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Alchen [17]
2 years ago
12

A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the outside diameter of the

tube if the stress is 120 MPa?
Engineering
2 answers:
Olegator [25]2 years ago
7 0

Answer:

119.35 mm

Explanation:

Given:

Inside diameter, d = 100 mm

Tensile load, P = 400 kN

Stress = 120 MPa

let the outside diameter be 'D'

Now,

Stress is given as:

stress = Load × Area

also,

Area of hollow pipe = \frac{\pi}{4}(D^2-d^2)

or

Area of hollow pipe = \frac{\pi}{4}(D^2-100^2)

thus,

400 × 10³ N = 120 × \frac{\pi}{4}(D^2-100^2)

or

D² = tex]\frac{400\times10^3+30\pi\times10^4}{30\pi}[/tex]

or

D = 119.35 mm

IgorC [24]2 years ago
3 0

Answer:

D =119.35 mm

Explanation:

given data:

inside diameter = 100 mm

load = 400 kN

stress = 120MPa

we know that load is given as

P = \sigma A  

where:

P=400kN = 400000N

\sigma = 120MPa

A =(\frac{1}{4} \pi D^2 - \frac{1}{4}\pi (100^2)

A=\frac{1}{4} \pi (D^2 - 10000)

putting all value in the above equation to get the required diameter value

400 =  120*\frac{1}{4} \pi (D^2 - 10000)

solving for

D =119.35 mm

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Answer:

A) Gray cast iron

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5 0
2 years ago
A 12-ft circular steel rod with a diameter of 1.5-in is in tension due to a pulling force of 70-lb. Calculate the stress in the
padilas [110]

Answer:

The stress in the rod is 39.11 psi.

Explanation:

The stress due to a pulling force is obtained dividing the pulling force by the the area of the cross section of the rod. The respective area for a cylinder is:

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Replacing the diameter the area results:

A= 17.76 in^2

Therefore the the stress results:

σ = 70/17.76 lb/in^2 = 39.11 lb/in^2= 39.11 psi

5 0
2 years ago
An electrical heater is a form of sensible heating process, and heats 0.1m/s of air from 15°C and 80% RH to 50°C? The barometric
lawyer [7]

Answer:

The heater load =35 KJ/kg

Explanation:

Given that

At initial condition

Temperature= 15°C

RH=80%

At final condition

Temperature= 50°C

We know that in sensible heating process humidity ratio remain constant.

Now from chart

At temperature= 15°C and RH=80%

h_1=38 \frac{KJ}{kg},v=0.8 \frac{m^3}{kg}

At  temperature= 50°C

h_2=73 \frac{KJ}{kg}

So\ the\ heater\ load =h_2-h_1

The heater load = 73 - 38 KJ/kg

The heater load =35 KJ/kg

3 0
2 years ago
The thermal efficiency of two reversible power cycles operating between the same thermal reservoirs will a)- depend on the mecha
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C ,, i’m pretty sure .
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Consider a very long, cylindrical fin. The temperature of the fin at the tip and base are 25 °C and 50 °C, respectively. The dia
nekit [7.7K]

Answer:

98°C

Explanation:

Total surface area of cylindrical fin = πr² + 2πrl , r = 0.015m; l= 0.1m; π =22/7

22/7*(0.015)² + 22/7*0.015*0.1 = 7.07 X 10∧-4 + 47.1 X 10∧-4 = (54.17 X 10∧-4)m²

Temperature change, t = (50 - 25)°C = 25°C = 298K

Hence, Temperature =  150 X (54.17 X 10∧-4) X 298/123 = 242.14/124 = 2.00K =

∴ Temperature change = 2.00K

But temperature, T= (373 - 2)K = 371 K

In °C = (371 - 273)K = 98°C         

7 0
3 years ago
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