Question

A glass bottle washing facility uses a well agi(Es) tated hot water bath at \(50^{\circ} \mathrm{C}\) with an open top that is placed on the ground. The bathtub is \(1 \mathrm{~m}\) high, \(2 \mathrm{~m}\) wide, and \(4 \mathrm{~m}\) long and is made of sheet metal so that the outer side surfaces are also at about \(50^{\circ} \mathrm{C}\). The bottles enter at a rate of 800 per minute at ambient temperature and leave at the water temperature. Each bottle has a mass of \(150 \mathrm{~g}\) and removes \(0.6 \mathrm{~g}\) of water as it leaves the bath wet. Makeup water is supplied at \(15^{\circ} \mathrm{C}\). If the average conditions in the plant are \(1 \mathrm{~atm}, 25^{\circ} \mathrm{C}\), and 50 percent relative humidity, and the average temperature of the surrounding surfaces is \(15^{\circ} \mathrm{C}\), determine (a) the amount of heat and water removed by the bottles themselves per second, \((b)\) the rate of heat loss from the top surface of the water bath by radiation, natural convection, and evaporation, \((c)\) the rate of heat loss from the side surfaces by natural convection and radiation, and \((d)\) the rate at which heat and water must be supplied to maintain steady operating conditions. Disregard heat loss through the bottom surface of the bath and take the emissivities of sheet metal and water to be \(0.61\) and \(0.95\), respectively.

Short answer

Question: Determine the rate at which heat and water must be supplied to maintain steady operating conditions in a bottle washing facility. Answer: To maintain steady operating conditions, the necessary heat input and water input must be calculated as follows: - Heat input: Q_dot_input = Q_dot_removed_by_bottles + Q_dot_radiation + Q_dot_convection_evaporation + Q_dot_radiation_sides + Q_dot_convection_sides - Water input: m_dot_water_input = m_dot_water_removed_by_bottles

Step-by-step solution

Calculate the amount of heat and water removed by the bottles

First, we need to determine the heat and water removed by the bottles as they leave the bath. The bottles are entering at \(25^{\circ} \mathrm{C}\) and leaving at \(50^{\circ} \mathrm{C}\). Given that there are 800 bottles per minute with a mass of \(150 \mathrm{~g}\) each and remove \(0.6 \mathrm{~g}\) of water as they leave the bath. We will convert the mass flow rate of bottles and water removed per second. To calculate the heat removed by the bottles: 1. Determine the heat capacity of the bottles: We can assume they are made of glass, which has a specific heat capacity of \(c_p \approx 840 \mathrm{J/(kg·K)}\) 2. Calculate the total mass flow rate of the bottles per second: m_dot = (number of bottles/minute * mass of bottle) / 60 3. Calculate the temperature difference: ΔT = T_exit - T_entry 4. Calculate the heat removed by the bottles: Q_dot = m_dot * c_p * ΔT To calculate the water removed by the bottles: 1. Calculate the total mass flow rate of water removed per second: m_dot_water = (number of bottles/minute * mass of water removed) / 60

Calculate the heat loss from the top surface of the water bath

Next, we need to calculate the heat loss from the top surface of the water bath by radiation, natural convection, and evaporation. First, determine the surface area of the top surface: A = 2 * 4 m² To calculate the heat loss due to radiation: 1. Determine the Stefan-Boltzmann constant: σ = 5.67×10^-8 W/(m²·K^4) 2. Calculate the emissivity of the water: ε_water = 0.95 3. Calculate the temperature of the water in Kelvin: T_water = 50 + 273.15 K 4. Calculate the temperature of the surrounding surfaces in Kelvin: T_surrounding = 15 + 273.15 K 5. Calculate the heat loss due to radiation: Q_dot_radiation = ε_water * σ * A * (T_water^4 - T_surrounding^4) To calculate the heat loss due to natural convection and evaporation, we can use the following empirical correlation: Q_dot_convection_evaporation = 230 * A * ΔT

Calculate the heat loss from the side surfaces of the water bath

Similarly, we need to calculate the heat loss from the side surfaces of the water bath by natural convection and radiation. First, determine the surface area of the side surfaces: A_side = (1 * 2 + 1 * 4) * 2 m² To Calculate the heat loss due to radiation: 1. Determine the emissivity of the sheet metal: ε_metal = 0.61 2. Calculate the heat loss due to radiation for side surfaces: Q_dot_radiation_sides = ε_metal * σ * A_side * (T_water^4 - T_surrounding^4) To calculate the heat loss due to natural convection for side surfaces, we can use a similar empirical correlation for the vertical plates: Q_dot_convection_sides = 8 * A_side * ΔT

