Heat and Mass Transfer The most commonly used method for the design of duct size is the Static regains method Equal friction method Velocity reduction method Dual or double method Static regains method Equal friction method Velocity reduction method Dual or double method ANSWER DOWNLOAD EXAMIANS APP
Heat and Mass Transfer Fourier's law of heat conduction is (where Q = Amount of heat flow through the body in unit time, A = Surface area of heat flow, taken at right angles to the direction of heat flow, dT = Temperature difference on the two faces of the body, dx = Thickness of the body, through which the heat flows, taken along the direction of heat flow, and k = Thermal conductivity of the body) k. (dx/dT) (dT/dx) k. (dT/dx) k. (dx/dT) k. k. (dx/dT) (dT/dx) k. (dT/dx) k. (dx/dT) k. ANSWER DOWNLOAD EXAMIANS APP
Heat and Mass Transfer The heat transfer by conduction through a thick cylinder (Q) is given by (where T₁ = Higher temperature, T₂ = Lower temperature, r₁ = Inside radius, r₂ = Outside radius, l = Length of cylinder, and k = Thermal conductivity) Q = = 2πlk/2.3 (T₁ - T₂) log (r₂/r₁) Q = [2π (T₁ - T₂)]/2.3 lk log (r₂/r₁) Q = 2.3 log (r₂/r₁)/[2πlk (T₁ - T₂)] Q = [2πlk (T₁ - T₂)]/2.3 log (r₂/r₁) Q = = 2πlk/2.3 (T₁ - T₂) log (r₂/r₁) Q = [2π (T₁ - T₂)]/2.3 lk log (r₂/r₁) Q = 2.3 log (r₂/r₁)/[2πlk (T₁ - T₂)] Q = [2πlk (T₁ - T₂)]/2.3 log (r₂/r₁) ANSWER DOWNLOAD EXAMIANS APP
Heat and Mass Transfer Stefan Boltzmann law is applicable for heat transfer by Convection Conduction Conduction and radiation combined Radiation Convection Conduction Conduction and radiation combined Radiation ANSWER DOWNLOAD EXAMIANS APP
Heat and Mass Transfer According to Stefan Boltzmann law, the total radiation from a black body per second per unit area is directly proportional to the Absolute temperature Square of the absolute temperature Fourth power of the absolute temperature Cube of the absolute temperature Absolute temperature Square of the absolute temperature Fourth power of the absolute temperature Cube of the absolute temperature ANSWER DOWNLOAD EXAMIANS APP
Heat and Mass Transfer Two balls of same material and finish have their diameters in the ratio of 2: 1 and both are heated to same temperature and allowed to cool by radiation. Rate of cooling by big ball as compared to smaller one will be in the ratio of 0.16736111111111 0.084027777777778 0.042361111111111 0.043055555555556 0.16736111111111 0.084027777777778 0.042361111111111 0.043055555555556 ANSWER DOWNLOAD EXAMIANS APP