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andrew-mc [135]

2 years ago

11

Consider a thin-walled cylindrical tube having a radius of 65 mm that is to be used to transport pressurized gas. If inside and

outside tube pressures are 100 and 1.0 atm (10.13 and 0.1013 MPa), respectively, compute the minimum required thickness for each of the following metal alloys. Assume a factor of safety of 3.5.

1 answer:

Marizza181 [45]2 years ago

6 0

Missing part of the question:

Find the missing table in the attached file. The table was used for the calculation. The below question (b) is also the missing part of the question.

(b) A tube constructed of which of the alloys will cost the least amount?

Answer:

The thickness for each of the metal alloys are given below.

(a)

Steel (plain) thickness = 6.25 mm

Steel (alloy) thickness = 2.28 mm

Cast iron thickness = 10.14 mm

Aluminum thickness = 8.30 mm

Magnesium thickness = 13.04 mm

(b) the alloys will cost the least amount is steel (alloy) which $33.8

Explanation:

Cost of each metal alloy

Steel (plain) = $34.43

Steel (alloy) = $33.80

Cast iron = $79.24

Aluminum = $73.22

Magnesium = $159

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In an air standard diesel cycle compression starts at 100kpa and 300k. the compression ratio is 16 to 1. The maximum cycle tempe

KIM [24]

Answer:

$\eta$ =0.60

Explanation:

Given :Take $\gamma$ =1.4 for air

$P_1=100 KPa ,T_1=300K$

$\frac{V_1}{V_2}$ =r ⇒ r=16

As we know that

$T_2=T_1(r^{\gamma-1})$

So $T_2=300\times 16^{\gamma-1}$

$T_2$ =909.42K

Now find the cut off ration $\rho$

$\rho=\frac{V_3}{V_2}$

$\frac{V_3}{V_2}=\frac{T_3}{T_2}$

$\rho=\frac{2031}{909.42}$

$\rho=2.23$

So efficiency of diesel engine

$\eta =1-\dfrac{\rho^\gamma-1}{\gamma\times r^{\gamma-1}(\rho-1)}$

Now by putting the all values

$\eta =1-\dfrac{2.23^{1.4}-1}{1.4\times 16^{1.4-1}(2.23-1)}$

So $\eta$ =0.60

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7 0

2 years ago

On a given day, a barometer at the base of the Washington Monument reads 29.97 in. of mercury. What would the barometer reading

Umnica [9.8K]

Answer:

The barometer reading will be 29.43 in

Explanation:

Using the formula of pressure variation

p2 - p1 = -yair * H

= 7.65 * $10^{-2} \frac{lb}{ft^{3} } * 500 ft\\$

= 38.5 $\frac{lb}{ft^{2} }$

According to the relationship between the pressure and the height of the mercury column

p = yHg * h --> where yHg and h is the barometer reading

yHg $(\frac{29.97}{12} ft)$ - yHg * h1 = 38.5 $\frac{lb}{ft^{2} }$

h1 = ( $\frac{29.97}{12} ft$ ) - $\frac{38.5 \frac{lb}{ft^{2} } }{847 \frac{lb}{ft^{3} } }$

$[(\frac{29.97}{12} ft) - 0.0455 ft] - 12 \frac{in}{ft} \\\\h1 = 29.43 in$

8 0

2 years ago

A ceramic matrix composite contains internal flaws as large as 0.001 cm in length. The plane strain fracture toughness of the co

murzikaleks [220]

Since the applied stress required for failure due to crack propagation is still higher than 550 MPa, the ceramic is expected to fail due to overload and not because of the flaws

Explanation:

<u>Plane -Strain Fracture toughness is calculated as</u>

$k_{IC}$ = $f$ б $\sqrt{\pi a}$

F=geometry factor of the flaw

б=Stress applied

$k_{IC}$ =Fracture toughness

a=Flaw size

<u>Given that </u>

Internal Flaw,a=0.001cm

Fracture Toughness $k_{IC}$ =45MPa $\sqrt{m}$

Tensile Strength б=550 MPa

Geometry Factor, $f$ =1

<u>Calculation</u>

An internal Flaw i s 0.001 cm

2a=0.001cm

a=0

6 0

2 years ago