Clothing made of several thin layers of fabric with trapped air in between, often called ski clothing, is commonly used in cold climates because it is light, fashionable, and a very effective thermal insulator. So, it is no surprise that such clothing has largely replaced thick and heavy old-fashioned coats. Consider a jacket made up of six layers of 0.1 mm thick synthetic fabric (k = 0.026W/m.K) with 1.2 mm thick air space (k = 0.026 W/m.K) between the fabric layers. Assuming the inner surface temperature of the jacket to be 25˚C and the surface area to be 1.25 m2 , determine the heat loss through the jacket when the temperature of the outdoors is -5˚C and the heat transfer co-efficient of outer surface is 25 W/m2 .K. What would be the thickness of a wool fabric (k = 0.035W/m.K) if the person has to achieve the same level of thermal comfort wearing a thick wool coat instead of a jacket. (30 points)

Answers

Answer 1

Answer:

Answer:

Q=127.66W

L=9.2mm

Explanation:

Heat transfer consists of the propagation of energy in the form of heat in different ways, these can be convection if it is through a fluid, radiation through electromagnetic waves and conduction through solid solids.

To solve any problem related to heat transfer, the general equation is used

Q = delta / R

Where

Q = heat

Delta = the temperature difference

R = is the thermal resistance by conduction, convection and radiation

to solve this problem we propose the previous equation

Q = delta / R

later we find R

$R=[tex]r=(6L1)/(AK1) +(5L2)/(AK2)+(1)/(Ah)$

$R=(6(0.0001))/((1.25)(0.026)) +(5(0.012))/((1.25)(0.026))+(1)/((25)(1.25)) =0.235 K/w$

Q=(25-(-5))/0.235=127.66W

part b

we use the same ecuation with Q=127.66

Q = delta / R

Δ $R=(L)/(KA) +(1)/(hA) \nR=(L)/((0.035)(1.25)) +(1)/((25)(1.25))\n R=22.85L+0.032\nQ=(T1-T2)/R\n\n127.66=(25-(-5))/(22.85L+0.032)\nsolving for L\nL=9.2mm$

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A new pipeline is installed to convey 2500 gal/min. If the pipeline can not exceed 6 ft/1,000ft of head loss what is the minimum standard diameter to convey the flowrate? Size both PVC (C=130) and steel (C=120) pipelines.

Answers

Answer:

Minimum standard diameter for the PVC pipe = 14.26 inches

Minimum standard diameter for the steel pipe = 14.70 inches

Explanation:

Head loss = 6/1000...................................................(1)

Head loss = hf/l

$Head loss = 10.44Q^(1.85) /C^(1.85) D^(4.8655)$ ............................(2)

Q = 2500 gal/min

a) Minimum standard diameter for PVC

C for PVC = 130

Equating (1) and (2) and putting C = 130

$6/1000 = 10.44* 2500^(1.85) /[130^(1.85) * D^(4.8655) ]\nD^(4.8655) = 10.44* 2500^(1.85) /[0.006*130^(1.85)]\nD = [10.44* 2500^(1.85) /[0.006*130^(1.85)]]^(1/4.8655) \nD = 14.26 inches$

b) Minimum standard diameter for steel

C for steel = 120

Equating (1) and (2) and putting C = 120

$6/1000 = 10.44* 2500^(1.85) /[120^(1.85) * D^(4.8655) ]\nD^(4.8655) = 10.44* 2500^(1.85) /[0.006*120^(1.85)]\nD = [10.44* 2500^(1.85) /[0.006*120^(1.85)]]^(1/4.8655) \nD = 14.70 inches$

What is the minimum hose diameter of an ideal vacuum cleaner that could lift a 14 kg dog off the floor?

Answers

The diameter would be 267km speed suction i think

Calculate the angle of banking on a bend of 100m radius so that vehicles can travel round the bend at 50km/hr without side thrust on the tyres.

Answers

Answer:

11.125°

Explanation:

Given:

Radius of bend, R = 100 m

Speed around the bend = 50 Km/hr = $(5)/(18)*50$ = 13.89 m/s

Now,

We have the relation

$\tan\theta=(v^2)/(gR)$

where,

θ = angle of banking

g is the acceleration due to gravity

on substituting the respective values, we get

$\tan\theta=(13.89^2)/(9.81*100)$

$\tan\theta=0.1966$

θ = 11.125°

What is the deflection equation for a simply supported beam with a uniformly distributed load?

Answers

Answer:

$\Delta _(max)=(5wL^4)/(384EI)$

Explanation:

Given that

Load is uniformly distributed load and beam is simply supported.

Ra + Rb= wL

Ra = Rb =wL / 2

Lets x is measured from left side,then the deflection of beam at any distance x is given as

$\Delta _x=(wx)/(24EI)(L^3-2Lx^2+x^3)$

The maximum deflection of beam will at x = L/2 (mid point )

$\Delta _(max)=(w* (L)/(2))/(24EI)(L^3-2L* \left ((L)/(2)\right)^2+\left((L)/(2)\right)^3)$

$\Delta _(max)=(5wL^4)/(384EI)$

For steels, we can assume the endurance limit measured by rotating beam tests is the 50% of the ultimate tensile strength (UTS) as long as the UTS is equal to or less than 200 ksi (1400 MPa).True/False

Answers

Answer:

True

Explanation:

For steels, we can assume the endurance limit measured by rotating beam tests is the 50% of the ultimate tensile strength (UTS) as long as the UTS is equal to or less than 200 kpsi (1400 MPa).

Also, It is a simplistic rule of thumb that, for steels having a UTS less than 160 kpsi, the endurance limit for the material will be approximately 45 to 50% of the UTS.

Benzene vapor at 480°C is cooled and converted to a liquid at 25°C in a continuous condenser. The condensate is drained into 1.75-m3 drums, each of which takes 2.0 minutes to fill. Calculate the rate (kW) at which heat is transferred from the benzene in the condenser.

Answers

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

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