Question

Long cylindrical steel rods (\(\rho=7833 \mathrm{kg} / \mathrm{m}^{3}\) and \(\left.c_{p}=0.465 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) of \(10-\mathrm{cm}\) diameter are heat treated by drawing them at a velocity of \(3 \mathrm{m} / \mathrm{min}\) through a 7 -m-long oven maintained at \(900^{\circ} \mathrm{C}\). If the rods enter the oven at \(30^{\circ} \mathrm{C}\) and leave at \(700^{\circ} \mathrm{C}\), determine ( \(a\) ) the rate of heat transfer to the rods in the oven and \((b)\) the rate of entropy generation associated with this heat transfer process.

Short answer

b) What is the rate of entropy generation associated with this heat transfer process?

Step-by-step solution

Calculate the mass flow rate

To calculate the mass flow rate, we need to first find the cross-sectional area and volume flow rate of the rods. The diameter of the rods is given as 10 cm, so the radius is 5 cm or 0.05 m. The cross-sectional area can be calculated using the formula for the area of a circle $$A = \pi r^2$$ Using the given radius, $$A = \pi (0.05)^2 \approx 0.007853\, \mathrm{m^²}$$ Now, we need to find the volume flow rate. The rods are moving at 3 m/min, which can be converted to m/s by dividing by 60 $$V = \frac{3\, \mathrm{m} / \mathrm{min}}{60\, \mathrm{s}} \approx 0.05\, \mathrm{m} / \mathrm{s}$$ The volume flow rate can be calculated as $$Q_{vol} = A \times V \approx 0.007853 \times 0.05 \approx 0.0003926\, \mathrm{m^3/s}$$ Finally, we can calculate the mass flow rate using the density of steel rods $$\dot{m} = \rho \times Q_{vol} \approx 7833 \times 0.0003926 \approx 3.074\, \mathrm{kg/s}$$

Calculate the rate of heat transfer

Now that we have the mass flow rate, we can calculate the rate of heat transfer using the specific heat capacity (\(c_p\)) and the temperature change. The temperature change is such that the rods enter at 30 °C and leave at 700 °C, so the overall change is $$(T_{out} - T_{in}) = 700 - 30 = 670\, °C$$ The rate of heat transfer (\(\dot{Q}\)) can be calculated using the following formula $$\dot{Q} = \dot{m} \times c_p \times \Delta T$$ Using the given values, $$\dot{Q} = 3.074 \times 0.465 \times 670 \approx 1007.98\, \mathrm{kJ/s} \approx 1008\, \mathrm{kJ/s}$$ So, the rate of heat transfer is 1008 kJ/s.

Calculate the rate of entropy generation

Finally, we'll calculate the rate of entropy generation. The entropy balance equation for an open system states: $$\frac{\Delta S_{gen}}{\Delta t} = \frac{\dot{Q}}{T_b}$$ where \(\Delta S_{gen}\) is the entropy generated, \(\Delta t\) is the time and \(T_b\) is the boundary temperature in Kelvin (oven temperature). First, we need to convert the given oven temperature in °C to Kelvin: $$T_b = 900 + 273.15 \approx 1173.15 K$$ Now, we can calculate the rate of entropy generation $$\frac{\Delta S_{gen}}{\Delta t} = \frac{1008 \times 10^3 \, \mathrm{J/s}}{1173.15 \, \mathrm{K}} \approx 859.79\, \mathrm{J/(s \cdot K)}$$ The rate of entropy generation is approximately 859.79 J/(s∙K). To summarize, (a) the rate of heat transfer to the rods in the oven is 1008 kJ/s, and (b) the rate of entropy generation associated with this heat transfer process is approximately 859.79 J/(s∙K).