Question

Overall, 80% of the energy used by the body must be eliminated as excess thermal energy and needs to be dissipated. The mechanisms of elimination are radiation, evaporation of sweat (2,430 kJ/kg), evaporation from the lungs (38 kJ/h), conduction, and convection.

A person working out in a gym has a metabolic rate of 2,500 kJ/h. His body temperature is 37°C, and the outside temperature 20°C. Assume the skin has an area of 2.0 m² and emissivity of 0.97. (σ = 5.6696  10^-8 W/m² · K⁴)

(a) At what rate is his excess thermal energy dissipated by radiation? (Enter your answer to at least one decimal place.)
W

(b) If he eliminates 0.39 kg of perspiration during that hour, at what rate is thermal energy dissipated by evaporation of sweat? (Enter your answer to at least one decimal place.)
W

(c) At what rate is energy eliminated by evaporation from the lungs? (Enter your answer to at least one decimal place.)  

W

(d) At what rate must the remaining excess energy be eliminated through conduction and convection?

W

Answer 1

Heat Transfer:

Heat Transfer rate is defined as a rate at which heat transfer occurs when there is a difference in temperature. The heat transfer always occurs from high temperature to low temperature.

There are three modes of heat transfer:

a) Conduction

b) Convection

c) Radiation

Given Data:

• Energy eliminated through evaporation of sweat is: qs = 2430 kJ / kg

• Energy eliminated through evaporation from lungs is: =38 kJ / h

• Metabolic rate is: qm = 2500 kJ / h

• Body Temperature is:

• Outside Temperature is:

• Area of skin is: A = 2.0 m2

• Emissivity is:

(a) The rate of energy dissipation by the radiation is calculated by the expression given below,

Here, A is the area, and

is the Stefan-Boltzmann constant and its value is 5.6696×10−8 Wm−2 K−4

Substitute the value in above expression and we get:

(b) The rate of thermal energy dissipated by the evaporation of sweat is calculated by the expression given below:

Here, m is the mass of sweat evaporated and t is time.

Substitute the values in above equation and we get:

Thus, the rate of thermal energy dissipated by evaporation of sweat is 263.25 W.

(c) The rate of thermal energy eliminated from the evaporation of the lung is calculated by the expression given below,

Here, tL is time.

Substitute the values in above expression, and we get:

Thus, the rate of thermal energy eliminated from evaporation of lungs is 10.56W.

(d) The total power to eliminate 80% of energy is,

The rate of remaining excess energy is:

Substitute all the values in above expression, and we get:

Thus, the rate of remaining excess energy be eliminated through conduction and convection is 77.65 W.