Consider a person standing in a room at 23°c. determine the total rate of heat transfer from this person if the exposed surface area and the skin temperature of the person are 1.7 m2 and 32°c, respectively, and the convection heat transfer coefficient is 3.5 w/m2⋅k. take the emissivity of the skin and the clothes to be 0.9, and assume the temperature of the inner surfaces of the room to be the same as the air temperature.

The total rate of heat transfer is 138.8894 W

Explanation:

Step 1: Given data

⇒ Temperature in the room Tr= 23° C

⇒ Exposed surface area As= 1.7 m²

⇒ Person's skin temperature Ts= 32 ° C

⇒ heat transfer coefficient h : 3.5 W/m²⋅K

⇒ Emissivity ε = 0.9

⇒ ( Stefan Boltzmann constant σ = 5.67×10^−8 W⋅m^−2⋅K^−4

Step 2: Calculate Qrad

Heat transfer Q = Qrad + Qconv   (Thermal properties of people are assumed to be constant. The heat transfer will take place due to both convective and radiation heat loss)

⇒ Qrad = Heat transfer from the person to the surroundings due to radiation

Q = ε*σ*As * (Ts ^4  - Tr ^4 ) + h* As*ΔT

⇒ Qrad =ε*σ*As * (Ts ^4  - Tr ^4)

     Qrad = 0.9 * (5.7 * 10 ^ -8) * 1.7 * ((32 + 273.15)^4  -(18 +273.15)^4)

Qrad = 85.3394 W

Step 3: Calculate Qconv

Qconv = The heat transfer from the person to the surroundings due to convection

Qconv =  h* As*ΔT

⇒ Q conv = 3.5 W/m²⋅K * 1.7 m² * (32 - 23)

Qconv = 3.5 * 1.7 * 9 = 53.55 W

Step 4: Calculate Q

Q = Qrad + Qconv = 85.3394 + 53.55 = 138.8894 W

The total rate of heat transfer is 138.8894 W

conservation of mass

mass before something happens = mass afterwards

the components of a vector are the parts of the vectors that are to each other.

b perpendicular