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 elastic modulus of the steel is 139062.5 N/in^2
Explanation:
Elastic modulus = stress ÷ strain
Load = 89,000 N
Area of square cross section of the steel bar = (0.8 in)^2 = 0.64 in^2
Stress = load/area = 89,000/0.64 = 139.0625 N/in^2
Length of steel bar = 4 in
Extension = 4×10^-3 in
Strain = extension/length = 4×10^-3/4 = 1×10^-3
Elastic modulus = 139.0625 N/in^2 ÷ 1×10^-3 = 139062.5 N/in^2
Answer:
The Poisson's Ratio of the bar is 0.247
Explanation:
The Poisson's ratio is got by using the formula
Lateral strain / longitudinal strain
Lateral strain = elongation / original width (since we are given the change in width as a result of compession)
Lateral strain = 0.15mm / 40 mm =0.00375
Please note that strain is a dimensionless quantity, hence it has no unit.
The Longitudinal strain is the ratio of the elongation to the original length in the longitudinal direction.
Longitudinal strain = 4.1 mm / 270 mm = 0.015185
Hence, the Poisson's ratio of the bar is 0.00375/0.015185 = 0.247
The Poisson's Ratio of the bar is 0.247
Please note also that this quantity also does not have a dimension
Answer:
Feedback control of arm movements using Neuro-Muscular Electrical Stimulation (NMES) combined with a lockable, passive exoskeleton for gravity compensation
Answer:
Both come from the sun
Both are reusable sources
and both don't cause pollution
Explanation:
