Problem 28 - University Physics with Modern Physics

Question

Conventional hot-water heaters consist of a tank of water maintained at a fixed temperature. The hot water is to be used when needed. The drawbacks are that energy is wasted because the tank loses heat when it is not in use and that you can run out of hot water if you use too much. Some utility companies are encouraging the use of on-demand water heaters (also known as flash heaters), which consist of heating units to heat the water as you use it. No water tank is involved, so no heat is wasted. A typical household shower flow rate is 2.5 gal/min (9.46 L/min) with the tap water being heated from 50\(^\circ\)F (10\(^\circ\)C) to 120\(^\circ\)F (49\(^\circ\)C) by the on-demand heater. What rate of heat input (either electrical or from gas) is required to operate such a unit, assuming that all the heat goes into the water?

Step-by-Step Solution

Answer

The rate of heat input required is 25.7 kW.

Step 1: Calculate Temperature Change

To find the rate of heat input, we first need to calculate the temperature change. The water is heated from \(50^\circ F\) to \(120^\circ F\). Therefore, the temperature change \(\Delta T\) is:\[\Delta T = 120 - 50 = 70^\circ F\]We'll need this in Celsius for further calculations:\[\Delta T_{\text{C}} = \frac{5}{9} \times 70 \approx 38.89^\circ C\].

Step 2: Convert Flow Rate to SI Units

The flow rate is given as 9.46 L/min. We need to convert this to cubic meters per second for standard units:\[1 L = 0.001 m^3, \quad 1 \, \text{min} = 60 \, \text{sec}\]\[\text{Flow rate} = 9.46 \, L/min \times 0.001 \, m^3/L \, \div \, 60 \, sec/min \approx 1.577 \times 10^{-4} \, m^3/s\]

Step 3: Use the Specific Heat Formula

The specific heat formula \(q = mc\Delta T\) can be used, where \(q\) is the heat added, \(m\) is the mass flow rate of water, \(c\) is the specific heat capacity, and \(\Delta T\) is the temperature change. For water, \(c = 4.186 \, kJ/kg\cdot{}^{\circ}C\).First, calculate the mass flow rate using the density of water (\(1000 \, kg/m^3\)):\[m = \text{density} \times \text{volumetric flow rate} = 1000 \, kg/m^3 \times 1.577 \times 10^{-4} \, m^3/s \approx 0.1577 \thinspace kg/s\].

Step 4: Compute Heat Input Rate

Now, substitute the \(m\), \(c\), and \(\Delta T\) into the heat input formula:\[q = 0.1577 \thinspace kg/s \times 4.186 \, kJ/kg\cdot{}^{\circ}C \times 38.89^\circ C \approx 25.7 \, kJ/s\]Therefore, the rate of heat input required is approximately 25.7 kW.

Key Concepts

On-demand water heaters

On-demand water heaters, also known as tankless or flash heaters, present an energy-efficient alternative to traditional storage water heaters. Unlike conventional models that keep a large tank of water continuously hot, wasting energy when not in use, on-demand heaters only use energy when hot water is needed.

Here's how they work:

• Water flows through the unit, triggering the heating element.
• Heat is applied directly to the water through either a gas burner or an electric element.
• Hot water is delivered almost instantaneously, avoiding standby losses from stored hot water.

One of the key advantages is the endless supply of hot water—perfect for large families. However, the initial purchase and installation cost can be higher, and they might require upgrades to electrical capacity or gas lines.

Specific heat capacity

Specific heat capacity is a fundamental property of materials and substances that tells us how much heat is required to change the temperature of a given amount of substance by one degree Celsius. For water, this value is particularly high, making it excellent for heating applications.

Water has a specific heat capacity of approximately 4.186 kJ/kg°C. This means:

• 1 kilogram of water requires 4.186 kilojoules of energy to increase its temperature by 1°C.
• High specific heat capacity allows water to absorb significant amounts of heat without a considerable temperature rise.
• This attribute is exploited in both heating and cooling applications across various sectors.

Thus, when calculating the energy required to heat water in systems like on-demand water heaters, this constant is critical.

Temperature change

When heating water using an on-demand water heater, one of the primary calculations involves figuring out the necessary temperature change the water must undergo. We usually begin with water at its initial temperature—often the temperature of groundwater, which in this exercise is 50°F (10°C)—and aim for a desirable endpoint temperature, here being 120°F (49°C).

To determine how much heat is needed, we compute the temperature change:

• In the Fahrenheit system: \[\Delta T = T_{final} - T_{initial} = 120 - 50 = 70^{\circ}F\]
• Converted to Celsius for scientific calculations, using the formula:\[\Delta T_{C} = \frac{5}{9} \times \Delta T_{F} = \frac{5}{9} \times 70 \approx 38.89^{\circ}C\]

Understanding this change allows us to compute the heat energy input needed, which is vital for determining the operational effectiveness of heating systems.

Flow rate conversion

In engineering, especially in calculating heating requirements, it's crucial to convert flow rates into standard units for accuracy. Such conversions enable uniformity in calculations, allowing engineers to apply established scientific formulae effectively.

For water heaters:

• Flow rate is typically given in gallons per minute (gal/min) or liters per minute (L/min).
• To fit into scientific equations, flow rates are usually converted into cubic meters per second (m³/s).
  • 1 Liter = 0.001 cubic meters (m³)
  • 1 minute = 60 seconds
• Example conversion (from the exercise):\[9.46 \, L/min \times 0.001 \, m^3/L \div 60 \, sec/min \approx 1.577 \times 10^{-4} \, m^3/s\]

This conversion is essential for deploying the specific heat formula in its standard SI unit form, ensuring calculations yield meaningful and applicable results.