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aliina [53]
3 years ago
6

A step-down transformer (turns ratio = 1:7) is used with an electric train to reduce the voltage from the wall receptacle to a v

alue needed to operate the train. When the train is running, the current in the secondary coil is 1.04 A. What is the current in the primary coil?
Engineering
1 answer:
SOVA2 [1]3 years ago
6 0

Answer:

wait lemme check it out

Explanation:

You might be interested in
Air expands through a turbine operating at steady state. At the inlet p1 = 150 lbf/in^2, T1 = 1400R and at the exit p2 = 14.8 lb
Paraphin [41]

Answer:

The power developed in HP is 2702.7hp

Explanation:

Given details.

P1 = 150 lbf/in^2,

T1 = 1400°R

P2 = 14.8 lbf/in^2,

T2 = 700°R

Mass flow rate m1 = m2 = m = 11 lb/s Q = -65000 Btu/h

Using air table to obtain the values for h1 and h2 at T1 and T2

h1 at T1 = 1400°R = 342.9 Btu/h

h2 at T2 = 700°R = 167.6 Btu/h

Using;

Q - W + m(h1) - m(h2) = 0

W = Q - m (h2 -h1)

W = (-65000 Btu/h ) - 11 lb/s (167.6 - 342.9) Btu/h

W = (-65000 Btu/h ) - (-1928.3) Btu/s

W = (-65000 Btu/h ) * {1hr/(60*60)s} - (-1928.3) Btu/s

W = -18.06Btu/s + 1928.3 Btu/s

W = 1910.24Btu/s

Note; Btu/s = 1.4148532hp

W = 2702.7hp

5 0
3 years ago
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
Assoli18 [71]

Answer: The exit temperature of the gas in deg C is 32^{o}C.

Explanation:

The given data is as follows.

C_{p} = 1000 J/kg K,   R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

P_{1} = 100 kPa,     V_{1} = 15 m^{3}/s

T_{1} = 27^{o}C = (27 + 273) K = 300 K

We know that for an ideal gas the mass flow rate will be calculated as follows.

     P_{1}V_{1} = mRT_{1}

or,         m = \frac{P_{1}V_{1}}{RT_{1}}

                = \frac{100 \times 15}{0.5 \times 300}  

                = 10 kg/s

Now, according to the steady flow energy equation:

mh_{1} + Q = mh_{2} + W

h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}

C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}

(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}

(T_{2} - T_{1}) = 5 K

T_{2} = 5 K + 300 K

T_{2} = 305 K

           = (305 K - 273 K)

           = 32^{o}C

Therefore, we can conclude that the exit temperature of the gas in deg C is 32^{o}C.

8 0
4 years ago
For an Otto cycle, plot the cycle efficiency as a function of compression ratio from 4 to 16.
Elza [17]

Assumptions:

  • Steady state.
  • Air as working fluid.
  • Ideal gas.
  • Reversible process.
  • Ideal Otto Cycle.

Explanation:

Otto cycle is a thermodynamic cycle widely used in automobile engines, in which an amount of gas (air) experiences changes of pressure, temperature, volume, addition of heat, and removal of heat. The cycle is composed by (following the P-V diagram):

  • Intake <em>0-1</em>: the mass of working fluid is drawn into the piston at a constant pressure.
  • Adiabatic compression <em>1-2</em>: the mass of working fluid is compressed isentropically from State 1 to State 2 through compression ratio (r).

        r =\frac{V_1}{V_2}

  • Ignition 2-3: the volume remains constant while heat is added to the mass of gas.
  • Expansion 3-4: the working fluid does work on the piston due to the high pressure within it, thus the working fluid reaches the maximum volume through the compression ratio.

         r = \frac{V_4}{V_3} = \frac{V_1}{V_2}

  • Heat Rejection 4-1: heat is removed from the working fluid as the pressure drops instantaneously.
  • Exhaust 1-0: the working fluid is vented to the atmosphere.

If the system produces enough work, the automobile and its occupants will propel. On the other hand, the efficiency of the Otto Cycle is defined as follows:

           \eta = 1-(\frac{1}{r^{\gamma - 1} } )

where:

           \gamma = \frac{C_{p} }{C_{v}} : specific heat ratio

Ideal air is the working fluid, as stated before, for which its specific heat ratio can be considered constant.

           \gamma = 1.4

Answer:

See image attached.

5 0
3 years ago
Explain why the following scenario fails to meet the definition of a project description.
s344n2d4d5 [400]

Answer:

The youth hockey training facility

Explanation:

7 0
3 years ago
A monatomic ideal gas undergoes a quasi-static process that is described by the function p(????)=p1+3(????−????1) , where the st
Alenkasestr [34]

A pure gas made up only of atoms. The noble gases argon, krypton, and xenon are some examples.

Concepts:

Perfect gas law: Work performed on the system: PV = nRT W = -∫PdV

Energy preservation formula: U = Q + W

Reasoning:

W = nRT ln(Vi/Vf) when the process is isothermal.

The temperature is said to be constant, and we are given n, Pfinal, and Vfinal.

Calculation information:

(A) A process that is isothermal has a constant temperature.

PV = nRT, and hence, constant

nRT = PV = 101000 Pa*25*10-3 m3

For a process that is isothermal, W = nRT ln(Vi/Vf).

W/(nRT)=3000 J/(101000 Pa*25*10-3 m3)=-1.19

(The gas produces -W of labor.)

Vi = (25*10-3 m3)/3.28 = 7.62*10-3 m3 = 7.62 L where Vf/Vi = exp(1.19) = 3.28 Vi (b) for a perfect gas PV = nRT. 101000 Pa*25*10-3 m3 = (8.31 J/K) T. T = 303.85 K.

To know more about process click here:

brainly.com/question/29310303

#SPJ4

5 0
1 year ago
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