Refrigerant-134a enters a 28-cm-diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. The refrigerant gains heat
1 answer:
Answer:
V = 0.30787 m³/s
m = 2.6963 kg/s
v2 = 0.3705 m³/s
v2 = 6.017 m/s
Explanation:
given data
diameter = 28 cm
steadily =200 kPa
temperature = 20°C
velocity = 5 m/s
solution
we know mass flow rate is
m = ρ A v
floe rate V = Av
m = ρ V
flow rate = V =
V = Av =
V =
V = 0.30787 m³/s
and
mass flow rate of the refrigerant is
m = ρ A v
m = ρ V
m = =
m = 2.6963 kg/s
and
velocity and volume flow rate at exit
velocity = mass × v
v2 = 2.6963 × 0.13741 = 0.3705 m³/s
and
v2 = A2×v2
v2 =
v2 =
v2 = 6.017 m/s
You might be interested in
Answer:
t = 59.37 s
Explanation:
Given data:
thermal diffusivity
theraml conductivity = k = 22 W/m.K
h = 300 W/ m^2.K
= 25 degree C = 298 k
= 60 degree C = 333 k
= 75 degree C = 348 L
diameter d = 0.1 m
characteristics length Lc = r/3 = = 0.0166
at Bi = 0.226
Ai = 0.982
-1.187 = - 0.02t
t = 59.37 s
Answer:
2)
a) to the right of the dipole E_total = kq [1 / (r + a)² - 1 / r²]
b)To the left of the dipole E_total = - k q [1 / r² - 1 / (r + a)²]
c) at a point between the dipole, that is -a <x <a
E_total = kq [1 / x² + 1 / (2a-x)²]
d) on the vertical line at the midpoint of the dipole (x = 0)
E_toal = 2 kq 1 / (a + y)² cos θ
Explanation:
2) they ask us for the electric field in different positions between the dipole and a point of interest. Using the principle of superposition.
This principle states that we can analyze the field created by each charge separately and add its value and this will be the field at that point
Let's analyze each point separately.
The test charge is a positive charge and in the reference frame it is at the midpoint between the two charges.
a) to the right of the dipole
The electric charge creates an outgoing field, to the right, but as it is further away the field is of less intensity
E₊ = k q / (r + a)²
where 2a is the distance between the charges of the dipole and the field is to the right
the negative charge creates an incoming field of magnitude
E₋ = -k q / r²
The field is to the left
therefore the total field is the sum of these two fields
E_total = E₊ + E₋
E_total = kq [1 / (r + a)² - 1 / r²]
we can see that the field to the right of the dipole is incoming and of magnitude more similar to the field of the negative charge as the distance increases.
b) To the left of the dipole
The result is similar to the previous one by the opposite sign, since the closest charge is the positive one
E₊ is to the left and E₋ is to the right
E_total = - k q [1 / r² - 1 / (r + a)²]
We see that this field is also directed to the left
c) at a point between the dipole, that is -a <x <a
In this case the E₊ field points to the right and the E₋ field points to the right
E₊ = k q 1 / x²
E₋ = k q 1 / (2a-x)²
E_total = kq [1 / x² + 1 / (2a-x)²]
in this case the field points to the right
d) on the vertical line at the midpoint of the dipole (x = 0)
In this case the E₊ field points in the direction of the positive charge and the test charge
in E₋ field the ni is between the test charge and the negative charge,
the resultant of a horizontal field in zirconium on the x axis (where the negative charge is)
E₊ = kq 1 / (a + y) 2
E₋ = kp 1 / (a + y) 2
E_total = E₊ₓ + E_{-x}
E_toal = 2 kq 1 / (a + y)² cos θ
e) same as the previous part, but on the negative side
E_toal = 2 kq 1 / (a + y)² cos θ
When analyzing the previous answer there is no point where the field is zero
The different configurations are outlined in the attached
3) We are asked to repeat part 2 changing the negative charge for a positive one, so in this case the two charges are positive
a) to the right
in this case the two field goes to the right
E_total = kq [1 / (r + a)² + 1 / r²]
b) to the left
E_total = - kq [1 / (r + a)² + 1 / r²]
c) between the two charges
E₊ goes to the right
E₋ goes to the left
E_total = kq [1 / x² - 1 / (2a-x)²]
d) between vertical line at x = 0
E₊ salient between test charge and positive charge
E_total = 2 kq 1 / (a + y)² sin θ
In this configuration at the point between the two charges the field is zero
