Engineering: What is the highest possible theoretical efficiency of a heat engine operating with a hot reservoir of furnace gases at 727°c when the cooling water available is at 27°c ?

What is the highest possible theoretical efficiency

kanga06

31.12.2021

$\eta_{max} = 70%$

Explanation:

given data:

maximum temperature = 727 degree C =727 +273 = 1000 K

Minimum temperature = 27 degree C = 27 +273 = 300 K

Maximum efficiency is given as $\eta$

$\eta = \frac{T_{max} - T_{min}}{T_{max}} *100%$

plugging all value from the above to get maximum efficiency

$\eta = \frac{1000-300}{1000} *100%$

$\eta_{max} = 70%$

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vinnay17

05.04.2022

D=1.0x10^-5 m^2/s

Explanation:

the data given by the exercise are as follows:

T=308 K

d=1 cm

PA=0.195 atm

pL=0.85 g/cm^3

The expression for binary gas-phase diffusion coefficient is equal to:

$D=\frac{\frac{pL*yB}{MA(\frac{t1^{2}-t2^{2} }{2} } }{c(yA1-yA2}$

$yB=\frac{1-0.805}{ln(\frac{1}{0.805}) }=0.89$

$C=\frac{P}{RT}=\frac{1}{82.06*308}=3.95x10^{-5} mol/cm^{2}$

substituting the values in the diffusion equation:

$D=9.6x10^{-6} m^{2}/s$

from Appendix J-1 from Welty:

298 K, D=9.62x10^-5 m^2/s

At 308 K, we have the following:

$D=9.62x10^{-5}(\frac{308}{298})^{3/2}=1.0x10^{-5}m^{2}/s$

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