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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

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

Step by step solution

01

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.
02

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)
03

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).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 cp, 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 cp 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 cp 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.

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Most popular questions from this chapter

Ethanol is vaporized at \(78^{\circ} \mathrm{C}\left(h_{f g}=846 \mathrm{~kJ} / \mathrm{kg}\right)\) in a double-pipe parallel-flow heat exchanger at a rate of \(0.03 \mathrm{~kg} / \mathrm{s}\) by hot oil \(\left(c_{p}=2200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters at \(120^{\circ} \mathrm{C}\). If the heat transfer surface area and the overall heat transfer coefficients are \(6.2 \mathrm{~m}^{2}\) and \(320 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively, determine the outlet temperature and the mass flow rate of oil using \((a)\) the LMTD method and \((b)\) the \(\varepsilon-\mathrm{NTU}\) method.

Saturated water vapor at \(40^{\circ} \mathrm{C}\) is to be condensed as it flows through the tubes of an air-cooled condenser at a rate of \(0.2 \mathrm{~kg} / \mathrm{s}\). The condensate leaves the tubes as a saturated liquid at \(40^{\circ} \mathrm{C}\). The rate of heat transfer to air is (a) \(34 \mathrm{~kJ} / \mathrm{s}\) (b) \(268 \mathrm{~kJ} / \mathrm{s}\) (c) \(453 \mathrm{~kJ} / \mathrm{s}\) (d) \(481 \mathrm{~kJ} / \mathrm{s}\) (e) \(515 \mathrm{~kJ} / \mathrm{s}\)

A counter-flow heat exchanger is stated to have an overall heat transfer coefficient of \(284 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) when operating at design and clean conditions. Hot fluid enters the tube side at \(93^{\circ} \mathrm{C}\) and exits at \(71^{\circ} \mathrm{C}\), while cold fluid enters the shell side at \(27^{\circ} \mathrm{C}\) and exits at \(38^{\circ} \mathrm{C}\). After a period of use, built-up scale in the heat exchanger gives a fouling factor of \(0.0004 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). If the surface area is \(93 \mathrm{~m}^{2}\), determine \((a)\) the rate of heat transfer in the heat exchanger and \((b)\) the mass flow rates of both hot and cold fluids. Assume both hot and cold fluids have a specific heat of \(4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\).

A counter-flow heat exchanger is used to cool oil \(\left(c_{p}=\right.\) \(2.20 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})\) from \(110^{\circ} \mathrm{C}\) to \(85^{\circ} \mathrm{C}\) at a rate of \(0.75 \mathrm{~kg} / \mathrm{s}\) by cold water \(\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the heat exchanger at \(20^{\circ} \mathrm{C}\) at a rate of \(0.6 \mathrm{~kg} / \mathrm{s}\). If the overall heat transfer coefficient is \(800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the heat transfer area of the heat exchanger is (a) \(0.745 \mathrm{~m}^{2}\) (b) \(0.760 \mathrm{~m}^{2}\) (c) \(0.775 \mathrm{~m}^{2}\) (d) \(0.790 \mathrm{~m}^{2}\) (e) \(0.805 \mathrm{~m}^{2}\)

Hot oil \(\left(c_{p}=2200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) is to be cooled by water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) in a 2 -shell-passes and 12 -tube-passes heat exchanger. The tubes are thin-walled and are made of copper with a diameter of \(1.8 \mathrm{~cm}\). The length of each tube pass in the heat exchanger is \(3 \mathrm{~m}\), and the overall heat transfer coefficient is \(340 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Water flows through the tubes at a total rate of \(0.1 \mathrm{~kg} / \mathrm{s}\), and the oil through the shell at a rate of \(0.2 \mathrm{~kg} / \mathrm{s}\). The water and the oil enter at temperatures \(18^{\circ} \mathrm{C}\) and \(160^{\circ} \mathrm{C}\), respectively. Determine the rate of heat transfer in the heat exchanger and the outlet temperatures of the water and the oil.

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