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GalinKa [24]
3 years ago
15

You can assume there is no pressure drop between the exit of the compressor and the entrance of the turbine. All the power from

the turbine runs the compressor. The turbine exit conditions are 780 K, 300 kPa. All components operate at the steady state, you can ignore kinetic and potential energy, as well as heat transfer, which is small compared to the power generated by the turbine. How much work is generated by the turbine?
Engineering
1 answer:
Eddi Din [679]3 years ago
6 0

Answer:

s6rt5x11j4fgu

j4

cf53yhu5

y4

hh

Explanation:

j

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The primary of an ideal transformer has 400 turns and its secondary has 200 turns. Neglecting electrical losses, if the power in
Vanyuwa [196]

Answer:

A.C. Voltage (transformers only work with A.C.) from input (primary) to output (secondary) is purely a function of the turns ratio.

Explanation:

6 0
2 years ago
Suppose you have a 9.00 V battery, a 2.00 μF capacitor, and a 7.40 μF capacitor. (a) Find the charge and energy stored if the ca
Andru [333]

Answer:

Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J

Q=9.4*10^-6*9=8.46*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *9.4*10^-6=3.81*10^-4J

Explanation:

<u>a)</u>

<u>Identify the unknown:  </u>

The charge and energy stored if the capacitors are connected in series  

<u>List the Knowns: </u>

Capacitance of the first capacitor: C_{1}= 2цF = 2 x 10-6 F

Capacitance of the second capacitor C_{2}= 7.4цF  = 7.4 x 10-6 F

Voltage of battery: V = 9 V  

<u>Set Up the Problem:   </u>

Capacitance of a series combination:  

\frac{1}{C_{s} } =\frac{1}{C_{1} } +\frac{1}{C_{2} } +\frac{1}{C_{3} }+............

\frac{1}{C_{s} } =\frac{1}{2} +\frac{1}{ 7.4} \\C_{s} =\frac{2*7.4}{2+7.4}=1.575 *10^-6 F\\

Capacitance of a series combination is given by:

C_{s}=\frac{Q}{V}

Then the charge stored in the series combination is:  

Q=C_{s} V

Energy stored in the series combination is:  

U_{c}=\frac{1}{2}  V^{2} C_{s}

<u>Solve the Problem:  </u>

Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J

<u>b)</u>

<u>Identify the unknown:  </u>

The charge and energy stored if the capacitors are connected in parallel  

<u>Set Up the Problem:  </u>

Capacitance of a parallel combination:

C_{p} =C_{1} +C_{2} +C_{3}

C_{p} =2+7.4=9.4*10^-6F

Capacitance of a parallel combination is given by

C_{p} =\frac{Q}{V}

Then the charge stored in the parallel combination is

Q=C_{p} V

Energy stored in the parallel combination is:  

U_{c}=\frac{1}{2} V^2C_{p}

<u>Solve the Problem: </u><em>  </em>

Q=9.4*10^-6*9=8.46*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *9.4*10^-6=3.81*10^-4J

5 0
3 years ago
Read 2 more answers
One kg of an idea gas is contained in one side of a well-insulated vessel at 800 kPa. The other side of the vessel is under vacu
laiz [17]

Answer:

Option C = internal energy stays the same.

Explanation:

The internal energy will remain the same or unchanged because this question has to do with a concept in physics or classical chemistry (in thermodynamics) known as Free expansion.

So, the internal energy will be equals to the multiplication of the change in temperature, the heat capacity (keeping volume constant) and the number of moles. And in free expansion the internal energy is ZERO/UNCHANGED.

Where, the internal energy, ∆U = 0 =quantity of heat, q - work,w.

The amount of heat,q = Work,w.

In the concept of free expansion the only thing that changes is the volume.

7 0
3 years ago
An eddy current separator is to separate aluminum product from an input streamshredded MSW. The feed rate to the separator is 2,
blsea [12.9K]

Answer:

<em>the % recovery of aluminum product is 80.5%</em>

<em>the % purity of the aluminum product is 54.7%</em>

<em></em>

Explanation:

feed rate to separator = 2500 kg/hr

in one hour, there will be 2500 kg/hr x 1 hr = 2500 kg of material is fed into the  machine

of this 2500 kg, the feed is known to contain 174 kg of aluminium and 2326 kg of rejects.

After the separation, 256 kg  is collected in the product stream.

of this 256 kg, 140 kg is aluminium.

% recovery of aluminium will be = mass of aluminium in material collected in the product stream ÷ mass of aluminium contained in the feed material

% recovery of aluminium = 140kg/174kg x 100% = <em>80.5%</em>

% purity of the aluminium product = mass of aluminium in final product ÷ total mass of product collected in product stream

% purity of the aluminium product = 140kg/256kg

x 100% = <em>54.7%</em>

8 0
3 years ago
Since you became discouraged not being able to find a job in the San Diego area, you enlarged the area in which you looked for a
nordsb [41]

Answer:

need points 48986

Explanation:

5 0
3 years ago
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