Calculate the rate at which heat and water must be supplied

Finally, we need to determine the rate at which heat and water must be supplied to maintain steady operating conditions. To calculate the necessary heat input: Q_dot_input = Q_dot_removed_by_bottles + Q_dot_radiation + Q_dot_convection_evaporation + Q_dot_radiation_sides + Q_dot_convection_sides To calculate the necessary water input, we simply need to balance the mass flow rate of water being removed by the bottles: m_dot_water_input = m_dot_water_removed_by_bottles Now we have calculated the amount of heat and water removed by the bottles, the rates of heat loss from the top surface by radiation, natural convection, and evaporation, the rates of heat loss from the side surfaces by natural convection and radiation, and the rate at which heat and water must be supplied to maintain steady operating conditions.

Key concepts

Radiation

Radiation is the process by which energy is emitted as electromagnetic waves or as moving subatomic particles, especially high-energy particles that cause ionization. In the context of heat transfer, radiation allows energy to be transferred through space without the need for a medium, which makes it distinct from conduction and convection. In our specific problem, the top and sides of the bathtub are losing heat to the environment via radiation.

When calculating heat loss due to radiation, key factors include the emissivity of the material and the temperatures of the surfaces involved. Emissivity is a measure of how effectively a surface emits energy; water and sheet metal have different emissivities, with water being more emissive (e_water = 0.95) compared to sheet metal (e_metal = 0.61). The formula to calculate the radiative heat loss is based on the Stefan-Boltzmann law: \[ Q_{dot ext{ radiation}} = \epsilon \cdot \sigma \cdot A \cdot (T_{object}^4 - T_{environment}^4) \]

• \(\sigma\) is the Stefan-Boltzmann constant, \( 5.67 \times 10^{-8} W/(m^2·K^4) \)
• \(A\) is the surface area
• \(T_{object}\) and \(T_{environment} \) are the absolute temperatures of the object and its surroundings, respectively

It's crucial to convert temperatures into Kelvin by adding e273.15 to the Celsius reading. This will ensure accurate calculations. Understanding how radiative heat transfer works is essential for determining energy efficiency, especially in systems like the bathtub discussed here.

Convection

Convection is the transfer of heat through the movement of fluids such as air or water. In natural convection, this movement is caused by differences in temperature and density within the fluid. In the described bathtub scenario, both the top and sides are subject to natural convection heat losses.

Factors impacting convection include the surface area and the temperature difference between the object and its surroundings. For the heat loss calculations in this example, empirical correlations are used to estimate the heat lost due to convection:

• The formula for the top surface: \[ Q_{dot ext{ convection ext{ and ext{ evaporation}}} = 230 \cdot A \cdot \Delta T \]
• The formula for side surfaces: \[ Q_{dot ext{ convection ext{ sides}}} = 8 \cdot A_{side} \cdot \Delta T \]

Here, \( \Delta T \) is the difference between the water temperature and the ambient temperature.

The calculation distinguishes between the top surface (where evaporation also occurs) and the side surfaces, noted by different coefficients (230 for the top and 8 for the sides). This distinction is due to the unique properties and mechanisms of heat loss experienced by horizontal versus vertical surfaces.
Convection is an important concept because it significantly affects how energy moves through systems, impacting both efficiency and temperature regulation.

Evaporation

Evaporation is the process in which liquid molecules escape into the gas phase from a liquid surface. It occurs due to the energy provided to the molecules, allowing them to break free from intermolecular forces. In our example, evaporation is notable over the open top surface of the water bath.

This process leads to both mass and energy loss. As water molecules evaporate, they take energy with them in the form of latent heat. This results in the cooling of the remaining liquid. In this system, each bottle leaving the bath brings a small amount of water with it, adding to the complexity of heat and mass transfer calculations.

• Evaporation-related energy loss can be combined with natural convection as was shown in the formula:\[ Q_{dot ext{ convection ext{ and ext{ evaporation}}} = 230 \cdot A \cdot \Delta T \]

It's crucial to evaluate evaporation when dealing with open systems—like the bathtub— to ensure the system can maintain the desired operating conditions. Recognizing the impacts of evaporation also helps design processes to replenish water as needed, thereby maintaining steady state operations and preventing efficiency losses.