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Luden [163]
3 years ago
8

The heat flux through a 1-mm thick layer of skin is 1.05 x 104 W/m2. The temperature at the inside surface is 37°C and the tempe

rature at the outside surface is 30°C.a. What is the thermal conductivity of the skin?b. A layer of clothing material with half the thermal conductivity of skin and twice the thickness of skin is placed on the outside surface of the skin. If the outside surface of the clothing is maintained at 30°C, what is the new heat flux from the skin and what is the temperature at the skin-insulation interface?
Engineering
1 answer:
miss Akunina [59]3 years ago
3 0

Answer:

a) Thermal conductivity of skin: k_{skin}=1.5W/mK

b) Temperature of interface: T_{interface}=35.6\°C

Heat flux through skin: \frac{Q}{A}=2100W/m^2

Explanation:

a)

k=\frac{QL}{A(T_{2}-T_{1})}

Where: k is thermal conductivity of a material, \frac{Q}{A} is heat flux through a material, L is the thickness of the material, T_{1} is the temperature on the first side and T_{2} is the temperature on the second side

k_{skin}=\frac{QL}{A(T_{2}-T_{1})}

k_{skin}=\frac{Q}{A}*\frac{L}{(T_{2}-T_{1})}

k_{skin}=1.05*10^{4}*\frac{1*10^{-3}}{(37-30)}

k_{skin}=1.5W/mK

b)

k_{insulation}=\frac{k_{skin}}{2}

k_{insulation}=\frac{1.5}{2}

k_{insulation}=0.75W/mK

The heat flux between both surfaces is constant, assuming the temperature is maintained at each surface.

\frac{Q}{A}=\frac{k(T_{2}-T_{1})}{L}

\frac{k_{skin}(T_{skin}-T_{interface})}{L_{skin}}=\frac{k_{insulation}(T_{interface}-T_{insulation})}{L_{insulation}}

\frac{1.5*(37-T_{interface})}{0.001}=\frac{0.75*(T_{interface}-30)}{0.002}

55500-1500T_{interface}=375T_{interface}-11250

1875T_{interface}=66750

T_{interface}=35.6\°C

\frac{Q}{A}=\frac{k_{skin}(T_{skin}-T_{interface})}{L_{skin}}

\frac{Q}{A}=\frac{1.5*(37-35.6)}{0.001}

\frac{Q}{A}=2100W/m^2

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A lagoon is designed to accommodate an input flow of 0.10 m^3/s of nonconservative pollutant with concentration 30 mg/L and deca
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Answer:

Volume of the lagoon required for the decay process must be larger than 86580 m³ = 8.658 × 10⁷ L

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The performance equation of a Mixed flow reactor is given from the material and component balance thus:

(V/F₀) = (C₀ - C)/((C₀)(-r)) (From the Chemical Reaction Engineering textbook, authored by Prof. Octave Levenspiel)

V = volume of the reactor (The lagoon) = ?

C₀ = Initial concentration of the reactant (the pollutant concentration) = 30 mg/L = 0.03 mg/m³

F₀ = Initial flow rate of reactant in mg/s = 0.10 m³/s × C₀ = 0.1 m³/s × 0.03 mg/m³ = 0.003 mg/s

C = concentration of reactant at any time; effluent concentration < 10mg/L, this means the maximum concentration of pollutant allowed in the effluent is 10 mg/L

For the sake of easy calculation, C = the maximum value = 10 mg/L = 0.01 mg/m³

(-r) = kC (Since we know this decay process is a first order reaction)

This makes the performance equation to be:

(kVC₀/F₀) = (C₀ - C)/C

V = F₀(C₀ - C)/(kC₀C)

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V = 0.003(0.03 - 0.01)/(2.31 × 10⁻⁶ × 0.03 × 0.01)

V = 86580 m³

Since this calculation is made for the maximum concentration of 10mg/L of pollutant in the effluent, the volume obtained is the minimum volume of reactor (lagoon) to ensure a maximum volume of 10 mg/L of pollutant is contained in the effluent.

The lower the concentration required for the pollutant in the effluent, the larger the volume of reactor (lagoon) required for this decay reaction. (Provided all the other parameters stay the same)

Hope this helps!

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