Question

(a) Calculate the rate of heat transfer by radiation from a car radiator at 110°C into a 50.0ºC environment, if the radiator has an emissivity of 0.750 and a\({\bf{1}}{\bf{.20}}\;{{\bf{m}}^{\bf{2}}}\)surface area. (b) Is this a significant fraction of the heat transfer by an automobile engine? To answer this, assume a horsepower of 200 hp (1.5 kW) and the efficiency of automobile engines as 25%.

Short answer

(a) The rate of heat transfer by radiation from car radiator is\({\bf{542}}{\bf{.6}}\;{\bf{J/s}}\).

(b) The fraction of heat transfer is not significant according to efficiency of car engine

Step-by-step solution

Determination of rate of heat transfer by radiation from car radiator(a)Given Data:

The area of car radiator is\(A = 1.20\;{{\rm{m}}^2}\)

The temperature of car radiator is\({T_1} = 110^\circ {\rm{C}} = 383\;{\rm{K}}\)

The temperature of environment is\({T_2} = 50^\circ {\rm{C}} = 323\;{\rm{K}}\)

The emissivity of car radiator is\(\varepsilon = 0.750\)

The efficiency of automobile engines is\(\eta = 25\% = 0.25\)

The power of car engine is: \(P = 1.5\;{\rm{kW}} = 1500\;{\rm{W}}\)

The radiant energy transfer is calculated by using the Stefan law. This law gives the energy transfer without any medium and heat transfer varies with the difference in temperatures.

The rate of heat transfer by radiation from car radiator is given as:

\(Q = \varepsilon \sigma A\left( {T_2^4 - T_1^4} \right)\)

Here, \(\sigma \) is Stefan’s constant and its value is \(5.67 \times {10^{ - 8}}\;{\rm{W}} \cdot {{\rm{m}}^{ - 2}} \cdot {{\rm{K}}^4}\)

\(\begin{aligned}{}Q &= \left( {0.750} \right)\left( {5.67 \times {{10}^{ - 8}}\;{\rm{W}} \cdot {{\rm{m}}^{ - 2}} \cdot {{\rm{K}}^4}} \right)\left( {1.20\;{{\rm{m}}^2}} \right)\left( {{{\left( {383\;{\rm{K}}} \right)}^4} - {{\left( {323\;{\rm{K}}} \right)}^4}} \right)\\Q &= 542.6\;{\rm{W}}\\Q &= \left( {542.6\;{\rm{W}}} \right)\left( {\frac{{1\;{\rm{J}}/{\rm{s}}}}{{1\;{\rm{W}}}}} \right)\\Q &= 542.6\;{\rm{J}}/{\rm{s}}\end{aligned}\)

Therefore, the rate of heat transfer by radiation from car radiator is \(542.6\;{\rm{J}}/{\rm{s}}\).

: Determination of fraction of heat transfer from automobile engine power(b)

The fraction of heat transfer from car radiator is given as:

\(\begin{aligned}{}{\eta _s} &= \frac{Q}{P}\\{\eta _s} &= \frac{{542.6\;{\rm{W}}}}{{1500\;{\rm{W}}}}\\{\eta _s} &= 0.362\\{\eta _s} &= 36.2\% \end{aligned}\)

As the fraction of heat transfer from car radiator by radiation is\(36.2\% \), which is more than the efficiency of automobile engines so the fraction of heat transfer is more than the significant fraction of heat transfer for car engine.

Therefore, the fraction of heat transfer is not significant according to efficiency of car engine.