A manufacturing facility requires saturated steam at \(120^{\circ} \mathrm{C}\) at a rate of \(1.2 \mathrm{~kg} / \mathrm{min}\). Design an electric steam boiler for this purpose under these constraints: \- The boiler will be in cylindrical shape with a heightto-diameter ratio of \(1.5\). The boiler can be horizontal or vertical. \- The boiler will operate in the nucleate boiling regime, and the design heat flux will not exceed 60 percent of the critical heat flux to provide an adequate safety margin. \- A commercially available plug-in type electrical heating element made of mechanically polished stainless steel will be used. The diameter of the heater cannot be between \(0.5 \mathrm{~cm}\) and \(3 \mathrm{~cm}\). \- Half of the volume of the boiler should be occupied by steam, and the boiler should be large enough to hold enough water for \(2 \mathrm{~h}\) supply of steam. Also, the boiler will be well insulated. You are to specify the following: (a) The height and inner diameter of the tank, \((b)\) the length, diameter, power rating, and surface temperature of the electric heating element, \((c)\) the maximum rate of steam production during short periods of overload conditions, and how it can be accomplished.

Short Answer

Expert verified
In this exercise, a design for an electric steam boiler was created for a manufacturing facility. The design constraints included the desired steam production rate, the heat flux, the electrical heating element, and the boiler's height-to-diameter ratio. The main steps in developing the design were: 1. Calculation of the boiler's volume required to hold enough water for a 2-hour supply of steam at the desired production rate. This was found to be 0.288 m³. 2. Derivation of the height and inner diameter of the boiler using the given height-to-diameter ratio. The boiler's height was found to be approximately 0.671 m, and the inner diameter was approximately 0.447 m. 3. Determination of the required heat flux and the corresponding power rating of the electric heating element. The required heat flux was calculated to be 5142.2 kJ/min, which corresponds to a power rating of 85.7 kW. 4. Calculation of the length, diameter, and surface temperature of the electric heating element. The heating element was found to have a length of approximately 9.09 m, a diameter of 0.035 m, and a surface temperature of about 156.8°C. 5. Determination of the maximum rate of steam production during short periods of overload and how it can be accomplished. The maximum rate of steam production was found to be 2 kg/min, achieved by increasing the power supplied to the electric heating element to operate at the critical heat flux. This design ensures that the electric steam boiler can produce steam at the required rate while adhering to the given constraints, making it suitable for use in the manufacturing facility.

Step by step solution

01

Determine the volume of the boiler required

Given that the boiler should be large enough to hold enough water for a 2-hour supply of steam, we first determine the mass of steam required for 2 hours. Using the steam production rate (1.2 kg/min): Mass of steam for 2 hours = 1.2 kg/min × 120 min = 144 kg Half of the boiler's volume will be occupied by steam, so the other half will be occupied by water. Therefore, corresponding to the mass of 144 kg of steam, we have 144 kg of water. Since the density of water is approximately \(1000~\mathrm{kg/m^3}\), the volume occupied by the water is: Volume of water = 144 kg ÷ 1000 kg/m³ = 0.144 m³
02

Calculate the height and inner diameter of the boiler

Since half the volume of the boiler is occupied by steam and the other half is occupied by water, the total volume would be twice the volume of water: Total volume of boiler = 0.144 m³ × 2 = 0.288 m³ The boiler is cylindrical with a height-to-diameter ratio of 1.5. Let the diameter = D and height = H. H = 1.5D Volume of the boiler = π(D/2)²H = 0.288 m³ Substituting the equation for H, we have: 0.288 m³ = π(D/2)²(1.5D) Solving for D, we get D ≈ 0.447 m. Therefore, the inner diameter of the tank is approximately 0.447 m. Consequently, the height of the tank is 1.5D ≈ 0.671 m.
03

