Answer:
11.8.4 Distillation Columns
Distillation columns present a hazard in that they contain large inventories of flammable boiling liquid, usually under pressure. There are a number of situations which may lead to loss of containment of this liquid.
The conditions of operation of the equipment associated with the distillation column, particularly the reboiler and bottoms pump, are severe, so that failure is more probable.
The reduction of hazard in distillation columns by the limitation of inventory has been discussed above. A distillation column has a large input of heat at the reboiler and a large output at the condenser. If cooling at the condenser is lost, the column may suffer overpressure. It is necessary to protect against this by higher pressure design, relief valves, or HIPS. On the other hand, loss of steam at the reboiler can cause underpressure in the column. On columns operating at or near atmospheric pressure, full vacuum design, vacuum breakers, or inert gas injection is needed for protection. Deposition of flammable materials on packing surfaces has led to many fires on opening of distillation column for maintenance.
Another hazard is overpressure due to heat radiation from fire. Again pressure relief devices are required to provide protection.
The protection of distillation columns is one of the topics treated in detail in codes for pressure relief such as APIRP 521. Likewise, it is one of the principal applications of trip systems.
Another quite different hazard in a distillation column is the ingress of water. The rapid expansion of the water as it flashes to steam can create very damaging overpressures.
Once for the water and once for the copper. Set up a table that accounts for each of the variables you know, and then identify the ones you need to obtain. Give me a moment or two and I will work this out for you.
Okay, so like I said before, you will need to use the equation twice. Now, keep in mind that when the copper is placed in the water (the hot into the cold), there is a transfer of heat. This heat transfer is measured in Joules (J). So, the energy that the water gains is the same energy that the copper loses. This means that for your two equations, they can be set equal to each other, but the copper equation will have a negative sign in front to account for the energy it's losing to the water.
When set equal to each other, the equations should resemble something like this:
(cmΔt)H20 = -(cmΔt)Cu
(Cu is copper).
Remember, Δt is the final temperature minus the initial temperature (T2-T1). We are trying to find T2. Since we are submerging the copper into the water, we can assume that the final temperature at equilibrium is the same for both the copper and the water. At a thermodynamic equilibrium, there is no heat transfer because both materials are at the same temperature.
T2Cu = T2H20
Now, the algebra for this part of the problem is a bit confusing, so make sure you keep track of your variables. If done right, the algebra should work out so you have this:
T2 = ((cmT1)Cu + (cmT1)H20) / ((cm)H20 + (cm)Cu)
Insert the values for the variables. Once you plug and chug, your final answer should be
26.8 degrees Celsius.
Answer:
Explanation:
The <em>purchase price</em> is what Janice invested for every share.
Since the stock was priced at $31.82 per share and she received a $1.11 dividend per share, her investment was:
- $31.82 - $1.11 = $30.71 per share ← answer
This price is the cost for Janice, over which she shall calculate their returns (gains or losses) on the future, when she sells the shares, for instance.
The total investment of Janice was the number of shares multipled by the purchase price:
- 40 shares × ($31.82 - $1.11)/ share
- 40 shares × ($30.71) / share = $1,228.40 (total investment)
