Answer:
im good thanks
Explanation:
good luck with your child
Answer:
Find attach the solution
Explanation:
Entropy production is almost the same in (b) and (c) as 
~ 
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:
119.35 mm
Explanation:
Given:
Inside diameter, d = 100 mm
Tensile load, P = 400 kN
Stress = 120 MPa
let the outside diameter be 'D'
Now,
Stress is given as:
stress = Load × Area
also,
Area of hollow pipe = 
or
Area of hollow pipe = 
thus,
400 × 10³ N = 120 ×
or
D² = tex]\frac{400\times10^3+30\pi\times10^4}{30\pi}[/tex]
or
D = 119.35 mm
