Answer:
The temperature of the strip as it exits the furnace is 819.15 °C
Explanation:
The characteristic length of the strip is given by;
![L_c = \frac{V}{A} = \frac{LA}{2A} = \frac{5*10^{-3}}{2} = 0.0025 \ m](https://tex.z-dn.net/?f=L_c%20%3D%20%5Cfrac%7BV%7D%7BA%7D%20%3D%20%5Cfrac%7BLA%7D%7B2A%7D%20%3D%20%5Cfrac%7B5%2A10%5E%7B-3%7D%7D%7B2%7D%20%3D%200.0025%20%5C%20m)
The Biot number is given as;
![B_i = \frac{h L_c}{k}\\\\B_i = \frac{80*0.0025}{21} \\\\B_i = 0.00952](https://tex.z-dn.net/?f=B_i%20%3D%20%5Cfrac%7Bh%20L_c%7D%7Bk%7D%5C%5C%5C%5CB_i%20%3D%20%5Cfrac%7B80%2A0.0025%7D%7B21%7D%20%5C%5C%5C%5CB_i%20%3D%200.00952)
< 0.1, thus apply lumped system approximation to determine the constant time for the process;
![\tau = \frac{\rho C_p V}{hA_s} = \frac{\rho C_p L_c}{h}\\\\\tau = \frac{8000* 570* 0.0025}{80}\\\\\tau = 142.5 s](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%5Cfrac%7B%5Crho%20C_p%20V%7D%7BhA_s%7D%20%3D%20%5Cfrac%7B%5Crho%20C_p%20L_c%7D%7Bh%7D%5C%5C%5C%5C%5Ctau%20%3D%20%5Cfrac%7B8000%2A%20570%2A%200.0025%7D%7B80%7D%5C%5C%5C%5C%5Ctau%20%3D%20142.5%20s)
The time for the heating process is given as;
![t = \frac{d}{V} \\\\t = \frac{3 \ m}{0.01 \ m/s} = 300 s](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bd%7D%7BV%7D%20%5C%5C%5C%5Ct%20%3D%20%5Cfrac%7B3%20%5C%20m%7D%7B0.01%20%5C%20m%2Fs%7D%20%3D%20300%20s)
Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;
![T(t) = T_{ \infty} + (T_i -T_{\infty})e^{-t/ \tau}\\\\T(t) = 930 + (20 -930)e^{-300/ 142.5}\\\\T(t) = 930 + (-110.85)\\\\T_{(t)} = 819.15 \ ^0 C](https://tex.z-dn.net/?f=T%28t%29%20%3D%20T_%7B%20%5Cinfty%7D%20%2B%20%28T_i%20-T_%7B%5Cinfty%7D%29e%5E%7B-t%2F%20%5Ctau%7D%5C%5C%5C%5CT%28t%29%20%3D%20930%20%2B%20%2820%20-930%29e%5E%7B-300%2F%20142.5%7D%5C%5C%5C%5CT%28t%29%20%3D%20930%20%2B%20%28-110.85%29%5C%5C%5C%5CT_%7B%28t%29%7D%20%3D%20819.15%20%5C%20%5E0%20C)
Therefore, the temperature of the strip as it exits the furnace is 819.15 °C
Answer:
(b) 56%
Explanation:
the maximum thermal efficiency is possible only when power cycle is reversible in nature and when power cycle is reversible in nature the thermal efficiency depends on the temperature
here we have given T₁ (Higher temperature)= 600+273=873
lower temperature T₂=110+273=383
Efficiency of power cycle is given by =1-![\frac{T2}{T1}](https://tex.z-dn.net/?f=%5Cfrac%7BT2%7D%7BT1%7D)
=1-![\frac{383}{873}](https://tex.z-dn.net/?f=%5Cfrac%7B383%7D%7B873%7D)
=1-0.43871
=.56
=56%
The first one is the answer
Answer:
250.7mw
Explanation:
Volume of the reservoir = lwh
Length of reservoir = 10km
Width of reservoir = 1km
Height = 100m
Volume = 10x10³x10³x100
= 10⁹m³
Next we find the volume flow rate
= 0.1/100x10⁹x1/3600
= 277.78m³/s
To get the electrical power output developed by the turbine with 92 percent efficiency
= 0.92x1000x9.81x277.78x100
= 250.7MW
Answer:
Time taken = 136.32 minutes
Explanation:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.