Carbon dioxide enters an adiabatic compressor at 100 kPa and 300K at a rate of 0.5 kg/s and leaves at 600 kPa and 450K. Neglecti
ng kinetic energy changes, determine a) the volume flow rate of the carbon dioxide at the compressor inlet (Ans. around 0.3 m3/s) and b) the power input to the compressor (Ans. around 70 kW).
2 answers:
Answer:
The answers to the question are
a) The volume flow rate of the carbon dioxide at the compressor inlet is 0.2834 m³/s ≈ 0.3 m³/s
b) The power input to the compressor is 73.35 kW ≈ 70 kW
Explanation:
We note the following
Mass flow rate = 0.5 kg/s
Inlet pressure = 100 pKa
Outlet pressure = 600 kPa
Inlet temperature = 300 K
Outlet temperature = 450 K
Molar mass of CO₂ = 44.01 g/mol
R Universal Gas Constant = 8.314 4621. J K−1 mol−1
a) Number of moles = =
= 11.361 moles
P·V= n·R·T ∴ V = =
= 0.2834 m³
Therefore the volume flow rate = 0.2834 m³/s ≈ 0.3 m³/s
b) Cp at 300 K = 0.846 kJ/(kg K)
Cp at 600 K = 0.978 kJ/(kg K)
Cv = 0.657
K = 1.289
While the power input to the compressor can be calculated by
m'×Cp×(T₂-T₁)
Where m' = mass flow rate = 0.5 kg/s
Therefore power = 0.5 kg/s×0.978 kJ/(kg K)×(450 K - 300 K)
= 73.35 kJ/s = 73.35 kW ≈ 70 kW
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Answer:
a) ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J, d) W = 150 10⁻¹² J
Explanation:
Let's find the capacitance of the capacitor
C =
C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³
C = 2.62 10⁻¹² F
for the initial data let's look for the accumulated charge on the plates
C =
Q₀ = C ΔV
Q₀ = 2.62 10⁻¹² 8.70
Q₀ = 22.8 10⁻¹² C
a) we look for the capacity for the new distance
C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³
C₁ = 1.04 10⁻¹² F
C₁ = Q₀ / ΔV₁
ΔV₁ = Q₀ / C₁
ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²
ΔV₁ = 21.9 V
b) initial stored energy
U₀ =
U₀ = (22.8 10⁻¹²)²/(2 2.62 10⁻¹²)
U₀ = 99.2 10⁻¹² J
c) final stored energy
U_f = (22.8 10⁻¹²) ² /(2 1.04 10⁻⁻¹²)
U_f = 249.9 10⁻¹² J
d) the work of separating the plates
as energy is conserved work must be equal to energy change
W = U_f - U₀
W = (249.2 - 99.2) 10⁻¹²
W = 150 10⁻¹² J
note that as the energy increases the work must be supplied to the system
Given data
The angle of inclination of the plane is theta = 30 degree
The applied force in the inclined plane is F = 94 N
The distance moved in the inclined plane is d = 2.30 m
The coefficient of kinetic friction is u_k = 0.280
The free-body diagram of the above configuration is shown below:
Here, the normal reaction force on the box is N, the acceleration due to gravity is denoted as g, the friction force on the box is F_f, and the mass of the box is denoted as m.
(a)
The expression for the work done by the pushing force is given as:
Substitute the value in the above equation.
Thus, the work done by the pushing force is 216.2 J.
(b)
The box is moving at the constant velocity, therefore, the pushing force will be equal to the frictional force and the component of the gravitational force in the inclined plane.
The expression for the normal reaction force is given as:
The expression for the mass of the box is given as:
Substitute the value in the above equation.
Thus, the mass of the box is 12.9 kg.
