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Zanzabum
3 years ago
6

Write a Nested While Loop that will increment the '*' from 1 to 10.

Engineering
1 answer:
andrey2020 [161]3 years ago
8 0

Answer:

The program is as follows:

i = 1

while(i<11):

   j = 1

   while(j<=i):

       print('*', end = '')

       j += 1

   i += 1

   print()

Explanation:

Initialize i to 1

i = 1

The outer loop is repeated as long as i is less than 11

while(i<11):

Initialize j to 1

   j = 1

The inner loop is repeated as long as j is less than or equal i

   while(j<=i):

This prints a *

       print('*', end = '')

This increments j and ends the inner loop

       j += 1

This increments i

   i += 1

This prints a blank and ends the inner loop

   print()

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For each function , sketch the Bode asymptotic magnitude and asymptotic phase plots.
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Answer:

attached below

Explanation:

a) G(s) = 1 / s( s+2)(s + 4 )

Bode asymptotic magnitude and asymptotic phase plots

attached below

b) G(s) = (s+5)/(s+2)(s+4)

phase angles = tan^-1 w/s , -tan^-1 w/s , tan^-1 w/4

attached below

c) G(s)= (s+3)(s+5)/s(s+2)(s+4)

solution attached below

5 0
3 years ago
Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat trans
aksik [14]

Answer:

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet, T_{i} = 20^{\circ}C = 273 + 20 = 293 K

Temperature at the outlet, T_{o} = 200{\circ}C = 273 + 200 = 473 K

Pressure at inlet, P_{i} = 80 kPa = 80\times 10^{3} Pa

Pressure at outlet, P_{o} = 800 kPa = 800\times 10^{3} Pa

Speed at the outlet, v_{o} = 20 m/s

Diameter of the tube, D = 10 cm = 10\times 10^{- 2} m = 0.1 m

Input power, P_{i} = 400 kW = 400\times 10^{3} W

Now,

To calculate the heat transfer, Q, we make use of the steady flow eqn:

h_{i} + \frac{v_{i}^{2}}{2} + gH  + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}

where

h_{i} = specific enthalpy at inlet

h_{o} = specific enthalpy at outlet

v_{i} = air speed at inlet

p_{s} = specific power input

H and H' = Elevation of inlet and outlet

Now, if

v_{i} = 0 and H = H'

Then the above eqn reduces to:

h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}

Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}                (1)

Also,

p_{s} = \frac{P_{i}}{ mass, m}

Area of cross-section, A = \frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}

Specific Volume at outlet, V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s

From the eqn:

P_{o}V_{o} = mRT_{o}

m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s

Now,

p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg

Also,

\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg

Now, using these values in eqn (1):

Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW

Now, rate of heat transfer, q:

q = mQ = 0.925\times 813.33 = 752.33 kW

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