Answer:
the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %
Explanation:
given data
pressure p1 = 1.4 MPa = 14 bar
temperature t1 = 32°C
exit pressure = 0.08 MPa = 0.8 bar
to find out
the quality of the refrigerant exiting the expansion valve
solution
we know here refrigerant undergoes at throtting process so
h1 = h2
so by table A 14 at p1 = 14 bar
t1 ≤ Tsat
so we use equation here that is
h1 = hf(t1) = 332.17 kJ/kg
this value we get from table A13
so as h1 = h2
h1 = h(f2) + x(2) * h(fg2)
so
exit quality = 
exit quality = 
so exit quality = 0.2337 = 23.37 %
the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %
Answer:
The appropriate solution is "1481.76 N".
Explanation:
According to the question,
Mass,
m = 540 kg
Coefficient of static friction,
= 0.28
Now,
The applied force will be:
⇒ 
By substituting the values, we get
C is your answers!!!!!$3&2)//
Answer:
fracture will occur as the value is less than E/10 (= 22.5)
Explanation:
If the maximum strength at tip Is greater than theoretical fracture strength value then fracture will occur and if the maximum strength is lower than theoretical fracture strength then no fracture will occur.
![\sigma_m = 2\sigma_o [\frac{a}{\rho_t}]^{1/2}](https://tex.z-dn.net/?f=%5Csigma_m%20%3D%202%5Csigma_o%20%5B%5Cfrac%7Ba%7D%7B%5Crho_t%7D%5D%5E%7B1%2F2%7D)
= 15 GPa
fracture will occur as the value is less than E/10 = 22.5
Answer:
Explanation:
Given that
Lattice constant = a
Radius of unit cell cell =R
Atom is in BCC structure.
In BCC unit cell (Body centered cube)
1.Eight atoms at eight corner of cube which have 1/8 part in each cube.
2.One complete atom at the body center of the cube
So the total number of atoms in the BCC
Z= 1/8 x 8 + 1 x 1
Z=2
In triangle ABD
In triangle ABC
AC=4R
BC=a
So
So the relationship between lattice constant and radius of unit cell
