Answer:
a) 358.8 KJ/kg
b) 0.0977 KJ/K- kg
c) 83.28%
Explanation:
N2 at 300 k. ( use the properties of N2 at 300 k (T1) )
Cp = 1.04 KJ/kg-k , Cv = 0.743 KJ/Kgk , R = 0.1297 KJ/kgk , y = 1.4 ,
Given data:
T2 = 645 k
P1 = 1 bar , P2 = 10.5 bar
<u>a)Determine the work input in KJ/Kg of N2 flowing </u>
Winput = h2 - h1 = Cp( T2 - T1 ) = 1.04 ( 645 - 300 ) = 358.8 KJ/kg
<u>b) Determine the rate of entropy in KJ/K- kg of N2 flowing </u>
Rate of entropy ( Δs ) = Cp*InT2/T1 - R*In P2/P1
= 1.04 * In (645/300) - 0.1297 * In ( 10.5 / 1 )
= 0.0977 KJ/K- kg
<u>c) Determine isentropic compressor efficiency </u>
Isentropic compressor efficiency = 83.28%
calculated using the relation below
( h'2 - h1 ) / ( h2 - h1 ) = ( T'2 - T1 ) / ( T2 - T1 )
where T'2 = 587.314
To develop the problem it is necessary to apply the concepts related to the ideal gas law, mass flow rate and total enthalpy.
The gas ideal law is given as,
![PV=mRT](https://tex.z-dn.net/?f=PV%3DmRT)
Where,
P = Pressure
V = Volume
m = mass
R = Gas Constant
T = Temperature
Our data are given by
![T_1 = 38\°C](https://tex.z-dn.net/?f=T_1%20%3D%2038%5C%C2%B0C)
![T_2 = 14\°C](https://tex.z-dn.net/?f=T_2%20%3D%2014%5C%C2%B0C)
![\eta = 97\%](https://tex.z-dn.net/?f=%5Ceta%20%3D%2097%5C%25)
![\dot{v} = 510m^3/kg](https://tex.z-dn.net/?f=%5Cdot%7Bv%7D%20%3D%20510m%5E3%2Fkg)
Note that the pressure to 38°C is 0.06626 bar
PART A) Using the ideal gas equation to calculate the mass flow,
![PV = mRT](https://tex.z-dn.net/?f=PV%20%3D%20mRT)
![\dot{m} = \frac{PV}{RT}](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D%20%3D%20%5Cfrac%7BPV%7D%7BRT%7D)
![\dot{m} = \frac{0.6626*10^{5}*510}{287*311}](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D%20%3D%20%5Cfrac%7B0.6626%2A10%5E%7B5%7D%2A510%7D%7B287%2A311%7D)
![\dot{m} = 37.85kg/min](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D%20%3D%2037.85kg%2Fmin)
Therfore the mass flow rate at which water condenses, then
![\eta = \frac{\dot{m_v}}{\dot{m}}](https://tex.z-dn.net/?f=%5Ceta%20%3D%20%5Cfrac%7B%5Cdot%7Bm_v%7D%7D%7B%5Cdot%7Bm%7D%7D)
Re-arrange to find ![\dot{m_v}](https://tex.z-dn.net/?f=%5Cdot%7Bm_v%7D)
![\dot{m_v} = \eta*\dot{m}](https://tex.z-dn.net/?f=%5Cdot%7Bm_v%7D%20%3D%20%5Ceta%2A%5Cdot%7Bm%7D)
![\dot{m_v} = 0.97*37.85](https://tex.z-dn.net/?f=%5Cdot%7Bm_v%7D%20%3D%200.97%2A37.85)
![\dot{m_v} = 36.72 kg/min](https://tex.z-dn.net/?f=%5Cdot%7Bm_v%7D%20%3D%2036.72%20kg%2Fmin)
PART B) Enthalpy is given by definition as,
![H= H_a +H_v](https://tex.z-dn.net/?f=H%3D%20H_a%20%2BH_v)
Where,
= Enthalpy of dry air
= Enthalpy of water vapor
Replacing with our values we have that
![H=m*0.0291(38-25)+2500m_v](https://tex.z-dn.net/?f=H%3Dm%2A0.0291%2838-25%29%2B2500m_v)
![H = 37.85*0.0291(38-25)-2500*36.72](https://tex.z-dn.net/?f=H%20%3D%2037.85%2A0.0291%2838-25%29-2500%2A36.72)
![H = 91814.318kJ/min](https://tex.z-dn.net/?f=H%20%3D%2091814.318kJ%2Fmin)
In the conversion system 1 ton is equal to 210kJ / min
![H = 91814.318kJ/min(\frac{1ton}{210kJ/min})](https://tex.z-dn.net/?f=H%20%3D%2091814.318kJ%2Fmin%28%5Cfrac%7B1ton%7D%7B210kJ%2Fmin%7D%29)
![H = 437.2tons](https://tex.z-dn.net/?f=H%20%3D%20437.2tons)
The cooling requeriment in tons of cooling is 437.2.
Answer:
A total = 0.04 m²
Explanation:
given data
mass = 75 kg
area = 0.05 m²
pressure = 3 kPa
solution
pressure when human walking on ground
p =
.................1
p =
p = 14.715 kPa
and
when walk on snow got extra pressure
P total = p + p (extra) ................2
P total = 14.715 + 3 = 17.715 kPa
and
total pressure is also
p total = ![\frac{mg}{A}](https://tex.z-dn.net/?f=%5Cfrac%7Bmg%7D%7BA%7D)
A total =
A total = 0.04 m²
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
The answer is B - a way of extracting natural gas shale formations.
Hydraulic fracturing is “a way of extracting natural gas shale formations.”