Determine the required heat flux and power rating

The boiler operates in the nucleate boiling regime. The design heat flux will not exceed 60% of the critical heat flux. To calculate the required heat flux, we first calculate the heat required to produce 1.2 kg/min of saturated steam at 120°C. 1. Heat required to raise the temperature of the water to its boiling point (100°C): Q1 = mcΔT = 1.2 kg/min × 4.186 kJ/kg.K × (100 - 25) = 376.92 kJ/min 2. Heat required to convert the water at its boiling point to steam: Q2 = mL = 1.2 kg/min × 2257 kJ/kg = 2708.4 kJ/min Therefore, the total heat required to produce 1.2 kg/min of saturated steam is Q1 + Q2 = 3085.32 kJ/min. To provide an adequate safety margin of 60%, the required heat flux can be determined: Required heat flux = (3085.32 kJ/min) ÷ 0.60 = 5142,2 kJ/min Consequently, the power rating of the heating element must be 5142.2 kJ/min = 85703 W or 85.7 kW.
04

Calculate the length, diameter, and surface temperature of the electric heating element

Let's assume that the heater diameter is 3.5 cm (the diameter is larger than the range of 0.5 cm to 3 cm). The heater area can be determined by the given heat flux and power rating: Heater Area = Power Rating ÷ Heat Flux = (85703 W) ÷ (5142.2 kJ/min × (1/60) min/s) ≈ 1 m² Now, we can calculate the length of the heating element as it is a cylinder: Length = Heater Area ÷ (π × Heater Diameter) = (1 m²) ÷ (π × 0.035 m) ≈ 9.09 m To calculate the surface temperature of the electric heating element, we can use the equation for the heat transfer coefficient (h) of nucleate boiling: h = (Q / A) / (Ts - Tf) Where Q is the power rating (85.7 kW), A is the heater area (1m²), Ts is the surface temperature of the heating element, and Tf is the fluid temperature (120°C). We know that h = C × (Ts - Tf)^n, where C and n are constants in the nucleate boiling regime. As per the literature, C = 2100 W/(m².K^n) and n = 1.3. Therefore, we have: 2100 W/(m².K^n) = (85703 W) ÷ [(Ts - 120) × 1 m²] Solving for Ts, we get Ts ≈ 156.8°C. So, the surface temperature of the electric heating element is approximately 156.8°C.
05

Determine the maximum rate of steam production during short periods of overload and how it can be accomplished

The maximum rate of steam production can be achieved by operating at the critical heat flux. We are given that the current design heat flux is at 60% of the critical heat flux: Critical heat flux = Design heat flux ÷ 0.60 = 5142.2 kJ/min ÷ 0.60 = 8570.3 kJ/min To achieve the maximum rate of steam production, the power supplied to the heating element must be increased accordingly. The new power rating P_max can be calculated as: P_max = Critical heat flux × (1/60) min/s = 8570.3 kJ/min × (1/60) min/s = 142838.3 W or 142.8 kW The maximum rate of steam production can be found by dividing the power supplied by the specific heat of steam: Max rate of steam production = P_max ÷ L_v = 142.8 kW ÷ 2257 kJ/kg = 2 kg/min Therefore, the maximum rate of steam production is 2 kg/min, which could be accomplished by increasing the power supplied to the electric heating element to operate at the critical heat flux.

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Most popular questions from this chapter

Saturated water vapor is condensing on a \(0.5 \mathrm{~m}^{2}\) vertical flat plate in a continuous film with an average heat transfer coefficient of \(7 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\). The temperature of the water is \(80^{\circ} \mathrm{C}\left(h_{f g}=2309 \mathrm{~kJ} / \mathrm{kg}\right)\) and the temperature of the plate is \(60^{\circ} \mathrm{C}\). The rate at which condensate is being formed is (a) \(0.0303 \mathrm{~kg} / \mathrm{s}\) (b) \(0.07 \mathrm{~kg} / \mathrm{s}\) (c) \(0.15 \mathrm{~kg} / \mathrm{s}\) (d) \(0.24 \mathrm{~kg} / \mathrm{s}\) (e) \(0.28 \mathrm{~kg} / \mathrm{s}\)

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.

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) \(515 \mathrm{~kJ} / \mathrm{s}\)

What is condensation? How does it occur?

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