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kaheart [24]
3 years ago
10

A certain type of D-cell battery that costs $0.50 is capable of producing 1.2 V and current of 0.1 A for a period of 75 hours. D

etermine the cost of the energy delivered by this battery per kilowatt hour.
Engineering
1 answer:
stealth61 [152]3 years ago
3 0

Answer:

$55.55/ kWh

Explanation:

Data provided in the question:

Cost of the D-cell battery = $0.50

Voltage produced by the battery, V = 1.2 V

Current produced = 0.1 A

Time = 75 hours

Now,

Energy produced by the battery = Voltage × Current × Time

= 1.2 × 0.1 × 75

= 9 Wh

= 9 × 10⁻³ kWh = 0.009 kWh

Therefore,

Cost of the energy delivered = Cost of the D-cell battery ÷ Energy produced

= $0.50 ÷ 0.009 kWh

= $55.55/ kWh

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Answer:

ηa=0.349

ηb=0.345

Explanation:

The enthalpy and entropy at state 3 are determined from the given pressure and temperature with data from table:

h_{3}=3658.8kJ/kg\\ s_{3}=7.1693kJ/kg

The quality at state 4 is determined from the condition  s_{4} =s_{3} and the entropies of the components at the condenser pressure taken from table:

 q_{4} =\frac{s_{4}-s_{liq50}  }{s_{evap,50} } \\=\frac{7.1693-1.0912}{6.5019}=0.935

The enthalpy at state 4 then is:  

h_{4} =h_{liq50} +q_{4} h_{evap,50}\\ (340.54+0.935*2304.7)kJ/kg\\=2495.43kJ/kg\\

Part A

In the case when the water is in a saturated liquid state at the entrance of the pump the enthalpy and specific volume are determined from A-5 for the given pressure:  

h_{1}=340.54kJ/kg\\ \alpha _{1}=0.00103m^3/kg

The enthalpy at state 2 is determined from an energy balance on the pump:

h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1}  )

    =346.67 kJ/kg

The thermal efficiency is then determined from the heat input and output in the cycle:  

=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.349

Part B  

In the case when the water is at a lower temperature than the saturation temperature at the condenser pressure we look into table and see the water is in a compressed liquid state. Then we take the enthalpy and specific volume for that temperature with data from  and the saturated liquid values:  

h_{1}=293.7kJ/kg\\ \alpha _{1}=0.001023m^3/kg

The enthalpy at state 2 is then determined from an energy balance on the pump:  

h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1}  )

    =299.79 kJ/kg  

The thermal efficiency in this case then is:  

=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.345

8 0
3 years ago
Stainless steel ball bearings (rho = 8085 kg/m3 and cp = 0.480 kJ/kg·°C) having a diameter of 1.2 cm are to be quenched in water
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Answer:

\dot Q = -2.341\,kJ

Explanation:

The rate of heat transfer from the balls to the air is:

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3 0
3 years ago
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A room is cooled by circulating chilled water through a heat exchanger located in the room. The air is circulated through the he
Zanzabum

Answer:

input power of the geothermal motor will be 0.4166 hP

Explanation:

We have given shaft output = 0.25 hP

hP is known as horse power which is unit of measuring the power of motor

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We have to fond the heat supply by the fan motor Q

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Answer:

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Explanation:

(a) The production schedule is 4000 hour per year, or 240,000 minutes. In that time, 133,333 1/3 parts can be produced. The production of 500,000 parts in a year will require the use of ...

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Answer:

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Explanation:

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