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 %
If a controlled input can transfer (alter) the control system's initial states to some other desired states in a finite amount of time, the control system is said to be controllable.
Using Kalman's test, we can determine whether a control system is controllable. The evolution model for the state variables (time-varying unknowns) and the observation model, which connects the observations to the state variables, make up the state space representation of a dynamical system. The capacity to move a system about in its full configuration space using just specific permitted actions is generally referred to as controllability. The precise definition changes slightly depending on the model type or framework used.
Learn more about control here-
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Answer:
work=281.4KJ/kg
Power=4Kw
Explanation:
Hi!
To solve follow the steps below!
1. Find the density of the air at the entrance using the equation for ideal gases
where
P=pressure=120kPa
T=20C=293k
R= 0.287 kJ/(kg*K)=
gas constant ideal for air
2.find the mass flow by finding the product between the flow rate and the density
m=(density)(flow rate)
flow rate=10L/s=0.01m^3/s
m=(1.43kg/m^3)(0.01m^3/s)=0.0143kg/s
3. Please use the equation the first law of thermodynamics that states that the energy that enters is the same as the one that must come out, we infer the following equation, note = remember that power is the product of work and mass flow
Work
w=Cp(T1-T2)
Where
Cp= specific heat for air=1.005KJ/kgK
w=work
T1=inlet temperature=20C
T2=outlet temperature=300C
w=1.005(300-20)=281.4KJ/kg
Power
W=mw
W=(0.0143)(281.4KJ/kg)=4Kw
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