Question

A refrigeration system is to cool bread loaves with an average mass of 350 g from 30 to \(-10^{\circ} \mathrm{C}\) at a rate of 1200 loaves per hour by refrigerated air at \(-30^{\circ} \mathrm{C}\). Taking the average specific and latent heats of bread to be \(2.93 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\) and \(109.3 \mathrm{kJ} / \mathrm{kg},\) respectively, determine \((a)\) the rate of heat removal from the breads, in \(\mathrm{kJ} / \mathrm{h} ;(b)\) the required volume flow rate of air, in \(m^{3} / h,\) if the temperature rise of air is not to exceed \(8^{\circ} \mathrm{C} ;\) and \((c)\) the size of the compressor of the refrigeration system, in \(\mathrm{kW}\), for a COP of 1.2 for the refrigeration system.

Short answer

(b) What is the required volume flow rate of air for the cooling process? (c) What is the size of the compressor for the refrigeration system in terms of power (kW)?

Step-by-step solution

Calculate the rate of heat removal

First, let's calculate the total heat to be removed from each bread loaf during the cooling process: $$Q_{total} = m_q \cdot (c_p \cdot \Delta T + L_f)$$ where \(m_q\) is the mass of a bread loaf, \(c_p\) is the specific heat capacity, \(\Delta T\) is the difference in temperature, and \(L_f\) is the latent heat. Substitute the given values and find the total heat removal Q_total for each bread loaf: $$Q_{total} = 0.350\,\text{kg} \cdot (2.93\frac{\text{kJ}}{\text{kg}\cdot^{\circ}\text{C}}\cdot(30-(-10))^{\circ}\text{C} + 109.3\frac{\text{kJ}}{\text{kg}})$$ Now, find the rate of heat removal Q_rate for 1200 bread loaves per hour: $$Q_{rate} = 1200 \cdot Q_{total}$$ Calculate the values and keep this result for later use.

Calculate the required volume flow rate of air

Next, we need to find the required volume flow rate of air. To do this, we will use the equation for the mass flow rate of air, given as follows: $$\dot{m} = \frac{Q_{rate}}{c_p \cdot \Delta T_{air}}$$ where \(\Delta T_{air}\) is the temperature rise of air and \(c_p\) refers to the specific heat capacity of air \(1.006 \frac{\mathrm{kJ}}{\mathrm{kg} \cdot ^{\circ}\mathrm{C}}\). The mass flow rate of air can also be related to the volume flow rate as: $$\dot{V} = \dot{m} \cdot \frac{1}{\rho} = \frac{Q_{rate}}{c_p \cdot \Delta T_{air} \cdot \rho}$$ We are given the temperature of the air stream rises by \(8^{\circ}\mathrm{C}\), and the air density at \(-30^{\circ}\mathrm{C}\) is approximately \(1.52\,\mathrm{kg} / \mathrm{m}^3\). Substitute these values and the heat removal rate from Step 1 and solve for the volume flow rate of air \(\dot{V}\) (in \(\mathrm{m}^3 / \mathrm{h}\)).

Calculate the size of the compressor

Lastly, we need to calculate the size of the compressor of the refrigeration system. We are given the COP (Coefficient of Performance) of the refrigeration system, which is defined as: $$\text{COP} = \frac{Q_{rate}}{W_{comp}}$$ Here, \(W_{comp}\) represents the power required by the compressor (in \(\mathrm{kW}\)). Using the COP given, we can solve for \(W_{comp}\): $$W_{comp} = \frac{Q_{rate}}{\text{COP}}$$ Substitute the heat removal rate from Step 1 and the given COP value, then calculate the compressor size in \(\mathrm{kW}\). Now we have calculated all the required values: (a) the rate of heat removal from the breads, (b) the required volume flow rate of air, and (c) the size of the compressor for the refrigeration system.