Question

In a manufacturing facility, 2-in-diameter brass balls $\left(k=64.1 \mathrm{Btw} / \mathrm{h} \cdot \mathrm{ft}{ }^{\circ} \mathrm{F}, \rho=532 \mathrm{lbm} / \mathrm{ft}^{3}\right.\(, and \)\left.c_{p}=0.092 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\( initially at \)250^{\circ} \mathrm{F}\( are quenched in a water bath at \)120^{\circ} \mathrm{F}$ for a period of \(2 \mathrm{~min}\) at a rate of \(120 \mathrm{balls}\) per minute. If the convection heat transfer coefficient is $42 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2},{ }^{\circ} \mathrm{F}\(, determine \)(a)$ the temperature of the balls after quenching and \((b)\) the rate at which heat needs to be removed from the water in order to keep its temperature constant at $120^{\circ} \mathrm{F}

Short answer

Answer: To find the final temperature of the brass balls after the quenching process and the rate at which heat needs to be removed from the water, follow these steps: 1. Gather given information and known constants. 2. Calculate the surface area and volume of a brass ball. 3. Calculate the total heat transferred from the balls to the water. 4. Determine the final temperature of the balls. 5. Calculate the rate at which heat needs to be removed from the water. After performing these calculations, you will be able to determine (a) the temperature of the balls after quenching and (b) the rate at which heat needs to be removed from the water in order to keep its temperature constant at \(120^{\circ} \mathrm{F}\).

Step-by-step solution

Gather given information and known constants

- Diameter of brass balls, d = 2 inches - Brass balls' initial temperature, \(T_{i} = 250^{\circ} \mathrm{F}\) - Water bath temperature, \(T_{w} = 120^{\circ} \mathrm{F}\) - Quenching time, t = 2 minutes - Rate of quenching, R = 120 balls per minute - Thermal conductivity of brass, k = 64.1 Btu / h ⋅ ft ⋅ \(^{\circ} \mathrm{F}\) - Density of brass, ρ = 532 lbm / \(ft^3\) - Specific heat of brass, \(c_{p} = 0.092 \frac{Btu}{lbm \cdot { }^{\circ} \mathrm{F}}\) - Convection heat transfer coefficient, h = 42 Btu / h ⋅ \(ft^2\) ⋅ \(^{\circ} \mathrm{F}\)

Calculate the surface area and volume of a brass ball

Since the diameter is given in inches, we need to first convert it to feet: \(d_{ft} = \frac{2}{12} = \frac{1}{6} ft\) Now, we can find the surface area (A) and volume (V) of the brass ball: \(A = 4 \pi r^2 = 4 \pi \left(\frac{d_{ft}}{2}\right)^2\) \(V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left(\frac{d_{ft}}{2}\right)^3\)

Calculate the total heat transferred from the balls to the water

First, we determine the mass (m) of one brass ball: \(m = \rho \cdot V\) Next, we find the combined heat absorbed by all balls in 2 minutes, using the formula for heat transfer through convection: \(Q = m \cdot c_{p} \cdot (T_{i} - T_{f})\) The total number of balls quenched in 2 minutes is: \(N = R \cdot t = 120 \cdot 2\) Now, we calculate the total heat absorbed by all balls: \(Q_{total} = N \cdot Q\)

Determine the final temperature of the balls

We know that the heat transfer rate through convection is given by formula: \(\dot{Q} = h \cdot A \cdot (T_{w} - T_{f})\) Using the total heat transferred and the convection formula, we can find the final temperature, \(T_{f}\): \(Q_{total} = \dot{Q} \cdot t_{hr}\), where \(t_{hr} = \frac{1}{30} h\) \(T_{f} = T_{w} - \frac{Q_{total}}{h \cdot A \cdot t_{hr}}\)

Calculate the rate at which heat needs to be removed from the water

The rate at which heat needs to be removed is the same as the heat transfer rate from the balls to the water bath: \(\dot{Q}_{removed} = \dot{Q} = h \cdot A \cdot (T_{w} - T_{f})\) Now, we have determined (a) the temperature of the balls after quenching and (b) the rate at which heat needs to be removed from the water in order to keep its temperature constant at \(120^{\circ} \mathrm{F}\).