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sergey [27]
3 years ago
11

An average person generates heat at a rate of 84 W while resting. Assuming one-quarter of this heat is lost from the head and di

sregarding radiation, determine the average surface temperature of the head when it is not covered and is subjected to winds at 10°C and 29 km/h. The head can be approximated as a 0.2-m-diameter sphere. Assume a surface temperature of 15ºC for evaluation of μs. The properties of air at 1 atm pressure and the free stream temperature of 10°C are k
Engineering
1 answer:
eimsori [14]3 years ago
6 0

Answer:

Average surface temperature = 11.4 C

Explanation:

Temperature of the surface = Incoming Temperature + (Heat generated by the Average person's head/ Coefficient of heat transfer x Surface Area of head)

Area = πD² = π(0.2)² = 0.1256 m²

h = (Thermal Conductivity/Diameter of Surface)(Nusselt Number)

Nu = 2 + [0.4Re^0.5 + 0.06Re^2/3] x Pr^0.4 x [μ∞/μs]^1/4

where Re is the Reynold Number, Pr = 0.7336 is the Prandtl Number, μ∞ = 1.778 x 10^-5 is the dynamic viscosity and μs = 1.802 x 10^-5 is the Dynamic Viscosity at 15 C

Re = VD/v where V is the velocity of wind, D is the diameter of the sphere and v is the kinematic viscosity

Re = (8.056 x 0.2)/(1.426 x 10^-5) = 1.13 x 10^6

Substituting value of Re into Nu equation

Nu = 2 + [0.4(1.13 x 10^6)^0.5 + 0.06(1.13 x 10^6)2/3] (0.7336 ^0.4){(1.778 x 10^-5)/(1.802 x 10^-5)^1/4}

Nu = 2 + [1076.1](0.8835)(0.9967) = 949.5

Putting Nu values in h

h = (0.02439/0.2)(949.5) = 115.8 W/m² C

putting h value in Ts equation

Ts = 10 + [(84/4)/(115.8 x 0.1256)] = 11.4 C

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Answer:

The correct answer is C) Trimetric

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The most suitable answer is a trimetric projection because, in this type of projection, we see that the projection of the three angles between the axes are not equal. Therefore, to generate a trimetric projection of an object, it is necessary to have three separate scales.

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Are routers better for internet connection rather than a WiFi modem?
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Answer:

Depends

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A program contains the following function definition: int cube(int number) { return number * number * number; } Write a stateme
Nonamiya [84]

Answer:

The statement can be written as

int result = cube(4);

Explanation:

A function is a block of reusable codes to perform some tasks. For example, the function in the question is to calculate the cube of a number.

A function can also operate on one or more input value (argument) and return a result. The <em>cube </em>function in the question accept one input value through its parameter <em>number </em>and the <em>number</em> will be multiplied by itself twice and return the result.  

To call a function, just simply write the function name followed with parenthesis (e.g. <em>cube()</em>). Within the parenthesis, we can include zero or one or more than one values as argument(s) (e.g. <em>cube(4)</em>).

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8 0
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An aluminum metal rod is heated to 300oC and, upon equilibration at this temperature, it features a diameter of 25 mm. If a tens
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Ammonia contained in a piston-cylinder assembly, initially saturated vapor at 0o F, undergoes an isothermal process during which
Rudik [331]

ANSWERS:

-P_{2(a)} =15.6lbf/in^2\\-P_{2(b)} =30.146lbf/in^2\\ T_{2(a)} =0^oF\\T_{2(b)} =0^oF\\x_{2(b)} =49.87percent

Explanation:

Given:

Piston cylinder assembly which mean that the process is constant pressure process P=C.

<u>AMMONIA </u>

state(1)

saturated vapor x_{1} =1

The temperature T_{1} =0^0 F

Isothermal process  T=C

a)

-V_{2} =2V_{1} ( double)

b)

-V_{2} =.5V_{2} (reduced by half)

To find the final state by giving the quality in lbf/in we assume the friction is neglected and the system is in equilibrium.

state(1)

using PVT data for saturated ammonia

-P_{1} =30.416 lbf/in^2\\-v_{1} =v_{g} =9.11ft^3/lb

then the state exists in the supper heated region.

a) from standard data

-v_{1(a)} =2v_{1} =18.22ft^3/lb\\-T_{1} =0^oF

at\\P_{x} =14lbf/in^2\\-v_{x} =20.289 ft^3/kg

at\\P_{y} =16 lbf/in^2\\-v_{y} =17.701ft^3/kg

assume linear interpolation

\frac{P_{x}-P_{2(b)}  }{P_{x}- P_{y} } =\frac{v_{x}-v_{1(a)}  }{v_{x}-v_{y}  }

P_{1(b)}=P_{x} -(P_{x} -P_{y} )*\frac{v_{x}- v_{1(b)} }{v_{x}-v_{y}  }\\ \\P_{1(b)} =14-(14-16)*\frac{20.289-18.22}{20.289-17.701} =15.6lbf/in^2

b)

-v_{2(a)} =2v_{1} =4.555ft^3/lb\\v_{g}

from standard data

-v_{f} =0.02419ft^3/kg\\-v_{g} =9.11ft^3/kg\\v_{f}

then the state exist in the wet zone

-P_{s} =30.146lbf/in^2\\v_{2(a)} =v_{f} +x(v_{g} -v_{f} )

x=\frac{v_{2(a)-v_{f} } }{v_{g} -v_{f} } \\x=\frac{4.555-0.02419}{9.11-0.02419} =49.87%

3 0
3 years ago
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