Answer:
X=0.194
T=-33.6C
Explanation:
Hello!
To solve this problem use the following steps!
1. We will call the expansion valve inlet 1 and exit 2
2.Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
3. Find the enthalpy of state 1 using pressure and temperature using thermodynamic tables
h1=Enthalpy(Ammonia;T=24C;P=1000kPa)=312.9KJ/kg
4. An expansion valve is a device which does not have heat or work exchange which means that the enthalpy of state one is equal to that of state 2, so using thermodynamic tables uses the pressure of state 2 and enthalpy of state 1 to find quality and temperature
x2=Quality(Ammonia;P=100kPa;h=h1=312.9KJ/kg)
=0.194
T2=Temperature(Ammonia;P=100kPa;h=h1=312.9KJ/kg)=-33.6C
Answer:
IT IS SIMPLE LIKE HENRY FORD MADE THIS OPTION POSSIBLE
Explanation:
Complete Question:
A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm if the tool moves at 400 mm per second.
Answer:
26 mins 40 secs
Explanation:
Reduction in depth, Δd = 20 mm
Depth of cut, 
Number of passes necessary for this reduction, 
n = 20/0.5
n = 40 passes
Tool width, w = 5 mm
Width of metal plate, W = 200 mm
For a reduction in the depth per pass, tool will travel W/w = 200/5 = 40 times
Speed of tool, v = 100 mm/s

minimum time required to reduce the depth of the plate by 20 mm:
number of passes * Time/pass
n * Time/pass
40 * 40
1600 = 26 mins 40 secs
Answer:
Work = 651,1011 kJ
Explanation:
Let´s take the car as a system in order to apply the first law of thermodynamics as follows:

Where

And considering that there is no mass transfer and that the only energy flows that interact with the system are the heat losses and the work needed to move the car we have:

Regarding the energy system we have the following:

By doing the calculations we have:
![E_{system,final}- E_{system,initial}=[0,1*900]_{internal}+[0,5*900(30^2-10^2)/1000)_{kinetic}+(900*10*(20-0)/1000)_{potential}\\E_{system,final}- E_{system,initial}=90+360+180=630kJ](https://tex.z-dn.net/?f=E_%7Bsystem%2Cfinal%7D-%20E_%7Bsystem%2Cinitial%7D%3D%5B0%2C1%2A900%5D_%7Binternal%7D%2B%5B0%2C5%2A900%2830%5E2-10%5E2%29%2F1000%29_%7Bkinetic%7D%2B%28900%2A10%2A%2820-0%29%2F1000%29_%7Bpotential%7D%5C%5CE_%7Bsystem%2Cfinal%7D-%20E_%7Bsystem%2Cinitial%7D%3D90%2B360%2B180%3D630kJ)
Consider that in the previous calculation, the kinetic and potential energy terms were divided by 1.000 to change the units from J to kJ.
Finally, the work needed to move the car under the required conditions is calculated as follows:

Consider that in the previous calculation, the heat loss was changed previously from BTU to kJ.