I believes it’s long body hair
Answer:
The electrical power is 96.5 W/m^2
Explanation:
The energy balance is:
Ein-Eout=0
![qe+\alpha sGs+\alpha skyGsky-EEb(Ts)-qc=0](https://tex.z-dn.net/?f=qe%2B%5Calpha%20sGs%2B%5Calpha%20skyGsky-EEb%28Ts%29-qc%3D0)
if:
Gsky=oTsky^4
Eb=oTs^4
qc=h(Ts-Tα)
![\alpha s=\frac{\int\limits^\alpha _0 {\alpha l Gl} \, dl }{\int\limits^\alpha _0 {Gl} \, dl }](https://tex.z-dn.net/?f=%5Calpha%20s%3D%5Cfrac%7B%5Cint%5Climits%5E%5Calpha%20_0%20%7B%5Calpha%20l%20Gl%7D%20%5C%2C%20dl%20%7D%7B%5Cint%5Climits%5E%5Calpha%20_0%20%7BGl%7D%20%5C%2C%20dl%20%7D)
![\alpha s=\frac{\int\limits^\alpha _0 {\alpha lEl(l,5800 } \, dl }{\int\limits^\alpha _0 {El(l,5800)} \, dl }](https://tex.z-dn.net/?f=%5Calpha%20s%3D%5Cfrac%7B%5Cint%5Climits%5E%5Calpha%20_0%20%7B%5Calpha%20lEl%28l%2C5800%20%7D%20%5C%2C%20dl%20%7D%7B%5Cint%5Climits%5E%5Calpha%20_0%20%7BEl%28l%2C5800%29%7D%20%5C%2C%20dl%20%7D)
if Gl≈El(l,5800)
![\alpha s=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))](https://tex.z-dn.net/?f=%5Calpha%20s%3D%281-0.2%29F%280-2%29%2B%281-0.7%29%281-F%280-2%29%29)
lt= 2*5800=11600 um-K, at this value, F=0.941
![\alpha s=(0.8*0.941)+0.3(1-0.941)=0.77](https://tex.z-dn.net/?f=%5Calpha%20s%3D%280.8%2A0.941%29%2B0.3%281-0.941%29%3D0.77)
The hemispherical emissivity is equal to:
![E=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))](https://tex.z-dn.net/?f=E%3D%281-0.2%29F%280-2%29%2B%281-0.7%29%281-F%280-2%29%29)
lt=2*333=666 K, at this value, F=0
![E=0+(1-0.7)(1)=0.3](https://tex.z-dn.net/?f=E%3D0%2B%281-0.7%29%281%29%3D0.3)
The hemispherical absorptivity is equal to:
![qe=EoTs^{4}+h(Ts-T\alpha )-\alpha sGs-\alpha oTsky^{4}=(0.3*5.67x10^{-8}*333^{4})+10(60-20)-(0.77-600)-(0.3*5.67x10^{-8}*233^{4})=96.5 W/m^{2}](https://tex.z-dn.net/?f=qe%3DEoTs%5E%7B4%7D%2Bh%28Ts-T%5Calpha%20%20%29-%5Calpha%20sGs-%5Calpha%20oTsky%5E%7B4%7D%3D%280.3%2A5.67x10%5E%7B-8%7D%2A333%5E%7B4%7D%29%2B10%2860-20%29-%280.77-600%29-%280.3%2A5.67x10%5E%7B-8%7D%2A233%5E%7B4%7D%29%3D96.5%20W%2Fm%5E%7B2%7D)
Answer:
second-law efficiency = 62.42 %
Explanation:
given data
temperature T1 = 1200°C = 1473 K
temperature T2 = 20°C = 293 K
thermal efficiency η = 50 percent
solution
as we know that thermal efficiency of reversible heat engine between same temp reservoir
so here
efficiency ( reversible ) η1 = 1 -
............1
efficiency ( reversible ) η1 = 1 -
so efficiency ( reversible ) η1 = 0.801
so here second-law efficiency of this power plant is
second-law efficiency =
second-law efficiency =
second-law efficiency = 62.42 %
Answer:
Detailed solution is given in the attached diagram
Answer:
I need some more point and i do not understand your question
Explanation: