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A steam trap is a device to purge steam condensate from a system without venting uncondensed steam. In one of the crudest trap types, the condensate collects and raises a float attached to a drain plug. When the float reaches a certain level, it "pulls the plug," opening the drain valve and allowing the liquid to discharge. The float then drops down to its original position and the valve closes, preventing uncondensed steam from escaping. (a) Suppose saturated steam at 25 bar is used to heat \(100 \mathrm{kg} / \mathrm{min}\) of an oil from \(135^{\circ} \mathrm{C}\) to \(185^{\circ} \mathrm{C}\). Heat must be transferred to the oil at a rate of \(1.00 \times 10^{4} \mathrm{kJ} / \mathrm{min}\) to accomplish this task. The steam condenses on the exterior of a bundle of tubes through which the oil is flowing. Condensate collects in the bottom of the exchanger and exits through a steam trap set to discharge when 1200 g of liquid is collected. How often does the trap discharge? (b) Especially when periodic maintenance checks are not performed, steam traps often fail to close completely and so leak steam continuously. Suppose a process plant contains 1000 leaking traps (not an unrealistic supposition for some plants) operating at the condition of Part (a), and that on the average 10\% additional steam must be fed to the condensers to compensate for the uncondensed steam venting through the leaks. Further suppose that the cost of generating the additional steam is \$7.50 per million Btu, where the denominator refers to the enthalpy of the leaking steam relative to liquid water at \(20^{\circ} \mathrm{C}\). Estimate the yearly cost of the leaks based on \(24 \mathrm{h} /\) day, 360 day/yr operation.

Step by step solution

01

Heat Transferred

Calculate the heat transferred by using the given rate \[1.00 \times 10^{4} \mathrm{kJ} / \mathrm{min}\]. The total heat transferred per minute will be \(Q = 1.00 \times 10^{4} \mathrm{kJ}\)

02

Heat Capacity

Now calculate the heat capacity per kg or the heat content represented by the mass of the condensate. From a standard steam table, at 25 bar, the heat of condensation or heat content of steam is 1940 kJ/kg. Thus, for 1200 g or 1.2 kg of steam, the heat content becomes \(GC = 1.2 \times 1940\) = 2328 kJ.

03

Frequency of Discharge

Use the heat transferred and heat capacity to find the frequency of discharge. To heat the oil at the given rate, the number of times condensate collected and discharged per minute can be found as \(n = Q/GC = 1.00 \times 10^{4}/2328 = 4.29\) discharges per min.

04

Additional Steam

Calculate the additional steam needed due to leaks. As given, 10% more steam is required to compensate for the steam lost through leaks. So the additional steam required per trap per minute is \(AS = 0.10 \times 1.2 = 0.12 kg/min\}

05

Total Leaked Steam

Calculate the total steam leaked per year. The total steam leaked per year for 1000 traps is \(TS = 1000 \times 0.12 \times 60 \times 24 \times 360 = 311040000 kg\).

06

Yearly Cost

Determine the yearly cost of the leaks. The extra cost due to the leaked steam can be calculated by \(EC = TS \times 1940 \times 0.0000075 = $4397800\)

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Key Concepts

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

Heat Transfer

Heat transfer is a key concept when discussing steam traps, as these devices play a crucial role in removing condensate without losing steam in a heating process. In our exercise, steam is used to heat oil from 135°C to 185°C at a heat transfer rate of \(1.00 \times 10^{4} \text{kJ/min}\). This means that the steam provides the necessary energy through condensation to elevate the oil’s temperature efficiently.

• Heat transfer occurs when there is a difference in temperature, allowing energy to move from the steam (hotter) to the oil (cooler) via conduction through the tube walls.
• The calculation of heat transferred, \(Q\), focuses on the amount of energy needed to change the temperature of a given mass of oil.
• This process ensures that the system operates effectively, only allowing condensate to leave via the steam trap, thus maintaining steam within the system effectively.

Understanding this flow of energy is crucial for efficient system design and operation, especially in industrial processes where energy efficiency leads to cost savings.

Condensation Process

The condensation process is where steam transforms back into the liquid phase as it transfers its latent heat to the oil. In a sealed heat exchanger, the steam is exposed to the oil-cooled tube bundle, causing the steam to condense.

• This phase change releases a significant amount of heat—known as the heat of condensation—which is used to heat the oil.
• As steam condenses, it loses energy and collects as liquid condensate at the bottom of the exchanger.
• The heat content per kilogram of the condensate is crucial in calculating system efficiency and discharge frequency of the steam trap. For steam at 25 bar, this value is 1940 kJ/kg.

Efficient condensation is vital to ensure the thermal energy is adequately transferred, preventing energy loss and making the process more economical.

Discharge Frequency Calculation

Calculating the discharge frequency of the steam trap is fundamental to ensuring that only condensate is removed, avoiding the loss of steam. This frequency dictates how often the float 'pulls the plug' to release the collected condensate.

• The formula \(n = \frac{Q}{GC}\) helps determine this frequency, where \(Q\) is the heat transfer rate (\(1.00 \times 10^{4} \text{kJ/min}\)) and \(GC\) is the heat content of the condensate collected per cycle (2328 kJ for 1.2 kg).
• So, the trap must discharge approximately 4.29 times per minute to balance the energy transfer and remove the excess condensate efficiently.
• These calculations help in designing the system for optimal performance by ensuring that steam is neither wasted nor trapped, impacting overall plant efficiency.

This aspect is critical in the long-term maintenance and operational planning of systems employing steam traps.