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Mariana [72]
3 years ago
10

Why do we feel chilled in the winter but comfortable in the summer under conditions for which the same room temperature is maint

ained by a heating or cooling system?
Engineering
1 answer:
pantera1 [17]3 years ago
4 0

Answer:

Explanation:

We feel chilled in summer but comfortable in the summer under the same temperature because of our thermal comfort.

Thermal comfort is a function of heat loss from the body and during winter there is excessive heat loss due to heat flux.

As room temperature is fixed so summer and winter comfort cannot be attributed due to heat loss by convection. Due to differences in heat flux radiations in summer and winter there is a chilled feeling associated with winters.

 

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B1) 20 pts. The thickness of each of the two sheets to be resistance spot welded is 3.5 mm. It is desired to form a weld nugget
kap26 [50]

Answer:

minimum current level required =  8975.95 amperes

Explanation:

Given data:

diameter = 5.5 mm

length = 5.0 mm

T = 0.3

unit melting energy = 9.5 j/mm^3

electrical resistance = 140 micro ohms

thickness of each of the two sheets = 3.5mm

Determine the minimum current level required

first we calculate the volume of the weld nugget

v = \frac{\pi }{4} * D^2 * l = \frac{\pi }{4} * 5.5^2 * 5 = 118.73 mm^3

next calculate the required melting energy

= volume of weld nugget * unit melting energy

= 118.73 * 9.5 = 1127.94 joules

next find the actual required electric energy

= required melting energy / efficiency

= 1127 .94 / ( 1/3 )  = 3383.84 J

TO DETERMINE THE CURRENT LEVEL REQUIRED  use the relation below

electrical energy =  I^2 * R * T

3383.84 / R*T = I^2

3383.84 / (( 140 * 10^-6 ) * 0.3 ) = I^2

therefore  8975.95 = I ( current )

4 0
3 years ago
A strain gauge with a 5 mm gauge length gives a displacement reading of 1.25 um. Calculate the stress value given by this displa
KengaRu [80]

Answer:

stress  = 50MPa

Explanation:

given data:

Length of strain guage is 5mm

displacement\delta = 1.25 \mu m =\frac{1.25}{1000} =  0.00125 mm

stress due to displacement in structural steel can be determined by using following relation

E =\frac{stress}{strain}

stress = E \times strain

where E is young's modulus of elasticity

E for steel is 200 GPa

stress = 200\times 10^3 *\frac{1.25*10^{-3}}{5}

stress  = 50MPa

7 0
4 years ago
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krok68 [10]
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5 0
3 years ago
You have just finished your OST takeoffs for a single-story home and found 175 LF of interior walls with 2x6 studs 14" OC. The h
zimovet [89]

Answer:

Total BF for the interior wall is 7.50BD

Explanation:

Given Data:

· Size of stud = 2” x 6”

· Height of Wall = 8 ft

· Top plates = 2

· Bottom Plate = 1

BF stands for board feet in lumber/wood terminology. It is the unit of volume.

1 BF (Board feet) = 1 ft x 1 ft x 1 inch

Since there are total three plates at top and bottom, we have to deduct their thickness from wall height to calculate height of stud.

Height of stud = 8’ – 3 x 2” = 7’6” = 7.5 ft

Board feet of one stud = 7.50 6/12 x 2 = 7.50 BD

Total BF for the interior wall is 7.50BD

7 0
3 years ago
3. When starting an automatic transmission
Alexxandr [17]

Answer:

It should be in Park or Neutral.

Explanation:

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4 years ago
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