness

\dfrac {PdK}{4\sigma}=t

\dfrac {400\times 540\times 3}{4\times 80\times 1000}=t

t=2.025 inches

Therefore thickness will be 2.025 inches.

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A cylindrical specimen of cold-worked steel has a Brinell hardness of 250.(a) Estimate its ductility in percent elongation.(b) I
koban [17]

Answer:

A) Ductility = 11% EL

B) Radius after deformation = 4.27 mm

Explanation:

A) From equations in steel test,

Tensile Strength (Ts) = 3.45 x HB

Where HB is brinell hardness;

Thus, Ts = 3.45 x 250 = 862MPa

From image 1 attached below, for steel at Tensile strength of 862 MPa, %CW = 27%.

Also, from image 2,at CW of 27%,

Ductility is approximately, 11% EL

B) Now we know that formula for %CW is;

%CW = (Ao - Ad)/(Ao)

Where Ao is area with initial radius and Ad is area deformation.

Thus;

%CW = [[π(ro)² - π(rd)²] /π(ro)²] x 100

%CW = [1 - (rd)²/(ro)²]

1 - (%CW/100) = (rd)²/(ro)²

So;

(rd)²[1 - (%CW/100)] = (ro)²

So putting the values as gotten initially ;

(ro)² = 5²([1 - (27/100)]

(ro)² = 25 - 6.75

(ro) ² = 18.25

ro = √18.25

So ro = 4.27 mm

6 0
4 years ago
A hollow aluminum alloy [G = 3,800 ksi] shaft having a length of 12 ft, an outside diameter of 4.50 in., and a wall thickness of
Lisa [10]

Answer:

Horse power = 167.84 hp

Explanation:

Horsepower is calculated using the formula;

P = T * w

See the attached file for the calculation

6 0
3 years ago
If the feedforward path of a control system contains at least one integrating element, then the output continues to change as lo
Thepotemich [5.8K]

Answer:

The attached system shows that there’s an integrator between the point where disturbance enters the system and error measuring element. A any time when R(s)=0 then

\frac {C(s)}{D(s)}=\frac {G(s)}{1+G_c(s)G(s)} and considering that E(s)=D(s)-G_c(s)C(s) then

\frac {E(s)}{D(s)}=1-(\frac {C(s)}{D(s)})G_c(s)

\frac {E(s)}{D(s)}=1-(\frac {G(s)}{1+G_c(s)D(s)})G_c(s)

\frac {E(s)}{D(s)}=\frac {1}{1+G_c(s)G(s)}

E(s)=\frac {D(s)}{1+G_c(s)G(s)}

For ramp disturbance d(t)=at

D(s)=\frac {a}{s^{2}} therefore, the steady state error is given by

e(\infty)= \lim_{s \to 0} s E(s)

e(\infty)= \lim_{s \to 0} s [\frac {D(s)}{1+G_c(s)G(s)}]

e(\infty)= \lim_{s \to 0} s [\frac {a}{s^{2}+s^{2}G_c(s)G(s)}]

e(\infty)= \lim_{s \to 0} s [\frac {a}{s+sG_c(s)G(s)}]

e(\infty)= \lim_{s \to 0} s [\frac {a}{sG_c(s)G(s)}]

Whenever G_c(s) has a double intergrator, the error e(\infty) becomes zero

3 0
3 years ago
Are engineers needed in today’s society ? Why or why not ? I need a short three paragraph essay !!! Please help me !!!
masha68 [24]
Of course they are needed because without them the society wouldn’t be as nice as it is right now and plus there would be no more buildings ! :)
8 0
3 years ago
Write IEEE floating point representation of the following decimal number. Show your work.<br> 1.25
lisabon 2012 [21]

Answer:

00111111101000000000000000000000

Explanation:

View Image

0   01111111   01000000000000000000000

The first bit is the sign bit. It's 0 for positive numbers and 1 for negative numbers.

The next 8-bits are for the exponents.

The first 0-126₁₀ (0-2⁷⁻¹) are for the negative exponent 2⁻¹-2⁻¹²⁶.

And the last 127-256₁₀ (2⁷-2⁸) are for the positive exponents 2⁰-2¹²⁶.

You have 1.25₁₀ which is 1.010₂ in binary. But IEEE wants it in scientific notation form. So its actually 1.010₂*2⁰

The exponent bit value is 127+0=127 which is 01111111 in binary.

The last 23-bits are for the mantissa, which is the fraction part of your number. 0.25₁₀ in binary is 010₂... so your mantissa will be:

010...00000000000000000000

6 0
3 years ago
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