Question

Stainless steel ball bearings $$\left(\rho=8085 \mathrm{kg} / \mathrm{m}^{3}\text { and }\right.$$ $$\left.c_{p}=0.480 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)$$ having a diameter of \(1.2 \mathrm{cm}\) are to be \(r\) quenched in water at a rate of 1400 per minute. The balls leave the oven at a uniform temperature of \(900^{\circ} \mathrm{C}\) and are exposed to air at \(30^{\circ} \mathrm{C}\) for a while before they are dropped into the water. If the temperature of the balls drops to \(850^{\circ} \mathrm{C}\) prior to quenching, determine ( \(a\) ) the rate of heat transfer from the balls to the air and \((b)\) the rate of exergy destruction due to heat loss from the balls to the air.

Short answer

Answer: The rate of heat transfer from the balls to the air is 245,616 J/min, and the rate of exergy destruction due to heat loss from the balls to the air is 94,284.15 J/min.

Step-by-step solution

List the known values

We are provided with the following information: 1. Density of the stainless steel balls: \(\rho = 8085 \,\mathrm{kg/m^3}\) 2. Specific heat capacity: \(c_p = 0.480 \,\mathrm{kJ/(kg \cdot ^{\circ} C)}\) 3. Diameter of each steel ball: \(d =1.2 \,\mathrm{cm}\) 4. Quenching rate: \(1400 \,\mathrm{balls/min}\) 5. Initial ball temperature: \(T_{i} = 900 \,^{\circ}\mathrm{C}\) 6. Temperature of the balls before quenching: \(T_{f} = 850 \,^{\circ}\mathrm{C}\) 7. Air temperature: \(T_{air} = 30^{\circ} \mathrm{C}\)

Calculate the mass of a single ball bearing

We can calculate the mass of a single ball with the provided information. 1. Calculate the volume of a single ball: \(V = \dfrac{4}{3}\pi r^{3}\), where \(r\) is the radius of the sphere. 2. Convert the diameter to meters: \(d = 1.2 \,\mathrm{cm} \times \dfrac{1 \,\mathrm{m}}{100 \,\mathrm{cm}} = 0.012 \,\mathrm{m}\). Then, calculate the radius: \(r = \dfrac{d}{2} = 0.006 \,\mathrm{m}\). 3. Calculate the volume: \(V = \dfrac{4}{3}\pi \left( 0.006 \,\mathrm{m} \right)^{3} = 9.051 \times 10^{-7} \,\mathrm{m^3}\). 4. Calculate the mass: \(m = \rho V = 8085 \,\mathrm{kg/m^3} \times 9.051 \times 10^{-7}\, \mathrm{m^3} = 0.0073 \,\mathrm{kg}\).

Calculate the rate of heat transfer

The rate of heat transfer can be determined using the mass, specific heat capacity, temperature difference, and quenching rate. 1. Calculate the temperature difference: \(\Delta T = T_{i} - T_{f} = 900^{\circ}\mathrm{C} - 850^{\circ}\mathrm{C} = 50^{\circ}\mathrm{C}\). 2. Convert specific heat capacity to J/(kg ⋅ °C): \(c_p = 0.480 \,\mathrm{kJ/(kg \cdot ^{\circ} C)} \times \dfrac{1000\, \mathrm{J}}{1\, \mathrm{kJ}} = 480 \,\mathrm{J/(kg ⋅ ^{\circ} C)}\). 3. Calculate the heat transfer rate for a single ball: \(q = mc_p\Delta T = 0.0073\, \mathrm{kg} \times 480\, \mathrm{J/(kg \cdot ^{\circ} C)} \times 50 \,^{\circ}\mathrm{C} = 175.44\, \mathrm{J}\). 4. Calculate the heat transfer rate for 1400 balls: \(Q = 1400\, \mathrm{balls/min} \times \dfrac{175.44\,\mathrm{J}}{\mathrm{ball}} \times 1\, \mathrm{min} = 245616 \, \mathrm{J/min}\). The rate of heat transfer from the balls to the air is \(245616 \,\mathrm{J/min}\).

Calculate the rate of exergy destruction due to heat loss

The rate of exergy destruction can be calculated using the heat transfer rate and the Carnot efficiency. 1. Calculate the Carnot efficiency: \(\eta_{C} = 1 - \dfrac{T_{air}}{T_{f}}\), where temperatures should be in Kelvin. Convert the temperatures: \(T_{air} = 30 + 273.15 = 303.15\, \mathrm{K}\) and \(T_{f} = 850 + 273.15 = 1123.15\, \mathrm{K}\). 2. Calculate the Carnot efficiency: \(\eta_{C} = 1 - \dfrac{303.15\,\mathrm{K}}{1123.15\, \mathrm{K}} = 0.7299\). 3. Calculate the rate of exergy destruction: \(E_{D} = \dfrac{Q \left( 1 - \eta_{C} \right)}{\eta_{C}} = \dfrac{245616\, \mathrm{J/min} \times \left( 1-0.7299\right)}{0.7299} = 94284.15\,\mathrm{J/min}\). The rate of exergy destruction due to heat loss from the balls to the air is \(94284.15\, \mathrm{J/min}\).