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Design the condenser of a steam power plant that has a thermal efficiency of 40 percent and generates \(10 \mathrm{MW}\) of net electric power. Steam enters the condenser as saturated vapor at \(10 \mathrm{kPa}\), and it is to be condensed outside horizontal tubes through which cooling water from a nearby river flows. The temperature rise of the cooling water is limited to \(8^{\circ} \mathrm{C}\), and the velocity of the cooling water in the pipes is limited to \(6 \mathrm{~m} / \mathrm{s}\) to keep the pressure drop at an acceptable level. Specify the pipe diameter, total pipe length, and the arrangement of the pipes to minimize the condenser volume.

Short Answer

Expert verified
Based on the given conditions and requirements for a steam power plant condenser, determine the pipe diameter and total pipe length that would result in a minimum condenser volume.

Step by step solution

01

Calculate the heat rejected in the condenser

Given the thermal efficiency of the steam power plant and the net electric power generated, we can determine the total heat input into the power plant and the heat rejected in the condenser using the following formulas: Thermal efficiency = \(\frac{Power\, output}{Heat\, input}\) Heat input = \( \frac{Power\, output}{Thermal\, efficiency}\) Heat rejected in the condenser = Heat input - Power output Heat input = \(\frac{10 \times 10^6 W}{0.40}\) Heat rejected in the condenser = Heat input - \(10 \times 10^6 W\).
02

Calculate the mass flow rate of the cooling water

As the cooling water flows in the condenser, its temperature increases by \(8^\circ C\). Using the heat rejected in the condenser (from Step 1), we can determine the mass flow rate of cooling water as: Mass flow rate of cooling water = \(\frac{Heat\, rejected\, in\, condenser}{Specific\, heat\_capacity\_water \times Temperature\, rise\, of\, water}\) Manifold cooling water enters the system at a maximum velocity of \(6 m/s\). Based on this constraint, we will calculate the mass flow rate of the cooling water.
03

Calculate the required pipe area

Given the mass flow rate of the cooling water (from Step 2) and the maximum allowable velocity, we can determine the required pipe area as: Required pipe area = \(\frac{Mass\, flow\, rate\, of\, cooling\, water}{Density\, of\, water \times Maximum\_allowable\_velocity}\)
04

Determine the pipe diameter

With the required pipe area, we can now determine the diameter of the pipes to be used in the condenser: Pipe diameter = \(2 \times \sqrt{\frac{Required\, pipe\, area}{\pi}}\)
05

Calculate the required heat transfer surface area

We use the average heat flux based on the outer surface of the tubes to calculate the total required heat transfer surface area: Required heat transfer surface area = \(\frac{Heat\, rejected\, in\, condenser}{Average\, heat\, flux}\)
06

Calculate the total pipe length

Knowing the required heat transfer surface area and the pipe diameter, we can determine the total pipe length as: Total pipe length = \(\frac{Required\, heat\, transfer\, surface\, area}{\pi \times Pipe\, diameter}\)
07

Design the arrangement of pipes

The objective is to minimize the condenser volume. Using the pipe diameter and total pipe length determined above, we can design the arrangement of the pipes to minimize the volume. Pipes can be arranged in parallel or series configurations to optimize the condenser size efficient heat transfer. The designed condenser has a specified pipe diameter, total pipe length, and optimized arrangement of pipes to ensure efficient heat transfer and minimum condenser volume.

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Key Concepts

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

Steam Power Plant Efficiency
When discussing steam power plants, efficiency is a crucial metric that tells us how well the plant converts heat into electrical energy. The thermal efficiency of a steam power plant is calculated by dividing the power output—electricity generated—by the total heat input from the fuel. In simpler terms, if a power plant has a high thermal efficiency, it means it's good at using the heat from fuel to generate electricity. This is important not only for cost savings but also for minimizing environmental impact.

