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FromTheMoon [43]
4 years ago
12

During January, at a location in Alaska winds at -20°C can be observed, However, several meters below ground the temperature rem

ains at 11°C. An inventor claims to have devised a power cycle working between these temperatures having a thermal efficiency of 10%.
Investigate this claim.

(a) What is the maximum thermal efficiency for these conditions?

(b) Is the inventor's claim possible?
Engineering
1 answer:
maw [93]4 years ago
4 0

Answer:

a) \eta_{th} = 10.910\,\%, b) Yes.

Explanation:

a) The maximum thermal efficiency is given by the Carnot's Cycle, whose formula is:

\eta_{th} =\left(1-\frac{253.15\,K}{284.15\,K}  \right) \times 100\,\%

\eta_{th} = 10.910\,\%

b) The claim of the inventor is possible since real efficiency is lower than maximum thermal efficiency.

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You find an unnamed fluid in the lab we will call Fluid A. Fluid A has a specific gravity of 1.65 and a dynamic viscosity of 210
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Answer:

1.2727 stokes

Explanation:

specific gravity of fluid A = 1.65

Dynamic viscosity = 210 centipoise

<u>Calculate the kinematic viscosity of Fluid A </u>

First step : determine the density of fluid A

Pa = Pw * Specific gravity =  1000 * 1.65 = 1650 kg/m^3

next : convert dynamic viscosity to kg/m-s

210 centipoise = 0.21 kg/m-s

Kinetic viscosity of Fluid A = dynamic viscosity / density of fluid A

                                            = 0.21 / 1650 = 1.2727 * 10^-4 m^2/sec

Convert to stokes = 1.2727 stokes

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3 years ago
When your fixing a car, what is the first thing you want to do?
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Answer:

Changing oil.

Explanation:

You need to regularly check and change your car’s oil to ensure smooth running of the vehicle and to prolong the lifespan of its engine.

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3 years ago
A heat pump operates on a Carnot heat pump cycle with a COP of 12.5. It keeps a space at 24°C by consuming 2.15 kW of power. Det
Vinil7 [7]

Answer:

a) T_{L} = 273.378\,K\,(0.228\,^{\textdegree}C), b) \dot Q_{H} = 26.875\,kW

Explanation:

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After some algebraic handling, the temperature of the cold reservoir is determined:

T_{H}-T_{L} = \frac{T_{H}}{COP_{HP}}

T_{L} = T_{H}\cdot \left(1-\frac{1}{COP_{HP}}  \right)

T_{L} = (297.15\,K)\cdot \left(1-\frac{1}{12.5}\right)

T_{L} = 273.378\,K\,(0.228\,^{\textdegree}C)

b) The heating load provided by the heat pump is:

\dot Q_{H} = COP_{HP}\cdot \dot W

\dot Q_{H} = (12.5)\cdot (2.15\,kW)

\dot Q_{H} = 26.875\,kW

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