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DIA [1.3K]
3 years ago
6

(TCO 1) Name one disadvantage of fixed-configuration switches over modular switches. a. Ease of management b. Port security b. F

ewer high-speed ports d. Easier to add additional portse. Increased delay
Engineering
1 answer:
frutty [35]3 years ago
5 0

Answer:

The answer is A.

Explanation:

Two main types of network switches, modular and fixed configuration switches, are used for connecting the devices with one another provided they are on the same network.

As the name suggests, modular switches can be configured according to your needs and specific situations where you need a different setup.

The one advantage fixed-configuration switches have over the modular switches is that they are easier to operate. You can't change anything for a different application but they are simpler to setup and use, you can just plug them in and start using. They are usually for the more casual end-user and home networks etc.

I hope this answer helps.

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Steam at 5 MPa and 400 C enters a nozzle steadily with a velocity of 80 m/s, and it leavesat 2 MPa and 300 C. The inlet area of
Korolek [52]

Answer:

a) the mass flow rate of the steam is  \mathbf{m_1 =6.92 \ kg/s}

b) the exit velocity of the steam  is \mathbf{V_2 = 562.7 \ m/s}

c) the exit area of the nozzle is  A_2 = 0.0015435 m²

Explanation:

Given that:

A steam with 5 MPa and 400° C enters a nozzle steadily

So;

Inlet:

P_1 = 5 MPa

T_1 = 400° C

Velocity V = 80 m/s

Exit:

P_2 = 2 MPa

T_2 = 300° C

From the properties of steam tables  at P_1 = 5 MPa and T_1 = 400° C we obtain the following properties for enthalpy h and the speed v

h_1 = 3196.7 \ kJ/kg \\ \\ v_1 = 0.057838 \ m^3/kg

From the properties of steam tables  at P_2 = 2 MPa and T_1 = 300° C we obtain the following properties for enthalpy h and the speed v

h_2 = 3024.2 \ kJ/kg  \\ \\ v_2= 0.12551 \ m^3/kg

Inlet Area of the nozzle = 50 cm²

Heat lost Q = 120 kJ/s

We are to determine the following:

a) the mass flow rate of the steam.

From the system in a steady flow state;

m_1=m_2=m_3

Thus

m_1 =\dfrac{V_1 \times A_1}{v_1}

m_1 =\dfrac{80 \ m/s \times 50 \times 10 ^{-4} \ m^2}{0.057838 \ m^3/kg}

m_1 =\dfrac{0.4 }{0.057838 }

\mathbf{m_1 =6.92 \ kg/s}

b) the exit velocity of the steam.

Using Energy Balance equation:

\Delta E _{system} = E_{in}-E_{out}

In a steady flow process;

\Delta E _{system} = 0

E_{in} = E_{out}

m(h_1 + \dfrac{V_1^2}{2}) = Q_{out} + m (h_2 + \dfrac{V_2^2}{2})

-  Q_{out} = m (h_2 - h_1 + \dfrac{V_2^2-V^2_1}{2})

-  120 kJ/s = 6.92 \ kg/s  (3024.2 -3196.7 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000  \ m^2/s^2})

-  120 kJ/s = 6.92 \ kg/s  (-172.5 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000  \ m^2/s^2})

-  120 kJ/s =  (-1193.7 \ kg/s  + 6.92\ kg/s ( \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000  \ m^2/s^2})

V_2^2 = 316631.29 \  m/s

V_2 = \sqrt{316631.29 \  m/s

\mathbf{V_2 = 562.7 \ m/s}

c) the exit area of the nozzle.

The exit of the nozzle can be determined by using the expression:

m = \dfrac{V_2A_2}{v_2}

making A_2 the subject of the formula ; we have:

A_2  = \dfrac{ m \times v_2}{V_2}

A_2  = \dfrac{ 6.92 \times 0.12551}{562.7}

A_2 = 0.0015435 m²

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