Explanation:
angular velocity is given by
w = 0.626
now tangential velocity is
V = rw
= 25 x 0.626
= 15.65 m/s
Refrigerant-134a enters a compressor at 180 kPa as a saturated vapor with a flow rate of 0.35 m3/min and leaves at 900 kPa. The
power supplied to the refrigerant during the compression process is 2.35 kW. What is the temperature of R-134a at the exit of the compressor
1 answer:
Answer:
52.5°C
Explanation:
The final enthalpy is determined from energy balance where initial enthalpy and specific volume are obtained from A-12 for the given pressure and state
mh1 + W = mh2
h2 = h1 + W/m
h1 + Wα1/V1
242.9 kJ/kg + 2.35.0.11049kJ/ 0.35/60kg
=287.4 kJ/kg
From the final enthalpy and pressure the final temperature is obtained A-13 using interpolation
i.e T2 = T1 + T2 -T1/h2 -h1(h2 - h1)
= 50°C + 60 - 50/295.15 - 284.79
(287.4 - 284.79)°C
= 52.5°C
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Answer:
μ = 0.125
Explanation:
To solve this problem, which is generally asked for the coefficient of friction, we will use the conservation of energy.
Let's start working on the ramp
starting point. Highest point of the ramp
Em₀ = U = m h y
final point. Lower part of the ramp, before entering the rough surface
= K = ½ m v²
as they indicate that there is no friction on the ramp
Em₀ = Em_{f}
m g y = ½ m v²
v =
we calculate
v = √(2 9.8 0.25)
v = 2.21 m / s
in the rough part we use the relationship between work and kinetic energy
W = ΔK = K_{f} -K₀
as it stops the final kinetic energy is zero
W = -K₀
The work is done by the friction force, which opposes the movement
W = - fr x
friction force has the expression
fr = μ N
let's write Newton's second law for the vertical axis
N-W = 0
N = W = m g
we substitute
-μ m g x = - ½ m v²
μ =
Let's calculate
μ =
μ = 0.125