Thermal relief refers to automatic release of fluids or gases from a system to a specified level. Pressure relief systems are intended to prevent pressures in process equipment from increasing to the point where a mechanical failure or rupture might take place, automatically releasing any materials within.
There are several most common causes of overpressure.
In previous post we learned what is conduction and its example on heat loss through a pipe wall with insulation. In this post, I want to share what is convection and its problem example.
Convection is heat transfer between a solid and an adjacent fluids develops as a result of fluid molecular movement. Cold molecules replace hot molecules as they leave the solid surface. A thin layer or film next to the solid surface is where most of the resistance to this type of heat transfer occurs. Even though the bulk fluid flow is extremely turbulent, this layer still exists.
Convective heat transport is governed by Newton’s law of cooling.
Q = h ∙A∙ ∆T
Where:
Q = heat transfer (Btu/hr)
h = heat transfer coefficient [Btu/(hr∙ ft2∙oF)]
A = area (ft2)
∆T = temperature difference (oF)
Convective heat transfer is divided into two types, natural or free convection and forced convection. Read More
In this post I want to share what is conduction and its problem example.
In contrast to general molecular motion or mixing, conduction describes the rate of heat transfer through materials as a function of vibrations and interactions between nearby molecules. Conduction always applies to solids and rarely to fluids.
There are several fundamental equations for steady heat conduction through some basic solid shapes, neglecting conditions of border:
In this post I want to share several basic types of fired equipment.
Fire equipment transfers heat produced by fuel combustion to the process stream. Natural gas is often selected as the fuel for gas processing equipment. The process stream includes a wide range of materials, such as heavier hydrocarbon, natural gas, water, amine solutions, glycol, and heat transfer oils.
Fire equipment can be categorized into two general types, direct fired heaters and firetube heaters.
In direct fired heaters, the combustion gases occupy the majority of the heater’s capacity and heat the process stream contained in pipes positioned in front of refractory walls (the radiant section) and in a bundle in the top portion (the convective section). Read More
In this post, I want to share a simple method to evaluate heat exchanger performance. One of the most helpful ways to evaluate the performance of heat exchanger is to determine its effectiveness by comparing the actual heat transfer rate to the maximum rate that is thermodynamically feasible. The simple formula is expressed below.
Where:
η = effectiveness
Q = actual heat transfer rate
Qmax = heat transfer rate which would be achieved if it were possible to bring the exit temperature of the stream with the lower heat capacity, to the inlet temperature of the other stream. Read More
In this post, I want to share how to estimate time required for heating and cooling.
The contents of a large batch reactor or storage tank frequently need to be heated or cooled. In this circumstance, the physical properties of the liquor may change throughout the process, as well as the overall transfer coefficient. When estimating the amount of time needed to heat or cool a batch of liquid, it is frequently possible to assume an average value for the transfer coefficient. Steam condensing, either in a coil or some type of hairpin tube heater, is a common method for heating the content of storage tank.
It is reasonable to assume that the overall transfer coefficient U is constant in the context of a storage tank filled with liquor having mass m and specific heat Cp and heated by steam condensing in a helical coil. The rate of heat transfer is given by: If T s is the temperature of the condensing steam, T1 and T2 are the initial and final temperatures of the liquor, A is the area of the heat transfer surface, and T is the temperature of the liquor at any time t, then:
The time t for heating from T1 to T2 can be determined using this equation. If the steam condenses in a reaction vessel’s jacket, the same analysis may be applied.
Heat losses during the heating or, for that matter, cooling operation are not considered in this analysis. The heat losses increase naturally as the temperature of the vessel’s contents rises, and at a certain point, the heat supplied to the vessel equals the heat losses, making further increases in the temperature of the vessel’s contents impossible.
By increasing the rate of heat transfer to the fluid, for example, by agitating the fluid, and by minimizing heat losses from the vessel by insulation, the heating-up time can be shortened.
The amount of agitation that can be achieved in a large vessel is constrained, thus one attractive alternative is to circulate the fluid through an external heat exchanger.
Let’s see example below on how to estimate time required for heating or cooling. Read More
In this post, I want to share a simple example on how to design thin-walled vessel under internal pressure. Please check my previous post about the method on how to estimate minimum thickness of vessel component (shell, flat end closures, domed head closures).
Estimate the thickness required for the component parts of the vessel shown in the diagram.
The vessel is to operate at a pressure of 16 bar (absolute) and design temperature of 300oC. The material of construction will be plain carbon steel. Welds will be fully radiographed. A corrosion allowance of 2 mm should be used. Read More
In this post, I want to share how to design thin-walled vessel under internal pressure. I will also share a simple example about the application in the next post.
For information, I do not have any experience of the calculation in my whole career until posted this post. The design of thin-walled vessel under internal pressure is usually job of mechanical engineer. But it is good for process engineers to understand it in general.
The design of thin-walled vessel under internal pressure, in general, will be divided into two parts: design of cylinder and spherical shells and design of heads and closures. Read More
In this post, I want to share how to do projects economic evaluation by using example. In this example, we will learn relationship between investment, sales, raw material cost, operating cost, to get the following parameters that usually used in projects economic evaluation:
It is proposed to build a plant to produce a new product. The estimated investment required is 12.5 million dollars and the timing of the investment will be:
In this example, I want to show you a simple example on how to do projects economic evaluation. As mentioned previously, there are several parameters that can be used to do projects economic evaluation, such as:
In this example we will use ROR and pay-back time.
Example on How to Do Projects Economic EvaluationA plant is producing 12,000 ton per year product. The overall yield is 70%, on a mass basis (kg of product per kg raw material). The raw material cost $12 per ton, and the product sells for $37 per ton. A process modification has been devised that will increase the yield to 75%. The additional investment required is $35,000, and the additional operating costs are negligible. Read More
