Plans normally use a reference line to indicate elevations on the ____

Following are several z-transforms. For each one, determine inverse z-transform using both the method based on the partial-fract

tigry1 [53]

Answer:

For now the answer to this question is only for partial fraction. Find attached.

6 0

3 years ago

You are asked to design a software package for an On-line Banking System (OBS). The OBS will take requests from its user on the

Anit [1.1K]

Answer:

From the statements described above,The proposed software package Online Banking System(OBS) shall provide the following services to the customer(account holder)

Receive all transaction requests from the customers trough online to perform banking activities

Offer a variety of services in banking like:

Opening an account with facilities of Savings,Checking and CD

Issuing checks and check clearance

Transfer of funds between savings and checking account

Establishing a CD(Certificate of Deposit)

Performing routine activities like

Deposit or withdraw funds from a checking account

Closing an account in one of the following operation

Complete closure of account

Closing only savings account

Withdraw CD (on Maturity of CD deposits) and close CD account

Unlike traditional Banking system, OBS offers all the operations through On-line mode over the internet thereby it eliminates manual intervention for the banking transactions from the Bank's side for most of the transactions except some cases like verification of new account applications or complaint redressal and technical support .In such cases ,There will be some users of the software with special privileges to access certain services to perform activities like user data verification and technical support.

Step:1 Identification of actors and software entities for the system to be implemented

The following are the actors identified in the system

Actors:

The users of the proposed system were classified as:

Customer : A user with an active account

Guest user:The user who applies for an account

Administrator:Maintainsand Controls OBS

Manager: Manages the transactions and Banking related issues

Technical Manager: Resolves technical issues of customers

Software entities:

Bank database :Maintains the data of all user accounts

Application server : Maintains the application software components and hosts the OBS website

Automatic Clearing House system : Processes transactions for checks

Step 2: Designing use cases

Use case diagrams are used to describe the system model by representing a set of use cases .Each use case represents a set of actions performed on the system by an actor.

The following are the two approaches of representing the system in a use case model:

1.Representing a use case for actions done by actors e.g. login , logout,message etc.,

2.Representing a use case for business process(Business use cases) e.g. transfer,registration etc.,

The following is the description of five possible business use cases for the requirements specified above

3 0

3 years ago

A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the

raketka [301]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Intertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy = ?

(b) rotor acceleration = ?

(c) change in torque angle = ?

(c) rotor speed = ?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

(c) change in torque angle (in elec deg) = 6.75 elec deg

(c) change in torque angle (in rmp/s) = 28.12 rpm/s

(c) rotor speed = 1505.62 rpm

Explanation:

(a) Find the stored energy in the rotor at synchronous speed.

The stored energy is given by

$E = G \times H$

Where G represents complex rated power and H is the inertia constant of turbo-generator.

$E = 100 \times 8 \\\\E = 800 \: MJ$

(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.

The rotor acceleration is given by

$P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}$

Where M is given by

$M = \frac{E}{180 \times f}$

$M = \frac{800}{180 \times 50}$

$M = 0.0889 \: MJ \cdot s/ elec \: \: deg$

So, the rotor acceleration is

$P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}$

$30 = 0.0889 \frac{d^2 \delta}{dt^2}$

$\frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}$

$\frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2$

(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.

The change in torque angle is given by

$\Delta \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2$

Where t is given by

$1 \: cycle = 1/f = 1/50 \\\\10 \: cycles = 10/50 = 0.2 \\\\t = 0.2 \: sec$

So,

$\Delta \delta = \frac{1}{2} \cdot 337.46 \cdot (0.2)^2$

$$ \Delta \delta = 6.75 \: elec \: deg$

The change in torque in rpm/s is given by

$\Delta \delta = \frac{337.46 \cdot 60}{2 \cdot 360\circ }$

$\Delta \delta =28.12 \: \: rpm/s$

The rotor speed in revolutions per minute at the end of this period (10 cycles) is given by

$Rotor \: speed = \frac{120 \cdot f}{P} + (\Delta \delta)\cdot t$

Where P is the number of poles of the turbo-generator.

$Rotor \: speed = \frac{120 \cdot 50}{4} + (28.12)\cdot 0.2$

$Rotor \: speed = 1500 + 5.62$

$$ Rotor \: speed = 1505.62 \:\: rpm$

4 0

3 years ago

Explain 4 things you can do with a Combination Square

sergij07 [2.7K]

Answer:

A combination square is a multi-use measuring instrument which is primarily used for ensuring the integrity of a 90° angle, measuring a 45° angle, measuring the center of a circular object, find depth, and simple distance measurements. It can also be used to determine level and plumb using its spirit level vial.

Explanation:

3 0

3 years ago

The time to failure for a gasket follows the Weibull distribution with ß = 2.0 and a characteristic life of 300 days. What is th

Aleks04 [339]

Answer:

64.11% for 200 days.

t=67.74 days for R=95%.

t=97.2 days for R=90%.

Explanation:

Given that

β=2

Characteristics life(scale parameter α)=300 days

We know that Reliability function for Weibull distribution is given as follows

$R(t)=e^{-\left(\dfrac{t}{\alpha}\right)^\beta}$

Given that t= 200 days

$R(200)=e^{-\left(\dfrac{200}{300}\right)^2}$

R(200)=0.6411

So the reliability at 200 days 64.11%.

When R=95 %

$0.95=e^{-\left(\dfrac{t}{300}\right)^2}$

by solving above equation t=67.74 days

When R=90 %

$0.90=e^{-\left(\dfrac{t}{300}\right)^2}$

by solving above equation t=97.2 days

7 0

3 years ago

Plans normally use a reference line to indicate elevations on the _____.