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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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ser-zykov [4K]

Answer:

Shearing strain will be 0.1039 radian

Explanation:

We have given change in length \Delta L=0.12inch

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We have to find the shearing strain

Shearing strain is given by

\alpha =tan^{-1}\frac{\Delta L}{L}=tan^{-1}\frac{0.12}{1.15}=5.9571^{\circ}

Shearing strain is always in radian so we have to change angle in radian

So 5.9571\times \frac{\pi }{180}=0.1039radian

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7 0
2 years ago
A 1 m3 rigid tank initially contains air whose density is 1.18kg/m3. The tank is connected to a high pressure supply line throug
Elanso [62]

To solve this problem it is necessary to apply the concepts related to density in relation to mass and volume for each of the states presented.

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3 years ago
Write two scnr.nextInt statements to get input values into birthMonth and birthYear. Then write a statement to output the month,
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Answer:

import java.util.Scanner;

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   public static void main(String[] args) {

       Scanner scnr = new Scanner(System.in);

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3 0
3 years ago
A well insulated turbine operates at steady state. Steam enters the turbine at 4 MPa with a specific enthalpy of 3015.4 kJ/kg an
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Answer:

power developed by the turbine = 6927.415 kW

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given data

pressure = 4 MPa

specific enthalpy h1 = 3015.4 kJ/kg

velocity v1 = 10 m/s

pressure = 0.07 MPa

specific enthalpy h2 = 2431.7 kJ/kg

velocity v2 = 90 m/s

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solution

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energy equation that is

h1 + \frac{v1}{2}  + q = h2 + \frac{v2}{2}  + w

put here value with

turbine is insulated so q = 0

so here

3015.4 *1000 + \frac{10^2}{2}  =  2431.7 * 1000 + \frac{90^2}{2}  + w

solve we get

w = 579700 J/kg = 579.7 kJ/kg

and

W = mass flow rate × w

W = 11.95 × 579.7

W = 6927.415 kW

power developed by the turbine = 6927.415 kW

7 0
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