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:
used for ordinary combustibles, such as wood, paper, some plastics, and textiles. This class of fire requires the heat-absorbing effects of water or the coating effects of certain dry chemicals.
Explanation:
The person that is correct based on the 2 statements from Tech A and Tech B is; Tech B
A mass flow sensor is defined as a sensor that is used to measure the mass flow rate of air entering a fuel-injected internal combustion engine and then sends a voltage that represents the airflow to the electronic control circuit.
However, for Tech A is incorrect and so the correct answer is that Tech B is right because his statement corresponds with the definition of mass flow sensor.
Read more about fuel injection engines at; brainly.com/question/4561445