For our exercise, a plant with a 40% thermal efficiency suggests that 40% of the heat energy is converted to electrical energy, with the remaining 60% rejected as waste heat, mostly through the condenser. Improving efficiency is a key goal in power plant design, and this is where condensers play a role. By optimizing condenser design, we can slightly increase the overall efficiency of the plant.
Condenser Heat Rejection Calculation
The condenser in a steam power plant is where the remaining heat is expelled from the system, allowing the steam to condense back into water. Calculating the heat rejection in a condenser involves understanding how much energy the condensing steam releases. We begin with the total heat input, which is based on the electrical power output and the efficiency of the plant.

Using the provided formulas and the exercise's given values, the heat rejected in the condenser is the difference between the total heat input and the power output. The calculation of heat rejection provides the basis for designing the condenser system, as it impacts the size and the amount of cooling water required to absorb the rejected heat. This leads to the subsequent steps where the specifics of the condenser design are fleshed out.
Cooling Water Mass Flow Rate
Cooling water is integral to condenser operation, absorbing the heat rejected by the steam and thereby condensing it back into water. The rate at which this cooling water flows through the condenser is termed as mass flow rate. The mass flow rate is closely tied to the amount of heat that needs to be absorbed; thus, our condenser heat rejection calculation directly influences it.

To determine the mass flow rate, you need the specific heat capacity of water—the amount of heat per unit mass required to raise the water temperature by one degree Celsius—and the predetermined temperature rise, which is limited to 8°C in the exercise. By dividing the heat rejected in the condenser by the product of the specific heat capacity of water and the temperature rise, we obtain the mass flow rate. This rate is crucial for choosing pipe diameters and lengths because it must align with the physical constraints such as maximum allowable velocity and space available.
Heat Transfer Surface Area
The efficiency of heat transfer in a condenser is largely determined by its heat transfer surface area, which is the total area available for heat to move from the steam to the cooling water. To calculate this, we need to consider the average heat flux through the surface area of the condenser pipes. In the exercise, we calculate the required surface area by dividing the heat to be rejected by this heat flux.

The larger the surface area, the more efficient the heat transfer—thus enhancing the condenser's performance. However, increasing surface area typically involves longer pipes or more of them, which can increase the cost and size of the condenser. Consequently, there's a design trade-off, as larger surface areas can lead to larger and potentially more expensive condensers. The challenge lies in finding a balance that offers sufficient heat transfer within the space and budget constraints.

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Most popular questions from this chapter

When a saturated vapor condenses on a vertical, isothermal flat plate in a continuous film, the rate of heat transfer is proportional to (a) \(\left(T_{s}-T_{\text {sat }}\right)^{1 / 4}\) (b) \(\left(T_{s}-T_{\mathrm{sat}}\right)^{1 / 2}\) (c) \(\left(T_{s}-T_{\text {sat }}\right)^{3 / 4}\) (d) \(\left(T_{s}-T_{\text {sat }}\right)\) (e) \(\left(T_{s}-T_{\mathrm{sat}}\right)^{2 / 3}\)

Water is to be boiled at atmospheric pressure in a mechanically polished steel pan placed on top of a heating unit. The inner surface of the bottom of the pan is maintained at \(110^{\circ} \mathrm{C}\). If the diameter of the bottom of the pan is \(30 \mathrm{~cm}\), determine \((a)\) the rate of heat transfer to the water and \((b)\) the rate of evaporation.

What is condensation? How does it occur?

Saturated water vapor at \(40^{\circ} \mathrm{C}\) is to be condensed as it flows through a tube at a rate of \(0.2 \mathrm{~kg} / \mathrm{s}\). The condensate leaves the tube as a saturated liquid at \(40^{\circ} \mathrm{C}\). The rate of heat transfer from the tube 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) \(515 \mathrm{~kJ} / \mathrm{s}\)

Saturated water vapor at a pressure of \(12.4 \mathrm{kPa}\) is condensed over an array of 100 horizontal tubes, each with a diameter of \(8 \mathrm{~mm}\) and a length of \(1 \mathrm{~m}\). The tube surfaces are maintained with a uniform temperature of \(30^{\circ} \mathrm{C}\). Determine the condensation rates of the tubes for \((a)\) a rectangular array of 5 tubes high and 20 tubes wide and \((b)\) a square array of 10 tubes high and 10 tubes wide. Compare and discuss the results of (a) and (b).

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