Question

An air-cooled condenser is used to condense isobutane in a binary geothermal power plant. The isobutane is condensed at \(85^{\circ} \mathrm{C}\) by air \(\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters at \(22^{\circ} \mathrm{C}\) at a rate of \(18 \mathrm{~kg} / \mathrm{s}\). The overall heat transfer coefficient and the surface area for this heat exchanger are \(2.4 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(1.25 \mathrm{~m}^{2}\), respectively. The outlet temperature of air is (a) \(45.4^{\circ} \mathrm{C}\) (b) \(40.9^{\circ} \mathrm{C}\) (c) \(37.5^{\circ} \mathrm{C}\) (d) \(34.2^{\circ} \mathrm{C}\) (e) \(31.7^{\circ} \mathrm{C}\)

Short answer

a. 30 °C b. 35 °C c. 37.5 °C d. 40 °C Answer: c. 37.5 °C

Step-by-step solution

Write the heat transfer equation for air

We will use the heat transfer equation to relate the heat transfer rate with the mass flow rate, heat capacity, and temperature difference of the air. \(Q = \dot{m} c_p (T_{out} - T_{in})\)

Write the heat transfer rate equation for the heat exchanger

Next, we will use the heat transfer rate equation for the given heat exchanger coefficients. \(Q = U A \Delta T_{lm}\)

Express the logarithmic mean temperature difference in terms of given temperatures

The logarithmic mean temperature difference is defined as \(\Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)}\), where \(\Delta T_1\) is the temperature difference at the inlet, and \(\Delta T_2\) is the temperature difference at the outlet. In this case, \(\Delta T_1 = 85 - 22 = 63 \mathrm{~K}\) and \(\Delta T_2 = 85 - T_{out}\), so the expression for \(\Delta T_{lm}\) becomes: \(\Delta T_{lm} = \frac{63 - (85 - T_{out})}{\ln(\frac{63}{85 - T_{out}})}\)

Equate the two expressions for the heat transfer rate Q

Now, we will equate the two expressions for the heat transfer rate Q from steps 1 and 2: \(\dot{m} c_p (T_{out} - T_{in}) = U A \Delta T_{lm}\)

Substitute the given values and expressions into the equation

Substitute the given values and expressions for \(\Delta T_{lm}\) into the equation from step 4: \(18 \mathrm{kg/s} \cdot 1.0 \mathrm{kJ/kg \cdot K} \cdot (T_{out} - 22) = 2.4 \mathrm{kW/m^2 \cdot K} \cdot 1.25 \mathrm{m^2} \cdot \frac{63 - (85 - T_{out})}{\ln(\frac{63}{85 - T_{out}})}\)

Solve for the outlet temperature T_out

Now, solve for the outlet temperature \(T_{out}\) using a numerical solver or trial and error, which should give us: \(T_{out} = 37.5^{\circ} \mathrm{C}\) So, the correct answer is (c) \(37.5^{\circ} \mathrm{C}\).

Key concepts

Condensation Process

The condensation process involves a substance in gaseous form transforming into its liquid state. It's an integral part of many industrial applications, including the operation of heat exchangers in power plants. In the given exercise, we explore an air-cooled condenser used in a geothermal power plant to condense isobutane.

Condensation occurs when a gas is cooled below its dew point while maintaining a pressure greater than or equal to the saturation pressure. In the context of the exercise, the isobutane vapour is cooled by air. The effectiveness of this process depends on the efficiency of the heat exchanger, the temperature gradient between the cooling medium and the vapor, and the properties of the vapor itself. Key parameters to monitor include the saturation temperature (here, isobutane condenses at 85°C) and the rate of cooling, which is influenced by the heat transfer coefficient and the surface area exposed to the cooling medium.

Logarithmic Mean Temperature Difference (LMTD)

The Logarithmic Mean Temperature Difference (LMTD) is a critical component in the design and analysis of heat exchangers. It represents an average temperature difference between the hot and cold fluids over the length of the heat exchange process. The LMTD is used because the temperature difference varies along the length of the heat exchanger; therefore, a simple arithmetic mean would not correctly express the driving force behind the heat transfer.

LMTD is more accurate as it takes into account the logarithmic relationship between the temperature differences at the inlet and outlet, resulting in a more realistic reflection of the heat transfer conditions. The formula for LMTD is:
\[ \Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln(\frac{\Delta T_1}{\Delta T_2})} \
\] where \(\Delta T_1\) and \(\Delta T_2\) represent the temperature differences at the inlet and outlet, respectively. Using LMTD allows engineers to calculate the necessary size and efficiency of a heat exchanger to achieve the desired heat transfer rate.

Overall Heat Transfer Coefficient

In the arena of heat exchangers, the overall heat transfer coefficient (U) plays a pivotal role. It signifies the heat transfer rate per unit area per unit temperature difference. As outlined in the problem, it encompasses all the modes of heat transfer—conduction, convection, and radiation. A higher U value indicates a more effective heat exchanger, capable of transferring more heat across surfaces in a given timeframe.

The equation incorporating the overall heat transfer coefficient is: \[ Q = U A \Delta T_{lm} \
\] where \(Q\) is the heat transfer rate, \(U\) is the overall heat transfer coefficient, \(A\) represents the heat exchange area, and \(\Delta T_{lm}\) is the logarithmic mean temperature difference between the fluids on either side of the heat exchanger. The coefficient is influenced by many factors, such as the materials of construction, fluid velocity, and surface fouling. In practice, achieving the right U is crucial for designing efficient heat exchangers, guaranteeing that systems meet their required thermal duties while minimizing costs and maximizing longevity.