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2 years ago

What is the output of the following program fragment. Choose appropriate data-types of variables to match output.

Engineering

1 answer:

Alexxx [7]2 years ago

3 0

Answer:

Explanation:

Given :

i=7;j=8;k=9;

printf(“%d”,(i+10)%k/j);

The printf function only displays the result

%d - is used for the display format

However, the actual calculation is in the expression: (i+10)%k/j

Given the variables :

i = 7 ;

j = 8 ;

k = 9 ;

(i + 10) = (7 + 10) = 17

(i + 10) % k = 17 % 9

% = remainder value after division

17 % 9 = (17 / 9) = 1 remainder 8

17 % 9 = 8

Hence,

(i + 10) % k = 8

(i + 10) % k / j

j = 8

8 / 8

= 1

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Think about a packed lunch and the

Mazyrski [523]

Answer:

They are combination structures made up of solid, and shell structures.

Explanation:

Thinking of a lunch pack which is shaped as a solid bag made up of different compartments to hold food, what comes to mind are;

1. The solid structure: This is seen in the form of the bag which has solid makeup.

2. The shell structure: This is seen in the different compartments holding the different types of foods. Shell structures are usually in the form of containers made to hold something in.

4 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 = ?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

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

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