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Georgia [21]
4 years ago
5

Find the requested quantities for the circuit. We used the mesh-current method to identify the meshes. We then identified the me

sh currents and wrote a KVL equation for each mesh and a constraint equation for the dependent source that defines its controlling variable in terms of the mesh currents. We solved these equations simultaneously for the unknown mesh currents and constrained current, and we checked the solution by verifying that the power in the circuit balances Now use the mesh-current values to calculate the voltage v0 and the total power generated in the circuit. Enter your answers directly on the figure.

Engineering
1 answer:
Irina18 [472]4 years ago
6 0

Answer:

Explanation:

The image that is supposed to be attached to the question is displayed in the diagram below.

Applying Nodal Analysis at node 1;

\dfrac{V_o -50}{12.5*10^{-3}} + \dfrac{V_o}{50*10^3}+\dfrac{V_o-7500 \ in}{10*10^3}=0

where;

in = \dfrac{V_o}{50*10^3}   (from the circuit)

= \dfrac{V_o-50}{12.5}+\dfrac{V_o}{50} + V_o -\dfrac{7500 *V_o }{\frac{50*10^3}{10}}=0

= V_o [ \dfrac{1}{12.5}+\dfrac{1}{50}+\dfrac{1}{10}-\dfrac{75}{500}] = \dfrac{50}{12.5}

= V_o[ \dfrac{500*500+12.5*5000+12.5*5000*5-75*12.5*500}{12.5*50*10*500}]= \dfrac{50}{12.5}

= V_o = 80 \ volts

in = \frac{80}{50*10^3}= 1.6 mA \\ \\ 7500*in = 120 volts  \\ \\ I = \frac{120-80}{10(10^3} =4*10^{-3} Amps \\ \\  \\ \\ P_{generated} = 75000*in*I \\ \\ P_{generated} = 120*4*10^{-3}  \\ \\  P_{generated} = 480  \ MW

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Explanation:

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Describe the importance of ferrite and austenite stabilizing elements in steels
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Answer:

The importance of ferrite and austenite stabilizing elements in steels .

Explanation:

Alloying -

The process which improves the properties of the steel by changing the chemical composition of the steel via adding some elements .

The properties can be improved by - Stabilizing Austenite and Stabilizing Ferrite .

Stabilizing austenite -

The process by which temperature is increased , in which Austenite exists .

Elements with the same crystal structure as of the austenite ( FCC ) raises its A4 value i.e. the temperature of the formation of austenite from its liquid phase and reduces the value of A3 .

Hence, the elements are -

Cobalt , Nickel , Manganese , Copper.

The examples of the Austenitic steels are -

Hadfield Steel ( 13% Mn , 1.2% Cr , 1% C ) and Austenitic Stainless steel.

Stabilizing ferrite –

The process by which temperature is decreased , in which austenite exists .

Elements with the same crystal structure as of the ferrite (BCC - Cubic body centered ) lowers its A4 value i.e. the temperature of the formation of austenite from its liquid phase and increases the value of A3 .These elements have lower solubility of carbon in austenite, that lead to increase in the amount of carbides in the steel.

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4 years ago
Refrigerant-134a enters an adiabatic compressor as saturated vapor at -24°C and leaves at 0.8 MPa and 60°C. The mass flow rate o
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Answer:

(a) The power input to the compressor: \dot{W}=73.07 kJ/s = 73.07 kW

(b) The volume flow rate of the refrigerant at the compressor inlet: \dot{v}=0.209 m^{3}/s

Explanation:

(a)

We need to check the values of enthalpy (as we have an open system) for both states, being the inlet, state 1 and the outlet, state 2. We will know these values by checking the vapor charts of R134a, I used the ones found in Thermodynamics of Cengel, 7th edition.

Then, our values are:

h_{1}=235.92kJ/kg\\h_{2}=296.81kJkg

Now we proceed to know the work with the following expression:

\dot{W}=\dot{m}(h_{2}-h_{1})

Now we replace values and our result is:

\dot{W}=73.07 kJ/s = 73.07 kW

(b)

To know the volume rate at the compressor inlet, we need to know the specific volume in that phase, as we have that is saturated and at -24°C, we can read our table:

\nu=0.1739m^{3}/kg

With our specific volume and the mass rate, we can calculate the volume rate:

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Hope this helps!

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Answer:

True

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=> f(x,y,z,t) = constant

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