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3 years ago

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Using the technique of bitwise ANDing, find the IP address of the network # (this could be a network or subnet) on which the mac

hine with an IP address and (sub)netmask of 175.99.138.0/21 resides. (Do not include any spaces in your answer.)

1 answer:

schepotkina [342]3 years ago

7 0

Answer:

175.99.136.0

Explanation:

175.99.138.0/21

Step 1: convert both the ip and the subnet mask to binary format

175.99.138.0 - 10101111.1100011.10001010.00000000

/21 : 255.255.248.0 - 11111111.11111111.11111000.00000000

step 2: perform the bitwise AND operation

In an AND operation, a result of 1 can only be gotten when both inputs are 1. Otherwise, the result will be 0.

10101111.1100011.10001010.00000000

11111111. 11111111. 11111000. 00000000

------------------------------------------------------

10101111. 1100011. 10001000. 00000000

step 3: convert the binary result back to base 10 number

10101111. 1100011. 10001000. 00000000

175.99.136.0

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Answer:

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Solution:

As per the question:

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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 = $\frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}$

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

Also,

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Now, using these values in eqn (1):

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

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