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3 years ago

Identify the physical mechanism that causes turbulent thermal conductivity.

Physics

1 answer:

myrzilka [38]3 years ago

5 0

Answer:

Option A is correct.

Eddies due to enhanced mixing of ﬂuid

Explanation:

Turbulent thermal conductivity is thermal conductivity that arises from the turbulent flow of fluids. It comes into play when a particukar fluid moves into turbulent regiom of flow where flow is no longer orderly and streamlines aren't discernable with the fluid layers all warping into one another forming vortices.

It is represented as K and is shown mathematically through the heat flux at turbulent flow

q = vCρT' = - K (∂T/∂y)

where

K = turbulent thermal conductivity

T' = the eddy temperature relative to the mean value,

C = Heat capacity the fluid

q = the rate of thermal energy transport by turbulent eddies.

The physical mechanism that cause turbulent thermal conductivity are similar to the causes of turbulent flow of fluids.

This includes sharp changes in fluid pressure and velocity of flow which is evident in eddies that come about in the enhanced mixing of fluids.

Hope this Helps!!!

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3 years ago

Read 2 more answers

A piston–cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and 400°C. The location of the stops

Ilya [14]

Answer:

(a) Compression work at the final state with a pressure of 1(MPa) is: 44.32(KJ), (b) Compression work at the final state with a pressure of 500(KPa): 110.37(KJ) and (c) temperaure of the final state in part b: T=151.83(°C).

Explanation:

Remember that the substance is steam so it's water (H2O) and the initial conditions are $P_{1} =1MPa, T_{1}=400^{0}C, m=0.6Kg$ and $v_{2} =0.4v_{1}$ from a saturated water table and the initial conditions we can determine that the state phase is superheated (see Table 1 attached) because the $T_{sat}=179.88^{0} C \leq T_{1}$ from the table 1 we get: $v_{1} =0.30661(m^{3}/Kg)$ . Now we have second conditions as: $P_{2}=1(MPa), T_{2}=250^{0}C$ so from the same table we can see the state still superheated and we get $v_{2}=0.23275(m^{3}/Kg)$ , knowing that it's a isobaric process we can find the compression's work as: $W_{b}=m*P(v_{2}-v_{1})=0.6*1000*(0.23275-0.30661)=-44.32(KJ)$ so the compressor's work is: 44.32(KJ). (b) Then the piston reaches the stop and there are two processes in this stage, so Process 1 is isobaric and: $W_{1}=m*P*(v_{2}-v_{1}) =0.6*1000*(0.4*0.30661-0.30661)=-110.38(KJ)$ and the second process is isochoric: $W_{2}=zero$ ,now $W_{b}=W_{1}+ W_{2} =110.38+0=110.38(KJ)$ . Finally to get the temperarure at the final state in part (b) we get: $v_{2} =0.4v_{1} =0.4*0.30661=0.122644(m^{3}/Kg), P_{2}=500(KPa)$ from table 2 (see attached) we compare $v_{f}$ and $v_{g}$ at the saturated water table and find the following: $v_{f}=0.001093(m^{3}/Kg)$ , so we know that the final state phase is a satured mixture and we get the temperature at the final state as: $T_{2} =T_{sat} =151.83^{0}C$ .

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rusak2 [61]

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