Snap rings, and bearings can be used to keep a gear on a shaft, hope this helps!!
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 %
Examples of quality assurance activities include process checklists, process standards, process documentation and project audit. Examples of quality control activities include inspection, deliverable peer reviews and the software testing process. You may like to read more about the quality assurance vs quality control.
Answer:
The given blanks can be filled as given below
Voltmeter must be connected in parallel
Explanation:
A voltmeter is connected in parallel to measure the voltage drop across a resistor this is because in parallel connection, current is divided in each parallel branch and voltage remains same in parallel connections.
Therefore, in order to measure the same voltage across the voltmeter as that of the voltage drop across resistor, voltmeter must be connected in parallel.
Answer:
T=151 K, U=-1.848*10^6J
Explanation:
The given process occurs when the pressure is constant. Given gas follows the Ideal Gas Law:
pV=nRT
For the given scenario, we operate with the amount of the gas- n- calculated in moles. To find n, we use molar mass: M=102 g/mol.
Using the given mass m, molar mass M, we can get the following equation:
pV=mRT/M
To calculate change in the internal energy, we need to know initial and final temperatures. We can calculate both temperatures as:
T=pVM/(Rm); so initial T=302.61K and final T=151.289K
Now we can calculate change of U:
U=3/2 mRT/M using T- difference in temperatures
U=-1.848*10^6 J
Note, that the energy was taken away from the system.
