Hi! bridges could have been collapse due to an error made by the engineers during construction.
Thank you so much!! You too!
Answer:
Not possible.
Explanation:
According to second law of thermodynamics, the maximum efficiency any heat engine could achieve is Carnot Efficiency η defined by:
![\eta=1-\frac{T_{cold}}{T_{hot}}](https://tex.z-dn.net/?f=%5Ceta%3D1-%5Cfrac%7BT_%7Bcold%7D%7D%7BT_%7Bhot%7D%7D)
Where
and
are temperature (in Kelvin) of heat source and heatsink respectively
In our case (I will be using K = 273+°C) :
![\eta=1-\frac{-27+273}{14+273}\\=0.1428](https://tex.z-dn.net/?f=%5Ceta%3D1-%5Cfrac%7B-27%2B273%7D%7B14%2B273%7D%5C%5C%3D0.1428)
In percentage, this is 14.28% efficiency, which is the <em>maximum</em> theoretical efficiency <em>any</em> heat engine could have while working between -27 and 14 °C temperature. Any claim of more efficient heat engine between these 2 temperature are violates the second law of thermodynamics. Therefore, the claim must be false.
Hey there ..
I think the answer is as they put brake ..
I am not sure .. just check the image also provided above .. ..
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Answer:
The answer is "
"
Explanation:
Given data:
Initial temperature of tank ![T_1 = 300^{\circ}\ C= 573 K](https://tex.z-dn.net/?f=T_1%20%3D%20300%5E%7B%5Ccirc%7D%5C%20C%3D%20573%20K)
Initial pressure of tank
Diameter of throat
Mach number at exit ![M = 2.8](https://tex.z-dn.net/?f=M%20%3D%202.8)
In point a:
calculating the throat area:
![A*=\frac{\pi}{4} \times d^2](https://tex.z-dn.net/?f=A%2A%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%20%5Ctimes%20d%5E2)
![=\frac{\pi}{4} \times 2^2\\\\=\frac{\pi}{4} \times 4\\\\=3.14 \ cm^2](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%20%5Ctimes%202%5E2%5C%5C%5C%5C%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%20%5Ctimes%204%5C%5C%5C%5C%3D3.14%20%5C%20cm%5E2)
Since, the Mach number at throat is approximately half the Mach number at exit.
Calculate the Mach number at throat.
![M*=\frac{M}{2}\\\\=\frac{2.8}{2}\\\\=1.4](https://tex.z-dn.net/?f=M%2A%3D%5Cfrac%7BM%7D%7B2%7D%5C%5C%5C%5C%3D%5Cfrac%7B2.8%7D%7B2%7D%5C%5C%5C%5C%3D1.4)
Calculate the exit area using isentropic flow equation.
![\frac{A}{A*}= (\frac{\gamma -1}{2})^{\frac{\gamma +1}{2(\gamma -1)}} (\frac{1+\frac{\gamma -1}{2} M*^2}{M*})^{\frac{\gamma +1}{2(\gamma -1)}}](https://tex.z-dn.net/?f=%5Cfrac%7BA%7D%7BA%2A%7D%3D%20%28%5Cfrac%7B%5Cgamma%20-1%7D%7B2%7D%29%5E%7B%5Cfrac%7B%5Cgamma%20%2B1%7D%7B2%28%5Cgamma%20-1%29%7D%7D%20%20%28%5Cfrac%7B1%2B%5Cfrac%7B%5Cgamma%20-1%7D%7B2%7D%20M%2A%5E2%7D%7BM%2A%7D%29%5E%7B%5Cfrac%7B%5Cgamma%20%2B1%7D%7B2%28%5Cgamma%20-1%29%7D%7D)
Here:
is the specific heat ratio. Substitute the values in above equation.
![\frac{A}{3.14}= (\frac{1.4-1}{2})^{-\frac{1.4+1}{2(1.4 -1)}} (\frac{1+\frac{1.4-1}{2} (1.4)^2}{1.4})^{\frac{1.4+1}{2(1.4-1)}} \\\\A=\frac{\pi}{4}d^2 \\\\10.99=\frac{\pi}{4}d^2 \\\\d = 3.74 \ cm](https://tex.z-dn.net/?f=%5Cfrac%7BA%7D%7B3.14%7D%3D%20%28%5Cfrac%7B1.4-1%7D%7B2%7D%29%5E%7B-%5Cfrac%7B1.4%2B1%7D%7B2%281.4%20-1%29%7D%7D%20%20%28%5Cfrac%7B1%2B%5Cfrac%7B1.4-1%7D%7B2%7D%20%281.4%29%5E2%7D%7B1.4%7D%29%5E%7B%5Cfrac%7B1.4%2B1%7D%7B2%281.4-1%29%7D%7D%20%5C%5C%5C%5CA%3D%5Cfrac%7B%5Cpi%7D%7B4%7Dd%5E2%20%5C%5C%5C%5C10.99%3D%5Cfrac%7B%5Cpi%7D%7B4%7Dd%5E2%20%5C%5C%5C%5Cd%20%3D%203.74%20%5C%20cm)
exit diameter is 3.74 cm
In point b:
Calculate the temperature at throat.
