Question

Figure P4.44 shows a solar collector panel with a surface area of \(2.97 \mathrm{~m}^{2}\). The panel receives energy from the sun at a rate of \(1.5 \mathrm{~kW}\). Thirty-six percent of the incoming energy is lost to the surroundings. The remainder is used to heat liquid water from \(40^{\circ} \mathrm{C}\) to \(60^{\circ} \mathrm{C}\). The water passes through the solar collector with a negligible pressure drop. Neglecting kinetic and potential energy effects, determine at steady state the mass flow rate of water, in \(\mathrm{kg} .\) How many gallons of water at \(60^{\circ} \mathrm{C}\) can eight collectors provide in a 30 -min time period?

Short answer

The mass flow rate of the water is 0.01148 kg/s for a single collector. Eight collectors provide 43.67 gallons of heated water in 30 minutes.

Step-by-step solution

- Calculate Useful Energy

Determine the useful energy received by the panel. Since 36% of the incoming energy is lost, only 64% is used effectively. Therefore, the useful power is: \[ P_{useful} = 0.64 \times 1.5 \text{ kW} = 0.96 \text{ kW} \] which converts to \[ P_{useful} = 960 \text{ watts} \].

- Calculate Energy Required to Heat Water

The energy required to heat the water can be found using the specific heat formula: \[ Q = \text{mass flow rate} \times \text{specific heat of water} \times \text{temperature change} \]. The specific heat of water \( C_p \) is approximately 4.18 kJ/(kg·°C). The temperature change \( \triangle T \) is from 40°C to 60°C, so: \[ \triangle T = 60 - 40 = 20 \text{ °C} \].

- Set Up the Power Equation

Since power is the rate of energy transfer and we have the useful power (960 W), set up the equation: \[ P_{useful} = \text{mass flow rate} \times C_p \times \triangle T \] Rearrange to solve for the mass flow rate \( \text{m} \): \[ \text{m} = \frac{P_{useful}}{C_p \times \triangle T} \]. Substituting the values: \[ \text{m} = \frac{960 \text{ W}}{4.18 \text{ kJ/(kg·°C)} \times 20 \text{ °C}} \].

- Perform the Calculation

Convert watts to kJ/s (since 1 W = 1 J/s and 1 kJ = 1000 J): \[ \text{m} = \frac{960 \text{ J/s}}{4.18 \text{ kJ/(kg·°C)} \times 20 \text{ °C}} \times \frac{1 \text{ kJ}}{1000 \text{ J}} \] Simplify and calculate the mass flow rate: \[ \text{m} = \frac{960}{4.18 \times 20 \times 1000} \text{ kg/s} \] \[ \text{m} = 0.01148 \text{ kg/s} \].

- Calculate Total Mass Flow for Eight Collectors

Multiply the mass flow rate by the number of collectors to find the total mass flow rate for eight collectors: \[ \text{m}_{total} = 0.01148 \text{ kg/s} \times 8 = 0.09184 \text{ kg/s} \].

- Calculate Volume Flow Rate for 30 minutes

Find the volume of water heated in 30 minutes. First, find the total time in seconds: \[ 30 \text{ minutes} = 30 \times 60 \text{ s} = 1800 \text{ s} \]. The total mass of water heated is: \[ \text{mass}_{total} = \text{m}_{total} \times 1800 \text{ s} = 0.09184 \text{ kg/s} \times 1800 \text{ s} = 165.312 \text{ kg} \]. Considering water’s density, where 1 kg = 1 liter, and 1 gallon = 3.78541 liters, calculate the final volume: \[ \text{volume}_{total} = \frac{165.312 \text{ liters}}{3.78541 \text{ liters/gallon}} \text{ gallons} \].

- Perform Final Calculation

Perform the division to find the volume in gallons: \[ \text{volume}_{total} = \frac{165.312}{3.78541} \text{ gallons} \] Simplifying, we get: \[ \text{volume}_{total} = 43.67 \text{ gallons} \].

Key concepts

Solar Energy Conversion

Solar energy conversion is the process of transforming sunlight into usable energy, typically in the form of heat or electricity. In our exercise, the solar collector panel is designed to capture sunlight and convert it into thermal energy, which is then used to heat water. The efficiency of this conversion is critical. In our example, 36% of the incoming solar energy is lost to the surroundings, meaning only 64% is effectively used to heat the water. This portion of energy is known as the useful energy. To calculate this, we multiply the total solar energy received by the efficiency (0.64), giving us the usable power in watts.

Thermodynamics

Thermodynamics is the branch of physics that deals with heat and temperature and their relation to energy and work. In this problem, we use the principles of thermodynamics to understand how the energy from the solar collector heats the water. The key formula used here is the heat transfer equation, given by: \[ Q = \text{mass flow rate} \times \text{specific heat} \times \text{temperature change} \]. This tells us that the amount of heat required to change the water’s temperature depends on its mass, the specific heat capacity of water, and the temperature difference it undergoes. Specific heat capacity (\(C_p\)) is the amount of heat required to raise the temperature of one kilogram of a substance by one degree Celsius, and for water, it is approximately 4.18 kJ/(kg·°C).

Energy Loss Calculation

Energy loss in solar energy systems is a crucial aspect to consider in their design and analysis. In the problem, 36% of the energy incoming from the sun is lost to the surroundings due to inefficiencies such as reflection, radiation, and convection. This means we are left with 64% of the initial energy, which is used to heat the water. To calculate the useful energy, we need to multiply the total energy received from the sun by the efficiency (1.5 kW * 0.64), resulting in 0.96 kW or 960 W of effective power. This step is vital as it helps us understand the actual usable energy in the system, which directly impacts how much water can be heated.

Mass Flow Rate Determination

Mass flow rate determination is critical in understanding how much water can be heated within a given time frame using the available energy. We start by setting up the power equation: \[ P_{useful} = \text{mass flow rate} \times \text{specific heat} \times \text{temperature change} \]. Rearranging this equation, we solve for mass flow rate (\text{m}): \[ \text{m} = \frac{P_{useful}}{C_p \times \triangle T} \]. Substituting the known values (960 W, 4.18 kJ/(kg·°C), and 20°C) converts the watts into kilojoules per second, giving us: \[ \text{m} = \frac{960}{4.18 \times 20} \text{ kg/s} \rightarrow \text{m} = 0.01148 \text{ kg/s} \], showing the mass flow rate. Further, multiplying this rate by the number of collectors (8) allows us to determine the total mass flow rate: \[ \text{m}_{total} = 0.01148 \text{ kg/s} \times 8 = 0.09184 \text{ kg/s} \]. Finally, converting this mass flow rate into volume over a set time period allows us to calculate how much heated water is produced, considering multiple collectors over 30 minutes, converting kilograms to liters and then to gallons.