Answer:
![\dot W_{out} = 399.47\,kW](https://tex.z-dn.net/?f=%5Cdot%20W_%7Bout%7D%20%3D%20399.47%5C%2CkW)
Explanation:
The turbine is modelled after the First Law of Thermodynamics:
![-\dot Q_{out} -\dot W_{out} + \dot m\cdot (h_{in}-h_{out}) = 0](https://tex.z-dn.net/?f=-%5Cdot%20Q_%7Bout%7D%20-%5Cdot%20W_%7Bout%7D%20%2B%20%5Cdot%20m%5Ccdot%20%28h_%7Bin%7D-h_%7Bout%7D%29%20%3D%200)
The work done by the turbine is:
![\dot W_{out} = \dot m \cdot (h_{in}-h_{out})-\dot Q_{out}](https://tex.z-dn.net/?f=%5Cdot%20W_%7Bout%7D%20%3D%20%5Cdot%20m%20%5Ccdot%20%28h_%7Bin%7D-h_%7Bout%7D%29-%5Cdot%20Q_%7Bout%7D)
The properties of the water are obtained from property tables:
Inlet (Superheated Steam)
![P = 10\,MPa](https://tex.z-dn.net/?f=P%20%3D%2010%5C%2CMPa)
![T = 520\,^{\textdegree}C](https://tex.z-dn.net/?f=T%20%3D%20520%5C%2C%5E%7B%5Ctextdegree%7DC)
![h = 3425.9\,\frac{kJ}{kg}](https://tex.z-dn.net/?f=h%20%3D%203425.9%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D)
Outlet (Superheated Steam)
![P = 1\,MPa](https://tex.z-dn.net/?f=P%20%3D%201%5C%2CMPa)
![T = 280\,^{\textdegree}C](https://tex.z-dn.net/?f=T%20%3D%20280%5C%2C%5E%7B%5Ctextdegree%7DC)
![h = 3008.2\,\frac{kJ}{kg}](https://tex.z-dn.net/?f=h%20%3D%203008.2%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D)
The work output is:
![\dot W_{out} = \left(1.1\,\frac{kg}{s}\right)\cdot \left(3425.9\,\frac{kJ}{kg} -3008.2\,\frac{kJ}{kg}\right) - 60\,kW](https://tex.z-dn.net/?f=%5Cdot%20W_%7Bout%7D%20%3D%20%5Cleft%281.1%5C%2C%5Cfrac%7Bkg%7D%7Bs%7D%5Cright%29%5Ccdot%20%5Cleft%283425.9%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D%20-3008.2%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D%5Cright%29%20-%2060%5C%2CkW)
![\dot W_{out} = 399.47\,kW](https://tex.z-dn.net/?f=%5Cdot%20W_%7Bout%7D%20%3D%20399.47%5C%2CkW)
Answer:
0.5m^2/Vs and 0.14m^2/Vs
Explanation:
To calculate the mobility of electron and mobility of hole for gallium antimonide we have,
(S)
Where
e= charge of electron
n= number of electrons
p= number of holes
mobility of electron
mobility of holes
electrical conductivity
Making the substitution in (S)
Mobility of electron
![8.9*10^4=(8.7*10^{23}*(-1.602*10^{-19})*\mu_e)+(8.7*10^{23}*(-1.602*10^{-19})*\mu_h)](https://tex.z-dn.net/?f=8.9%2A10%5E4%3D%288.7%2A10%5E%7B23%7D%2A%28-1.602%2A10%5E%7B-19%7D%29%2A%5Cmu_e%29%2B%288.7%2A10%5E%7B23%7D%2A%28-1.602%2A10%5E%7B-19%7D%29%2A%5Cmu_h%29)
![0.639=\mu_e+\mu_h](https://tex.z-dn.net/?f=0.639%3D%5Cmu_e%2B%5Cmu_h)
Mobility of hole in (S)
![2.3*10^5 = (7.6*10^{22}*(-1.602*10^{-19})*\mu_e)+(1*10^{25}*(-1.602*10^{-19}*\mu_h))](https://tex.z-dn.net/?f=2.3%2A10%5E5%20%3D%20%287.6%2A10%5E%7B22%7D%2A%28-1.602%2A10%5E%7B-19%7D%29%2A%5Cmu_e%29%2B%281%2A10%5E%7B25%7D%2A%28-1.602%2A10%5E%7B-19%7D%2A%5Cmu_h%29%29)
![0.1436 = 7.6*10^{-3}\mu_e+\mu_h](https://tex.z-dn.net/?f=0.1436%20%3D%207.6%2A10%5E%7B-3%7D%5Cmu_e%2B%5Cmu_h)
Then, solving the equation:
(1)
(2)
We have,
Mobility of electron ![\mu_e = 0.5m^2/V.s](https://tex.z-dn.net/?f=%5Cmu_e%20%3D%200.5m%5E2%2FV.s)
Mobility of hole is ![\mu_h = 0.14m^2/V.s](https://tex.z-dn.net/?f=%5Cmu_h%20%3D%200.14m%5E2%2FV.s)
Answer:
minimum flow rate provided by pump is 0.02513 m^3/s
Explanation:
Given data:
Exit velocity of nozzle = 20m/s
Exit diameter = 40 mm
We know that flow rate Q is given as
![Q = A \times V](https://tex.z-dn.net/?f=Q%20%3D%20A%20%5Ctimes%20V)
where A is Area
![A =\frac{\pi}{4} \times (40\times 10^{-3})^2 = 1.256\times 10^{-3} m^2](https://tex.z-dn.net/?f=A%20%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%20%5Ctimes%20%2840%5Ctimes%2010%5E%7B-3%7D%29%5E2%20%3D%201.256%5Ctimes%2010%5E%7B-3%7D%20m%5E2)
![Q = 1.256\times 10^{-3} \times 20 = 0.02513 m^3/s](https://tex.z-dn.net/?f=Q%20%3D%201.256%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%2020%20%3D%200.02513%20m%5E3%2Fs)
minimum flow rate provided by pump is 0.02513 m^3/s
Answer:
The settlement that is expected is 1.043 meters.
Explanation:
Since the pre-consolidation stress of the layer is equal to the effective stress hence we conclude that the soil is normally consolidated soil
The settlement due to increase in the effective stress of a normally consolidated soil mass is given by the formula
![\Delta H=\frac{H_oC_c}{1+e_o}log(\frac{\bar{\sigma_o}+\Delta \bar{\sigma }}{\bar{\sigma_o}})](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Cfrac%7BH_oC_c%7D%7B1%2Be_o%7Dlog%28%5Cfrac%7B%5Cbar%7B%5Csigma_o%7D%2B%5CDelta%20%5Cbar%7B%5Csigma%20%7D%7D%7B%5Cbar%7B%5Csigma_o%7D%7D%29)
where
'H' is the initial depth of the layer
is the Compression index
is the inital void ratio
is the initial effective stress at the depth
is the change in the effective stress at the given depth
Applying the given values we get
![\Delta H=\frac{8\times 0.3}{1+0.87}log(\frac{154+28}{154})=1.04](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Cfrac%7B8%5Ctimes%200.3%7D%7B1%2B0.87%7Dlog%28%5Cfrac%7B154%2B28%7D%7B154%7D%29%3D1.04)
Answer:
diesel engine
Explanation:
because diesel is stronger than petrol