![\frac{T*}{T}=(1+\frac{\Gamma-1}{2} M*^2)^{-1}\\\\\frac{T*}{573}=(1+\frac{1.4-1}{2} (1.4)^2)^{-1}\\\\T*=411.41 \ K](https://tex.z-dn.net/?f=%5Cfrac%7BT%2A%7D%7BT%7D%3D%281%2B%5Cfrac%7B%5CGamma-1%7D%7B2%7D%20M%2A%5E2%29%5E%7B-1%7D%5C%5C%5C%5C%5Cfrac%7BT%2A%7D%7B573%7D%3D%281%2B%5Cfrac%7B1.4-1%7D%7B2%7D%20%281.4%29%5E2%29%5E%7B-1%7D%5C%5C%5C%5CT%2A%3D411.41%20%5C%20K)
Calculate the velocity at exit.
Here: R is the gas constant.
![V*=1.4 \times \sqrt{1.4 \times 287 \times 411.41}\\\\=569.21 \ \frac{m}{s}](https://tex.z-dn.net/?f=V%2A%3D1.4%20%5Ctimes%20%5Csqrt%7B1.4%20%5Ctimes%20287%20%5Ctimes%20411.41%7D%5C%5C%5C%5C%3D569.21%20%5C%20%5Cfrac%7Bm%7D%7Bs%7D)
Calculate the density of air at inlet
![\rho_1 =\frac{P_1}{RT_1}\\\\=\frac{400}{ 0.287 \times 573}\\\\=2.43\ \frac{kg}{m^3}](https://tex.z-dn.net/?f=%5Crho_1%20%3D%5Cfrac%7BP_1%7D%7BRT_1%7D%5C%5C%5C%5C%3D%5Cfrac%7B400%7D%7B%200.287%20%5Ctimes%20573%7D%5C%5C%5C%5C%3D2.43%5C%20%20%5Cfrac%7Bkg%7D%7Bm%5E3%7D)
Calculate the density of air at throat using isentropic flow equation.
![\frac{\rho}{\rho_1}=(1+\frac{\Gamma -1}{2} M*^2)^{-\frac{1}{\Gamma -1}} \\\\\frac{\rho *}{2.43}=(1+\frac{1.4-1}{2} (1.4)*^2)^{-\frac{1}{1.4-1}} \\\\\rho*= 1.045 \ \frac{kg}{m^3}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Crho%7D%7B%5Crho_1%7D%3D%281%2B%5Cfrac%7B%5CGamma%20-1%7D%7B2%7D%20M%2A%5E2%29%5E%7B-%5Cfrac%7B1%7D%7B%5CGamma%20-1%7D%7D%20%5C%5C%5C%5C%5Cfrac%7B%5Crho%20%2A%7D%7B2.43%7D%3D%281%2B%5Cfrac%7B1.4-1%7D%7B2%7D%20%281.4%29%2A%5E2%29%5E%7B-%5Cfrac%7B1%7D%7B1.4-1%7D%7D%20%5C%5C%5C%5C%5Crho%2A%3D%201.045%20%5C%20%5Cfrac%7Bkg%7D%7Bm%5E3%7D)
Calculate the mass flow rate.
![m= \rho* \times A* \times V*\\\\= 1.045 \times 3.14 times 10^{-4} \times 569.21\\\\= 0.186 \frac{kg}{s}](https://tex.z-dn.net/?f=m%3D%20%5Crho%2A%20%5Ctimes%20A%2A%20%5Ctimes%20V%2A%5C%5C%5C%5C%3D%201.045%20%5Ctimes%203.14%20times%2010%5E%7B-4%7D%20%5Ctimes%20569.21%5C%5C%5C%5C%3D%200.186%20%5Cfrac%7Bkg%7D%7Bs%7D)