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tatuchka [14]
3 years ago
7

Calculate the maximum load that a 7075 series aluminum alloy bar (with a T6 temper heat treatment) can support without permanent

ly deforming. The bar has a square cross section whose edge length is 8.1 mm.
Engineering
1 answer:
Aleksandr [31]3 years ago
6 0

Answer:

The maximum load the bar can withstand = 35.43 KN

Explanation:

Ultimate tensile strength of the given aluminium bar \sigma = 540 M pa

Cross section area of the bar = 8.1^{2}  = 65.61 mm^{2}

We know that the ultimate strength of the bar is calculated from

\sigma = \frac{P_{max} }{A}

540 = \frac{P_{max} }{65.61}

P_{max} = 540 × 65.61

P_{max} =  35.43 KN

Therefore the maximum load the bar can withstand = 35.43 KN

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An electrical current of 700 A flows through a stainlesssteel cable having a diameter of 5 mm and an electricalresistance of 610
KatRina [158]

Answer:

778.4°C

Explanation:

I = 700

R = 6x10⁻⁴

we first calculate the rate of heat that is being transferred by the current

q = I²R

q = 700²(6x10⁻⁴)

= 490000x0.0006

= 294 W/M

we calculate the surface temperature

Ts = T∞ + \frac{q}{h\pi Di}

Ts = 30+\frac{294}{25*\frac{22}{7}*\frac{5}{1000}  }

Ts=30+\frac{294}{0.3928} \\

Ts =30+748.4\\Ts = 778.4

The surface temperature is therefore 778.4°C if the cable is bare

6 0
2 years ago
A steel bar 110 mm long and having a square cross section 22 mm on an edge is pulled in tension with a load of 89,000 N, and exp
iragen [17]

Answer:

Elastic modulus of steel  = 202.27 GPa

Explanation:

given data

long = 110 mm = 0.11 m

cross section 22 mm  = 0.022 m

load = 89,000 N

elongation = 0.10 mm = 1 × 10^{-4} m

solution

we know that Elastic modulus is express as

Elastic modulus = \frac{stress}{strain}    ................1

here stress is

Stress = \frac{Force}{area}       .................2

Area = (0.022)²

and  

Strain = \frac{extension}{length}       .............3

so here put value in equation 1 we get

Elastic modulus = \frac{89,000\times 0.11}{0.022^2 \times 1 \times 10^{-4} }  

Elastic modulus of steel = 202.27 × 10^9 Pa

Elastic modulus of steel  = 202.27 GPa

3 0
3 years ago
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
Assoli18 [71]

Answer: The exit temperature of the gas in deg C is 32^{o}C.

Explanation:

The given data is as follows.

C_{p} = 1000 J/kg K,   R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

P_{1} = 100 kPa,     V_{1} = 15 m^{3}/s

T_{1} = 27^{o}C = (27 + 273) K = 300 K

We know that for an ideal gas the mass flow rate will be calculated as follows.

     P_{1}V_{1} = mRT_{1}

or,         m = \frac{P_{1}V_{1}}{RT_{1}}

                = \frac{100 \times 15}{0.5 \times 300}  

                = 10 kg/s

Now, according to the steady flow energy equation:

mh_{1} + Q = mh_{2} + W

h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}

C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}

(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}

(T_{2} - T_{1}) = 5 K

T_{2} = 5 K + 300 K

T_{2} = 305 K

           = (305 K - 273 K)

           = 32^{o}C

Therefore, we can conclude that the exit temperature of the gas in deg C is 32^{o}C.

8 0
3 years ago
A negative pressure respirator brings fresh air to you through a hose<br>A) True<br>B)False​
madreJ [45]

Answer:FALSE

Explanation: A negative pressure respirator is a respiratory system which is known to have a low air pressure inside the mask when compared to the air pressure on the outside during Inhalation.

Most of the personal protective equipment (PPE) which are in use in various industries are examples of Negative pressure respirator device,any leak or damage done to the device will allow the inflows of harmful and toxic Air into the person's respiratory system. AIR SUPPLY SYSTEMS ARE KNOWN TO SUPPLY FRESH UNCONTAMINATED AIR THROUGH AIR STORED INSIDE COMPRESSED CYLINDERS OR OTHER SOURCES AVAILABLE.

8 0
3 years ago
Read 2 more answers
1. ELECTRICAL SHOCK
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3 0
3 years ago
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