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3 years ago

11

A thermodynamicist claims to have developed a heat pump with a COP of 1.7 when operating with thermal energy reservoirs at 273 K

and 293 K. Is this claim valid?

1 answer:

Vinil7 [7]3 years ago

3 0

Answer:

cfcghvjvct

Explanation:

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Technician A says that after replacing a power steering hose, the system should be flushed, refilled, and bled. Technician B say

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The. Answer will be D

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4 years ago

Salvage ethnography is the effort to ensure that ethnography remains an important part of anthropology. recording of linguistic

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Answer:

D

Explanation:

D

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4 years ago

2–25 Consider a medium in which the heat conduction equation is given in its simplest form as

barxatty [35]

Answer:

d) Is the thermal conductivity of the medium constant or variable.

Explanation:

As we know that

Heat equation with heat generation at unsteady state and with constant thermal conductivity given as

$\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}+\dfrac{d^2T}{dz^2}+\dfrac{\dot{q}_g}{K}=\dfrac{1}{\alpha }\dfrac{dT}{dt}$

With out heat generation

$\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}+\dfrac{d^2T}{dz^2}=\dfrac{1}{\alpha }\dfrac{dT}{dt}$

In 2 -D with out heat generation with constant thermal conductivity

$\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}=\dfrac{1}{\alpha }\dfrac{dT}{dt}$

Given equation

$\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}=\dfrac{1}{a }\dfrac{dT}{dt}$

So we can say that this is the case of with out heat generation ,unsteady state and with constant thermal conductivity.

So the option d is correct.

d) Is the thermal conductivity of the medium constant or variable.

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3 years ago

How can you relate entropy to renewable and non-renewable energy?

djyliett [7]

Answer:

rrbtnhipsdjmskmbbylu.

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4 years ago

Two very long concentric cylinders of diameters D1 = 0.42 m and D2 = 0.5 m are maintained at uniform temperatures of T1 = 950 K

MrRa [10]

Answer:

Q=33.34 KW/m

Explanation:

Given that

D₁=0.42 m

A₁= π D₁ L

For unit length

A₁= π D₁ = 0.42 π m²

D₂=0.5 m

A₂= 0.5 π m²

ε₁= 1 ,ε₂= 0.55

T₁=950 K ,T₂ = 500 K

$Q=\dfrac{\sigma (T_1^4-T_2^4)}{\dfrac{1-\varepsilon _1}{A_1\epsilon _1}+\dfrac{1-\varepsilon _2}{A_2\epsilon _2}+\dfrac{1}{A_1F_{12}}}$

F₁₁+F₁₂= 1

F₁₁= 0

So, F₁₂= 1

$Q=A_1\dfrac{\sigma (T_1^4-T_2^4)}{\dfrac{1-\varepsilon _1}{\epsilon _1}+A_1\dfrac{1-\varepsilon _2}{A_2\epsilon _2}+1}$

$Q=0.42\pi \dfrac{5.67\times 10^{-8}(950^4-500^4)}{\dfrac{1-1}{1}+0.42\pi\times \dfrac{1-0.55}{0.5\pi\times 0.55}+1}$

Q=33.34 KW/m

3 0

4 years ago