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The condenser of a steam power plant operates at a pressure of \(4.25 \mathrm{kPa}\). The condenser consists of 100 horizontal tubes arranged in a \(10 \times 10\) square array. The tubes are \(8 \mathrm{~m}\) long and have an outer diameter of \(3 \mathrm{~cm}\). If the tube surfaces are at \(20^{\circ} \mathrm{C}\), determine \((a)\) the rate of heat transfer from the steam to the cooling water and (b) the rate of condensation of steam in the condenser.

Short Answer

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Question: In a steam power plant condenser, steam condenses on the outer surface of 100 tubes, each having an outer diameter of 3 cm and a length of 8 m. The heat transfer coefficient of the steam film is given, and the temperature difference between the steam and the tube surface is 80°C. Determine (a) the rate of heat transfer between the steam and the cooling water and (b) the rate of condensation of steam in the condenser. Answer: a) To determine the rate of heat transfer (Q), first, find the surface area (A) of the condenser tubes using the given dimensions and the equation, \(A = N \times 2\pi r L\). Then, calculate Q using the heat transfer coefficient (h) and A with the equation: \(Q = h \times A \times \Delta T\). Provide the value of Q in Watts. b) To find the rate of steam condensation (m_dot), divide the heat transfer rate (Q) by the heat of vaporization of steam (h_fg) using the equation: \(m_\mathrm{dot} = \frac{Q}{h_\mathrm{fg}}\). Provide the value of m_dot in kg/s.

Step by step solution

01

Calculate the heat transfer coefficient (h)

To find the rate of heat transfer between the steam and the cooling water in the condenser, we need to determine the heat transfer coefficient (h). The value of h will be based on the condensing steam film, properties of the steam, and the surface geometry of the tubes.
02

Calculate the surface area (A) of the condenser tubes

We are given the dimensions of the tubes in the condenser. Each tube has an outer diameter of \(3\mathrm{~cm}\) and a length of \(8\mathrm{~m}\). To calculate the total surface area of all 100 tubes, we can use the equation for the surface area of a cylinder: \(A = N \times 2\pi r L\) Where \(N = 100\) is the total number of tubes, \(r = 1.5\mathrm{~cm}\) is the radius, and \(L = 8\mathrm{~m}\) is the length of the tubes. \(2\pi r L\) represents the surface area of a single tube. Calculate the value of A in square meters.
03

Calculate the rate of heat transfer (Q) using the heat transfer coefficient and surface area

Once we have the values of the heat transfer coefficient (h) and surface area (A), we can use the following equation to find the rate of heat transfer (Q) from the steam to the cooling water: \(Q = h \times A \times \Delta T\) Here, \(\Delta T = 100 - 20 = 80\,^{\circ}\mathrm{C}\) is the temperature difference between the steam and the tube surface. Calculate the rate of heat transfer (Q) in Watts.
04

Determine the rate of steam condensation (m_dot)

To calculate the rate of steam condensation (m_dot), we need to divide the heat transfer rate (Q) by the heat of vaporization of steam (h_fg). The heat of vaporization of steam can be found in steam tables at the given pressure. \(m_\mathrm{dot} = \frac{Q}{h_\mathrm{fg}}\) By plugging in the values for Q and h_fg, we can determine the rate of steam condensation (m_dot) in kg/s. The results of the calculations in Steps 3 and 4 will be the final answers for parts (a) and (b) of the exercise.

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Key Concepts

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

Heat Transfer Coefficient
The heat transfer coefficient (h) is a crucial factor in thermodynamics and heat transfer processes. It quantifies the heat transferred between a solid surface and a fluid per unit surface area per unit temperature difference. Higher values of h indicate more efficient heat transfer.

To calculate the heat transfer coefficient for a steam condenser, several factors are considered, including the properties of the steam (like viscosity and thermal conductivity), the flow characteristics, and the surface geometry of the condenser tubes. In practical scenarios, h is determined through experimentation or using correlations from the literature. Understanding the value of h can help optimize the design and operation of a steam power plant's condenser by maximizing heat exchange while minimizing energy consumption.

