Question

Cold water \(\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters a counterflow heat exchanger at \(18^{\circ} \mathrm{C}\) at a rate of \(0.7 \mathrm{~kg} / \mathrm{s}\) where it is heated by hot air \(\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the heat exchanger at \(50^{\circ} \mathrm{C}\) at a rate of \(1.6 \mathrm{~kg} / \mathrm{s}\) and leaves at \(25^{\circ} \mathrm{C}\). The maximum possible outlet temperature of the cold water is (a) \(25.0^{\circ} \mathrm{C}\) (b) \(32.0^{\circ} \mathrm{C}\) (c) \(35.5^{\circ} \mathrm{C}\) (d) \(39.7^{\circ} \mathrm{C}\) (e) \(50.0^{\circ} \mathrm{C}\)

Short answer

Answer: The maximum possible outlet temperature of the cold water is 32.0°C.

Step-by-step solution

Calculate heat gained by the cold water and heat lost by the hot air

In a heat exchanger, the heat gained by the cold medium is equal to the heat lost by the hot medium. We can calculate the heat gained by the cold water (Q_cw) and the heat lost by the hot air (Q_ha) using the following formula: Q = mass_flow_rate * c_p * ΔT Where Q is the heat transfer, mass_flow_rate is the mass flow rate of the medium, c_p is the heat capacity of the medium, and ΔT is the change in temperature of the medium. First, let's calculate the heat gained by the cold water (Q_cw): Q_cw = mass_flow_rate_cw * c_p_cw * ΔT_cw Where mass_flow_rate_cw = 0.7 kg/s, c_p_cw = 4.18 kJ/kg·K, and ΔT_cw = T_out_cw - 18°C. Now, let's calculate the heat lost by the hot air (Q_ha): Q_ha = mass_flow_rate_ha * c_p_ha * ΔT_ha Where mass_flow_rate_ha = 1.6 kg/s, c_p_ha = 1.0 kJ/kg·K, and ΔT_ha = 50°C - 25°C.

Apply energy conservation principle

As the heat gained by the cold water is equal to the heat lost by the hot air, we can equate the above equations: mass_flow_rate_cw * c_p_cw * ΔT_cw = mass_flow_rate_ha * c_p_ha * ΔT_ha From the given data, we can plug in the values: 0.7 * 4.18 * (T_out_cw - 18) = 1.6 * 1.0 * (50 - 25)

Solve for the maximum possible outlet temperature of the cold water (T_out_cw)

Now, we will solve this equation to find the maximum possible outlet temperature of the cold water (T_out_cw): 2.924 * (T_out_cw - 18) = 40 T_out_cw - 18 = 13.69189 T_out_cw = 31.69189 ≈ 32.0°C Therefore, the maximum possible outlet temperature of the cold water is 32.0°C, which corresponds to option (b).

Key concepts

Counterflow Heat Exchanger

In a counterflow heat exchanger, two fluids flow in opposite directions, allowing for efficient heat transfer. This design improves the efficiency of the heat exchange process because the temperature gradient between the fluids remains more consistent throughout the exchanger. For example, in our exercise, cold water and hot air are the two fluids exchanging heat. The cold water is heated up by the hot air as both fluids pass each other in opposite directions. This setup can achieve higher temperatures of the heated fluid close to the incoming temperature of the hot fluid, which is an inherent benefit over other designs, like parallel flow heat exchangers.

Another benefit of a counterflow heat exchanger is that it can approach the temperature of the hot fluid more closely than other types. In the scenario provided, achieving a significant temperature rise in the cold water would be primarily due to the effectiveness of the counterflow design.

Heat Capacity

Heat capacity, often denoted as c_p, is the amount of thermal energy required to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin, as the scale increment is the same). It is a fundamental property that depends on the material's physical composition. In the context of our exercise, the cold water has a c_p of 4.18 kJ/kg·K, meaning each kilogram of water requires 4.18 kJ of heat to raise its temperature by one Kelvin. Conversely, the hot air has a c_p of 1.0 kJ/kg·K. These values play a pivotal role in determining how much thermal energy is transferred between substances in the heat exchanger.

Mass Flow Rate

Mass flow rate is a measure of the amount of mass passing through a cross-section per unit of time, usually denoted in kilograms per second (kg/s). It is a crucial factor in heat exchanger calculations, as it determines the potential amount of heat that can be transferred between the hot and cold fluids. In our exercise, the mass flow rate of cold water is 0.7 kg/s, while the hot air has a flow rate of 1.6 kg/s. When we multiply the mass flow rate by the fluid's heat capacity and the temperature change (∆T), it gives us the rate of heat transfer for that fluid, which allows us to apply the principle of thermal energy conservation.

Thermal Energy Conservation

Thermal energy conservation is a principle stating that, in the absence of external work being done or mass transfer, the total amount of heat energy within a closed system remains constant. In the context of a heat exchanger, this means that the heat lost by one fluid must be equal to the heat gained by the other fluid. Our exercise required us to apply this principle by equating the heat gained by the cold water to the heat lost by the hot air.

The formula involves the mass flow rate, heat capacity, and temperature change (∆T) for both fluids. Since energy is being conserved and no heat is lost to the environment, the equation from the cold water's perspective must balance with the one from the hot air's perspective. This understanding allows us to solve for unknown variables such as the maximum possible outlet temperature of the cold water.