Questions assignment about Viral Pharma

2 questions
Please answer the first and second questionsPlease answer the first and second questionsPlease answer the first and second questionsPlease answer the first and second questions

IS 215 Autumn 2020 1 | P a g e
Assignment #6
Chapter 3
Questions:
VIRAL PHARMA
At Viral Pharma, your boss says that he can estimate the total project time based on his personal
experience. You are trying to convince him that he should use project management techniques to
handle a complex project. To prove your point, you decide to use a simple example of a R&D process
with eight steps. You create a hypothetical work breakdown structure, as follows:
• Prepare the laboratory space (3 days), and then set up the laboratory station equipment
framework (3 days).
• Equip the laboratory stations (5 days), and then connect the utilities (e.g. power, network,
water) (3 days).
• When the utilities are connected, start two tasks at once: finish the laboratory interior work (5
days) and set up an appointment for the final FDA laboratory inspection (15 days).
• When the laboratory interior work is done, start two more tasks at once: stocking chemical
storeroom (7 days) and loading dock expansion (3 days).
• When the storeroom and loading dock are done, do the laboratory deep cleansing (2 days).
• Finally, when the deep cleansing is done and the final FDA laboratory inspection has occurred,
arrange the laboratory opening ceremony (2 days).
Now you ask your boss to review the tasks, estimate the total time, and write the answer on a piece of
paper. You look at the paper and see that his guess is wrong.
Questions: (you can use Microsoft Project or any other project management tools to answer the
following questions. DePaul provides access to Microsoft Project to students for free:
https://offices.depaul.edu/information-services/services/Software/Pages/Software-for-PersonalComputers.aspx)
1. What is the correct total time?
2. What is the critical path?
3. Create a Gantt chart that shows the WBS.
4. Create a PERT/CPM chart.

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15-5
ADIABATIC FLAME TEMPERATURE
In the absence of any work interactions and any changes in kinetic or potential energies, the
chemical energy released during a combustion process either is lost as heat to the surroundings
or is used internally to raise the temperature of the combustion products. The smaller the heat
loss, the larger the temperature rise. In the limiting case of no heat loss to the surroundings (Q =
0), the temperature of the products reaches a maximum, which is called the adiabatic
flameor adiabatic combustion temperature of the reaction (Fig. 15-24).
FIGURE 15-24
The temperature of a combustion chamber becomes maximum when combustion is complete and
no heat is lost to the surroundings (Q = 0).
The adiabatic flame temperature of a steady-flow combustion process is determined from Eq. 15-
11 by setting Q = 0 and W = 0. It yields
(15-16)
or
(15-17)
Once the reactants and their states are specified, the enthalpy of the reactants Hreact can be easily
determined. The calculation of the enthalpy of the products Hprod is not so straightforward,
however, because the temperature of the products is not known prior to the calculations.
Therefore, the determination of the adiabatic flame temperature requires the use of an iterative
technique unless equations for the sensible enthalpy changes of the combustion products are
available. A temperature is assumed for the product gases, and the Hprod is determined for this
temperature. If it is not equal to Hreact, calculations are repeated with another temperature. The
adiabatic flame temperature is then determined from these two results by interpolation. When the
oxidant is air, the product gases mostly consist of N2, and a good first guess for the adiabatic
flame temperature is obtained by treating the entire product gases as N2.
In combustion chambers, the highest temperature to which a material can be exposed is limited
by metallurgical considerations. Therefore, the adiabatic flame temperature is an important
consideration in the design of combustion chambers, gas turbines, and nozzles. The maximum
temperatures that occur in these devices are considerably lower than the adiabatic flame
temperature, however, since the combustion is usually incomplete, some heat loss takes place,
and some combustion gases dissociate at high temperatures (Fig. 15-25). The maximum
temperature in a combustion chamber can be controlled by adjusting the amount of excess air,
which serves as a coolant.
FIGURE 15-25
The maximum temperature encountered in a combustion chamber is lower than the theoretical
adiabatic flame temperature.
Note that the adiabatic flame temperature of a fuel is not unique. Its value depends on (1) the
state of the reactants, (2) the degree of completion of the reaction, and (3) the amount of air used.
For a specified fuel at a specified state burned with air at a specified state, the adiabatic flame
temperature attains its maximum value when complete combustion occurs with the theoretical
amount of air.
Page 781
FIGURE 15-26
Schematic for Example 15-8.
SOLUTIONLiquid octane is burned steadily. The adiabatic flame temperature is to be
determined for different cases.
Assumptions1 This is a steady-flow combustion process. 2 The combustion chamber is
adiabatic. 3 There are no work interactions. 4 Air and the combustion gases are ideal
gases. 5 Changes in kinetic and potential energies are negligible.
Analysis(a) The balanced equation for the combustion process with the theoretical amount of air
is
The adiabatic flame temperature relation Hprod = Hreact in this case reduces to
since all the reactants are at the standard reference state and = 0 for O2 and N2. The
and h values of various components at 298 K are
Substance
KJ/kmolKJ/kmol
C8H18( ) −249,950 —
O2 0 8682
N2 0 8669
H2O(g) −241,820 9904
CO2 −393,520 9364
Substituting, we have
which yields
It appears that we have one equation with three unknowns. Actually we have only one
unknown—the temperature of the products Tprod–since h = h(T) for ideal gases. Therefore, we
have to use an equation solver such as EES or a trial-and-error approach to determine the
temperature of the products.
Page 782
A first guess is obtained by dividing the right-hand side of the equation by the total number of
moles, which yields 5,646,081/(8 + 9 + 47) = 88,220 kJ/kmol. This enthalpy value corresponds
to about 2650 K for N2, 2100 K for H2O, and 1800 K for CO2. Noting that the majority of the
moles are N2, we see that Tprod should be close to 2650 K, but somewhat under it. Therefore, a
good first guess is 2400 K. At this temperature,
This value is higher than 5,646,081 kJ. Therefore, the actual temperature is slightly under 2400
K. Next we choose 2350 K. It yields
which is lower than 5,646,081 kJ. Therefore, the actual temperature of the products is between
2350 and 2400 K. By interpolation, it is found to be Tprod = 2395 K.
(b) The balanced equation for the complete combustion process with 400 percent theoretical air
is
By following the procedure used in (a), the adiabatic flame temperature in this case is
determined to be Tprod = 962 K.
Notice that the temperature of the products decreases significantly as a result of using excess air.
(c) The balanced equation for the incomplete combustion process with 90 percent theoretical air
is
Following the procedure used in (a), we find the adiabatic flame temperature in this case to
be Tprod = 2236 K.
DiscussionNotice that the adiabatic flame temperature decreases as a result of incomplete
combustion or using excess air. Also, the maximum adiabatic flame temperature is achieved
when complete combustion occurs with the theoretical amount of air.
Points
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Symbolic Execution

