Question

The wind chill, which is experienced on a cold, windy day, is related to increased heat transfer from exposed human skin to the surrounding atmosphere. Consider a layer of fatty tissue that is 3mm thick and whose interior surface is maintained at a temp of 36 C. On a calm day the convection heat transfer coefficient at the outer surface is 25 W/m?K, but with 30 km/h wind it reaches 65 W/m?K, in both cases the ambient air temp is -15 C. (A) What is the ratio of the heat loss per unit area from the skin for the calm day to that for the windy day?(B) What will be the skin outer surface temperature for the calm day? For the windy day? (C) What temperature would the air have to assume on the calm day to produce the same heat loss occurring with the air temperature at -15 C on the windy day?

Answer 1

Answer:

sdfdsfdsfdsfsfdsfdsfdsfdsfsxg

Explanation:fafdf

Answer 2

The given and known information:

• The layer of fatty tissues thickness, $L = 3mm$

• The temperature at which the tissue's Interior surface is maintained, $Ts,i=36$°$C$

• The convection coefficients in a calm and windy day, $hc=25\frac{W}{m^{2}K }, hw = 65\frac{W}{m^{2}K }$

• The ambient air temperature, $T\alpha =-15$°$C$

• Thermo conductivity from thermodynamic tables, $k=0.2\frac{W}{mK}$

a) 0.5524

b) Calm:22.09°C

b) Windy: 10.82°C

c) -56.32°C

The interior surface of the skin to the air transfers heat in the two types:

• Conduction
• Convection

The total thermal resistance can be set up as:

$R_{tot} =R_{cond}+R_{conv}\\\\R_{tot}=\frac{L}{kA} +\frac{1}{hA}\\\\R^{''} _{tot} = \frac{L}{k}+\frac{1}{h}$

Step-by-Step Explanation:

The answers for the sub-parts are solved below in the given order:

a) The ratio as-

$q^{''}_{c} = \frac{Ts,i-T\alpha }{\frac{L}{k}+\frac{1}{hc}} \\\\\\q^{''}_{c} = \frac{36+15 }{\frac{0.003}{0.2}+\frac{1}{25}}\\\\\\q^{''}_{c} = 927.27\frac{W}{m^{2} }$

$q^{''}_{w} = \frac{Ts,i-T\alpha }{\frac{L}{k}+\frac{1}{hw}} \\\\\\q^{''}_{w} = \frac{36+15 }{\frac{0.003}{0.2}+\frac{1}{65}}\\\\\\q^{''}_{w} = 1678.48\frac{W}{m^{2} }$

$\frac{q^{''}_{c}}{q^{''}_{w}} = \frac{927.27\frac{W}{m^{2} } }{1678.48\frac{W}{m^{2} } }\\\\\\\frac{q^{''}_{c}}{q^{''}_{w}}=0.5524$

b) The required temperature-

$q_{conv} =\frac{Ts,o-T\alpha }{\frac{1}{h} }\\\\\\q^{''} = q_{c,conv}\\\\\\=927.27\frac{W}{m^{2} }\\\\\\927.27=\frac{Ts,o+15}{\frac{1}{25}} \\\\\\Ts,o=22.09$

$q_{conv} =\frac{Ts,o-T\alpha }{\frac{1}{h} }\\\\\\q^{''} = q_{w,conw}\\\\\\=1678.48\frac{W}{m^{2} }\\\\\\1678.48=\frac{Ts,o+15}{\frac{1}{65}} \\\\\\Ts,o=10.82$

c) The air temperature on a calm day -

$q^{''}_{w} = q^{''}_{c,2} =1678.48\frac{W}{m^{2} } \\\\\q^{''}_{c,2} = \frac{Ts,i-T\alpha ,2}{\frac{L}{k}+\frac{1}{h} }\\\\1678.48 = \frac{36-T\alpha ,2}{\frac{0.003}{0.02}+\frac{1}{25}}\\\\T\alpha ,2= -56.32$

Learn more about the computations, refer to the link below:

https://brainly.com/question/13334824