Once the value of h is determined, it can be used to calculate the rate of heat transfer from the steam to the cooling water. This provides insights into the performance of the condenser and helps maintain the desired condensation rate.
Surface Area Calculation
Calculating the surface area of condenser tubes accurately is essential for heat transfer calculations. In a steam power plant condenser, the tubes are typically modeled as cylinders because this shape defines the surface through which heat exchange occurs. For each cylinder, the surface area (A) can be calculated using the formula:

\(A = N \times 2\pi r L\)

where N is the number of tubes, r is the radius of a single tube, and L is the length of the tube. It is essential to convert all measurements to consistent units, such as meters, to ensure accuracy. The total surface area directly influences the rate of heat transfer (Q) because the more significant the area, the more contact the steam and cooling water have, allowing for better heat transfer.
Steam Power Plant Condenser
The condenser is a pivotal part of a steam power plant. Its primary function is to condense exhaust steam from the turbine, changing its phase from vapor back into a liquid, which is then returned to the boiler as feedwater. This condensed steam releases latent heat, which is removed by the cooling water system.

The efficiency of a condenser is vital for the plant’s overall thermal efficiency. It depends on factors such as surface area and material of the tubes, cooling water flow rate, and the heat transfer coefficient. By designing a condenser with sufficient surface area and selecting materials with high thermal conductivity, plant operators can enhance the system's efficiency and reliability.

Maintaining a condenser at an optimal performance level requires regular monitoring and the use of accurate calculation methods, like those for the heat transfer coefficient and surface area, to ensure that heat is effectively transferred from the steam to the cooling water.
Rate of Steam Condensation
The rate of steam condensation (often denoted as \(m_\mathrm{dot}\)) is a measurement of how much steam changes phase into water in a given time frame within the condenser of a steam power plant. It is calculated by dividing the rate of heat transfer (Q) by the heat of vaporization (\(h_\mathrm{fg}\)), the heat required to change a unit mass of the substance from liquid to gas without a temperature change.

\(m_\mathrm{dot} = \frac{Q}{h_\mathrm{fg}}\)

The calculation assumes that all the heat removed by the cooling water is used for the phase change of steam only, and that there is no superheat or subcooling involved. By accurately determining the rate of steam condensation, one can assess the efficiency of the condenser and make adjustments to the cooling system as necessary. This parameter is also critical for ensuring that steam flows within the plant are controlled and balanced for optimal operation.

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Most popular questions from this chapter

Steam condenses at \(50^{\circ} \mathrm{C}\) on a \(0.8-\mathrm{m}\)-high and \(2.4-\mathrm{m}-\) wide vertical plate that is maintained at \(30^{\circ} \mathrm{C}\). The condensation heat transfer coefficient is (a) \(3975 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(5150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(8060 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(11,300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(14,810 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (For water, use \(\rho_{l}=992.1 \mathrm{~kg} / \mathrm{m}^{3}, \mu_{l}=0.653 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\), \(\left.k_{l}=0.631 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p l}=4179 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}, h_{f g \oplus T_{\text {sat }}}=2383 \mathrm{~kJ} / \mathrm{kg}\right)\)

Discuss some methods of enhancing pool boiling heat transfer permanently.

A 1-mm-diameter nickel wire with electrical resistance of \(0.129 \mathrm{\Omega} / \mathrm{m}\) is submerged horizontally in water at atmospheric pressure. Determine the electrical current at which the wire would be in danger of burnout in nucleate boiling.

Water is boiled at atmospheric pressure by a horizontal polished copper heating element of diameter \(D=0.5\) in and emissivity \(\varepsilon=0.05\) immersed in water. If the surface temperature of the heating element is \(788^{\circ} \mathrm{F}\), determine the rate of heat transfer to the water per unit length of the heating element.

Water is to be boiled at sea level in a 30 -cm-diameter mechanically polished AISI 304 stainless steel pan placed on top of a \(3-\mathrm{kW}\) electric burner. If 60 percent of the heat generated by the burner is transferred to the water during boiling, determine the temperature of the inner surface of the bottom of the pan. Also, determine the temperature difference between the inner and outer surfaces of the bottom of the pan if it is 6-mm thick. Assume the boiling regime is nucleate boiling. Is this a good assumption?

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