Question

Chickens with an average mass of \(2.2 \mathrm{kg}\) and average specific heat of \(3.54 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\) are to be cooled by chilled water that enters a continuous-flow-type immersion chiller at \(0.5^{\circ} \mathrm{C}\). Chickens are dropped into the chiller at a uniform temperature of \(15^{\circ} \mathrm{C}\) at a rate of 500 chickens per hour and are cooled to an average temperature of \(3^{\circ} \mathrm{C}\) before they are taken out. The chiller gains heat from the surroundings at a rate of \(200 \mathrm{kJ} / \mathrm{h}\). Determine \((a)\) the rate of heat removal from the chickens, in \(\mathrm{kW}\), and \((b)\) the mass flow rate of water, in \(\mathrm{kg} / \mathrm{s},\) if the temperature rise of water is not to exceed 3.

Short answer

1. The rate of heat removal from the chickens is 12.963 kW. 2. The mass flow rate of water required to cool the chickens, without exceeding a 3°C temperature rise, is 1.0406 kg/s.

Step-by-step solution

Calculate the heat removal rate from the chickens

First, we will find the amount of heat removed from the chickens. To do this, we will use the heat transfer equation for the chickens: $$Q_{chickens} = mc(T_{final} - T_{initial})$$ where \(m\) is the mass of the chickens, \(c\) is their specific heat, \(T_{final}\) is the final temperature, and \(T_{initial}\) is the initial temperature. Given: - The average mass of a chicken (\(m_{chicken}\)) = 2.2 kg - The specific heat of a chicken (\(c_{chicken}\)) = 3.54 kJ/kg°C - The initial temperature of the chickens (\(T_{initial}\)) = 15°C - The final temperature of the chickens (\(T_{final}\)) = 3°C - The rate at which the chickens are dropped into the chiller = 500 chickens/hour First, calculate the total mass of the chickens per hour: $$m_{total} = 500 \text{ chickens/hour} \times 2.2 \text{ kg/chicken} = 1,100 \text{ kg/hour}$$ Now, we can find the heat transfer from the chickens: $$Q_{chickens} = 1,100 \text{ kg/h} \times 3.54 \text{ kJ/kg°C} \times (3 \text{ °C} - 15 \text{ °C}) = -46,668 \text{ kJ/h}$$ Since the heat removal from the chickens is negative, we will take the absolute value (loss of heat): $$Q_{chickens} = 46,668 \text{ kJ/h}$$ To convert this to a rate in kW, divide by 3,600 seconds: $$Q_{chickens} = \frac{46,668 \text{ kJ/h}}{3,600 \text{ s/h}} = 12.963 \text{ kW}$$ Therefore, the rate of heat removal from the chickens is \(12.963\) kW.

Calculate the mass flow rate of water

Next, we will determine the mass flow rate of water required to cool the chickens. For this, we need to consider the heat gained from the surroundings as well. The heat transfer equation for the water is: $$Q_{water} = mwc_w(T_{water, final} - T_{water, initial})$$ Given: - The initial temperature of the water (\(T_{water, initial}\)) = 0.5°C - The maximum allowed temperature rise of water (\(\Delta T_{water}\)) = 3°C - The heat gained from the surroundings (\(Q_{surroundings}\)) = 200 kJ/h Taking the total heat transfer to be equal to the sum of the heat removal from the chickens and the heat gained from the surroundings, we have: $$Q_{total} = Q_{chickens} + Q_{surroundings}$$ Converting the heat gained from the surroundings to kW: $$Q_{surroundings} = \frac{200 \text{ kJ/h}}{3,600 \text{ s/h}} = 0.0556\text{ kW}$$ Now, we have: $$Q_{total} = 12.963 \text{ kW} + 0.0556 \text{ kW} = 13.0186 \text{ kW}$$ The final temperature of the water can be calculated as: $$T_{water, final} = T_{water, initial} + \Delta T_{water} = 0.5 \text{ °C} + 3 \text{ °C} = 3.5 \text{ °C}$$ Now, we have everything needed to solve for the mass flow rate of water (\(mw\)): $$13.0186 \text{ kW} = mw \times 4.18 \text{ kJ/kg°C} \times (3.5 \text{ °C} - 0.5 \text{ °C})$$ Solving for \(mw\): $$mw = \frac{13.0186 \text{ kW}}{4.18 \text{ kJ/kg°C} \times 3 \text{ °C}} = \frac{13.0186 \times 10^3 \text{ W}}{4.18 \times 10^3 \text{ J/kg°C} \times 3 \text{ °C}} = 1.0406 \text{ kg/s}$$ Therefore, the mass flow rate of water required is \(1.0406\) kg/s.