Lab08: Advanced Binary analysis

Tracking Symbolic State using Tree Diagram and proving recheability using z3 Constraint Solver

Lab Description: In this lab will utilize Tree Diagram to show how symbolic state evolve in a program and then use Z3 to prove reachability.

_________________________________________________

LAB EXERCISE:

 

Task 1: Tree Diagram – Consider the code above, create a tree diagram that shows how the symbolic state evolves for every path through this code (similar to the one we did in class). This statement 2**z stands for 2z. Note that the last two statements in this code are executed at the end of each code path, regardless of which branches were taken. However, the value of z in these last statements depends on which path was taken before. To capture this behavior in the tree, you have these two options:

  1. Create a private copy of the last two statements for each path in your diagram
  2. Merge all paths back together at these last statements and model the symbolic value of z with a conditional if-then-else expression that depends on the taken path.

Task 2: Use Z3 to figure out which of the calls to foo, bar, and baz are reachable in the listing from Task1.

 

What to submit

Submit a copy of your tree diagram and screenshot of your Z3 code including the output of the models.

 

COSC 650                                                                     Project Fall 20

  1. Run Wireshark to capture packets. Use a Mozilla Firefox browser to make a request to the Web server W at http://www.example.com.

Choose any 5 HTTP fields of interest in the each of the following captured HTTP packets sent during the connection to the server W at the above site.

  • HTTP GET request
  • HTTP 200 OK response

Briefly discuss why HTTP includes each of the fields you chose in the above packets.

 

  1. Using Java sockets, write a client C, a second client C1, and a server S that do the following:
    • The client C sends a GET request over TCP to the same Web server W as in Part 1 above. The request should be identical to the request that was sent to W by the Firefox browser
    • The server S runs as localhost and listens for client requests on UDP port 12331
    • After the client C receives all the HTTP data from the server W, it closes the TCP socket connection to server W
    • It then sends all the data it received from server W to the server S listening as localhost on UDP port 12331
    • The client C then prints the message: DATA SENT and closes the UDP socket to server S
    • The server S prints all the HTTP data it received that was sent by the client C
    • The second client C1 also connects to the server S on UDP port 12331 and sends it a message that has the following line:

CLIENT C1 SYSTEM TIME IS (it inserts the current system time)

  • When the server S gets the above message from the client C1, it prints this message
  • The server S then sends a message to client C1 that has the message: TIME ACK

When the client C1 receives the above message from the server S, it prints this message and closes the socket

2.10 The server S should create separate threads to handle client C and client C1 (both clients may be connected to the server at the same time).

 

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