Question

A cylindrical hot water tank (height \(1[\mathrm{~m}]\), radius \(22[\mathrm{~cm}]\) ) is heated during the day from \(20\left[{ }^{\circ} \mathrm{C}\right]\) to \(80\left[{ }^{\circ} \mathrm{C}\right]\). (a) Check that the heat capacity would be enough for two baths of \(120[\mathrm{~L}]\) at \(50\left[{ }^{\circ} \mathrm{C}\right]\) or seven showers \((40[\mathrm{~L}])\) at the same temperature, assuming the temperature of tap water is \(20\left[{ }^{\circ} \mathrm{C}\right]\). (b) The tank is insulated on all sides with \(d=10\) \([\mathrm{cm}]\) urethane foam. The air temperature is \(10\left[{ }^{\circ} \mathrm{C}\right]\). Estimate the loss by conduction for the first hour, assuming all insulated surfaces are flat (a very good approximation) and ignore the temperature decrease with time. Look at your results and argue why this is an acceptable assumption. Use \(h=10\left[\mathrm{Wm}^{-2} \mathrm{~K}^{-1}\right]\) for convection, ignore radiation.

Short answer

The tank can sufficiently provide heat for two baths but not for seven showers. The heat loss in one hour is \(114.912\, [\mathrm{kJ}]\), which is relatively small, validating the initial assumptions.

Step-by-step solution

Calculate the Total Heat Capacity of the Tank

The volume of the cylindrical tank is calculated using the formula \(V = \pi r^2 h\), where \(r = 0.22\,[\mathrm{~m}]\) and \(h = 1\,[\mathrm{~m}]\). The total volume is \(V = \pi \times (0.22)^2 \times 1 \approx 0.1521\,[\mathrm{~m}^3}\). Convert this to liters, \(0.1521\,[\mathrm{~m}^3} = 152.1\,[\mathrm{~L}]\). The specific heat capacity of water is \(4.18\,[\mathrm{~kJ/kg⋅K}]\) and the density is \(1\,[\mathrm{~kg/L}]\). The total heat capacity \(Q_{total} = 4.18 \times 152.1 \times (80 - 20) = 38139.6\,[\mathrm{~kJ}]\).

Calculate Heat Demands for Baths and Showers

For a single bath at \(120\,[\mathrm{~L}]\) and temperature \(50\,[^{\circ} \mathrm{C}]\): Heat required \(Q_{bath} = 4.18 \times 120 \times (50 - 20) = 15048\,[\mathrm{~kJ}]\). Two baths require \(2 \times 15048 = 30096\,[\mathrm{~kJ}]\). For one shower at \(40\,[\mathrm{~L}]\): \(Q_{shower} = 4.18 \times 40 \times (50 - 20) = 5026\,[\mathrm{~kJ}]\). Seven showers need \(7 \times 5026 = 35182\,[\mathrm{~kJ}]\).

Check If Heat Capacity Is Sufficient

The total heat available is \(38139.6\,[\mathrm{kJ}]\). Heat needed for two baths is \(30096\,[\mathrm{kJ}]\), which is less than the total heat, hence, the tank can provide enough thermal energy for two baths. However, seven showers require \(35182\,[\mathrm{kJ}]\), which exceeds the tank's capacity, so it cannot provide enough energy for seven showers under these conditions.

Calculate Heat Loss During the First Hour

Estimate the heat loss due to conduction through the insulation using the formula \(Q_{loss} = \frac{kA(T_{hot} - T_{air})}{d} \, \Delta t\). Use \(k = 0.03\, [\mathrm{~W/m⋅K}]\) for urethane foam, surface area \(A = 2\pi rh + 2\pi r^2 \approx 1.52\, [\mathrm{~m^2}]\), temperature difference \(\Delta T = 70\, [\mathrm{~K}]\), and \(d = 0.10\, [\mathrm{~m}]\). The heat loss \(Q_{loss} = \frac{0.03 \times 1.52 \times 70}{0.10} = 31.92\, [\mathrm{~W}]\); for one hour (3600 seconds), \(31.92 \times 3600 = 114912\, [\mathrm{~J}]\) or \(114.912\, [\mathrm{~kJ}]\).

Validate Temperature Assumption

The calculated heat loss is \(114.912\,[\mathrm{kJ}]\), which is a small fraction of the total heat capacity \(38139.6\,[\mathrm{kJ}]\). This small percentage loss during one hour justifies the assumption of no significant temperature drop during this period, allowing a constant temperature assumption for short-term analysis.

Key concepts

Thermal Energy

Thermal energy plays a crucial role in our daily lives, especially in the context of heating and cooling processes. It refers to the energy possessed by an object due to the motion of its particles. This energy is typically associated with temperature changes in the object. When you heat water in a tank, you are essentially increasing its thermal energy.

This energy is measured in terms of heat capacity, a property that signifies how much energy is needed to change an object’s temperature. In our water tank scenario, the heat capacity helps us understand how much thermal energy is required to heat the water from 20°C to 80°C.

To align thermal energy with practical application, note that the specific heat capacity of water is quite high (4.18 kJ/kg⋅K). This means that water can absorb or release large amounts of energy without huge temperature swings, making it ideal for thermal storage applications like bathing or shower usage. It allows us to calculate if there is enough heat stored to accommodate two baths or several showers at desired temperatures.

Insulation

Insulation is a key factor in controlling heat loss from systems like water tanks. The ability of materials to resist heat flow is termed insulation. For effective insulation, we often use materials that have low thermal conductivity, as they hinder the transfer of thermal energy.

In our example, urethane foam is used as the insulating material. With a relatively low thermal conductivity (k = 0.03 W/m⋅K), it reduces the heat loss from the heated water in the tank to the surrounding environment, which is at a lower temperature.

The thickness of the insulation matters too. A thicker layer of insulation (here, 10 cm) effectively reduces heat transfer, minimizing thermal energy loss through conduction. Good insulation ensures that the water in the tank remains hot for longer periods, making it more efficient by reducing the need for frequent reheating.

Convection

Convection is a mode of heat transfer that involves the movement of fluid, whether it be a liquid or a gas. It’s a process that occurs when warmer areas of a liquid or gas rise to cooler areas in the fluid. This movement helps to transfer heat and often sets up a cycle, called a convection current, which further facilitates heat dispersion.

In our water tank, convection within the water is crucial for distributing the heat evenly throughout the tank. This means that even if heat is applied at the bottom of the tank, convection currents help ensure the entire volume reaches a uniform temperature.

Externally, convection also plays a role when considering heat loss to the air surrounding the tank. However, for this scenario, convection heat loss, represented by a convective heat transfer coefficient (h = 10 W/m²K), is considered separately from conduction. Good design of systems accounts for this to minimize overall heat loss.

Conduction

Conduction refers to the transfer of thermal energy through a material without any movement of the material itself. It occurs due to the vibration and movement of particles at the microscopic level, passing kinetic energy from high energy areas to lower energy areas.

In the water tank scenario, conduction is critical in understanding insulation performance. The heat transferred through the foam insulation from the hot water inside the tank to the cooler air outside is primarily through conduction.

The efficiency of this process can be quantified using the formula: \[ Q_{loss} = \frac{kA(T_{hot} - T_{air})}{d} \, \Delta t \] where every variable aids in calculating how much heat is lost over a particular duration (in this case, an hour).

This understanding helps in ensuring the insulation is sufficient to maintain water temperatures with minimal energy loss. Conduction is a key consideration when designing efficient thermal systems to minimize energy consumption.