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2 years ago

6

: A cyclical load of 1500 lb is to be exerted at the end of a 10 in. long aluminium beam (see Figure below). The bar must surviv

e for at least 10° cycles. What is the minimum diameter of the bar?

1 answer:

scZoUnD [109]2 years ago

4 0

Answer:

the minimum diameter of the bar is 1.634 in

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A light bulb is switched on and within a few minutes its temperature becomes constant. Is it at equilibrium or steady state.

EleoNora [17]

Answer:

The temperature attains equilibrium with the surroundings.

Explanation:

When the light bulb is lighted we know that it's temperature will go on increasing as the filament of the bulb has to constantly dissipates energy during the time in which it is on. Now this energy is dissipated as heat as we know it, this heat energy is absorbed by the material of the bulb which is usually made up of glass, increasing it's temperature. Now we know that any object with temperature above absolute zero has to dissipate energy in form of radiations.

Thus we conclude that the bulb absorbs as well as dissipates it's absorbed thermal energy. we know that this rate is dependent on the temperature of the bulb thus it the temperature of the bulb does not change we can infer that an equilibrium has been reached in the above 2 processes i.e the rate of energy absorption equals the rate of energy dissipation.

Steady state is the condition when the condition does not change with time no matter whatever the surrounding conditions are.

6 0

3 years ago

Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost m

Ilia_Sergeevich [38]

Answer:

ordinary bulb total cost is $39.54

fluorescent bulb total cost is $13.05

amount save = 39.54 - 13.05 = $26.49

resistance = 626.1 ohm

Explanation:

in the 1st part

bulb on time = 3 year = 4380 hours

life of bulb = 750 h

so number of bulb required = $\frac{4380}{750}$

number of bulb required = 6

cost of 6 bulb is = 6 × 0.75 = $4.5

so

cost of operation is = 100 × 4380 × $\frac{0.08}{1000}$

cost of operation = $35.04

so total cost will be = $4.5 + $35.04 = $39.54

and

when compare with florescent bulb

time = 3 year = 4380 h

life of bulb = 10000 h

so number of bulb required = $\frac{4380}{10000}$

number of bulb required = 0.43 = 1

cost of 6 bulb is = 1 × 5 = $5

so

cost of operation is = 23 × 4380 × $\frac{0.08}{1000}$

cost of operation = $8.05

so total cost will be = $5 + $8.05 = $13.05

in part 2nd

total amount save while compare bulb is

amount save = 39.54 - 13.05 = $26.49

and in part 3rd

resistance of bulb is

resistance = $\frac{v^2}{P}$

resistance = $\frac{120^2}{23}$

resistance = 626.1 ohm

6 0

2 years ago

Consider a 2-shell-passes and 8-tube-passes shell-and-tube heat exchanger. What is the primary reason for using many tube passes

Maru [420]

Answer:

See explanation

Explanation:

Solution:-

- The shell and tube heat exchanger are designated by the order of tube and shell passes.

- A single tube pass: The fluid enters from inlet, exchange of heat, the fluid exits.

- A multiple tube pass: The fluid enters from inlet, exchange of heat, U bend of the fluid, exchange of heat, .... ( nth order of pass ), and then exits.

- By increasing the number of passes we have increased the "retention time" of a specific volume of tube fluid; hence, providing sufficient time for the fluid to exchange heat with the shell fluid.

- By making more U-turns we are allowing greater length for the fluid flow to develop with " constriction and turns " into turbulence. This turbulence usually at the final passes allows mixing of fluid and increases the heat transfer coefficient by:

U ∝ v^( 0.8 ) .... ( turbulence )

- The higher the velocity of the fluids the greater the heat transfer coefficient. The increase in the heat transfer coefficient will allow less heat energy carried by either of the fluids to be wasted ; hence, reduced losses.

Thereby, increases the thermal efficiency of the heat exchanger ( higher NTU units ).

5 0

2 years ago

Ben İngiliz oldum düzelte bilirmiyim

Lelu [443]

Answer:

What laguange is that?

Explanation:

7 0

2 years ago

a circular pile, 19 m long is driven into a homogeneous sand layer. The piles width is 0.5 m. The standard penetration resistanc

Elena L [17]

Answer:

Point force (Qp) = 704 kn/m²

Explanation:

Given:

length = 19 m

Width = 0.5 m

fs = 4

Vicinity of the pile = 25

Find:

Point force (Qp)

Computation:

Point force (Qp) = fs²(l+v)

Point force (Qp) = 4²(25+19)

Point force (Qp) = 16(44)

Point force (Qp) = 704 kn/m²

5 0

2 years ago