Air enters the compressor of an ideal gas refrigeration cycle at 7∘C and 35 kPa and the turbine at 37∘C and 160 kPa. The mass fl
ow rate of air through the cycle is 0.2 kg/s. Assuming variable specific heats for air, determine
(a) the rate of refrigeration, (b) the net power input, and (c) the coefficient of performance.
(a) the rate of refrigeration, (b) the net power input, and (c) the coefficient of performance.
1 answer:
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Answer:
The inside temperature, is approximately 248 °C.
Explanation:
The parameters given are;
Temperature of the air = 20°C
Carbon steel surface temperature 250°C
Area of surface = 50 cm × 75 cm = 0.5 × 0.75 = 0.375 m²
Convection heat transfer coefficient = 25 W/(m²·K)
Heat lost by radiation = 300 W
Assumption,
Air temperature = 20 °C
Hot plate temperature = 250 °C
Thermal conductivity K = 65.2 W/(m·K)
Steady state heat transfer process
One dimensional heat conduction
We have;
Newton's law of cooling;
q = h×A×( -
) + Heat loss by radiation
= 25×0.325×(250 - 20) + 300
= 2456.25 W
The rate of energy transfer per second is given by the following relation;
Thermal conductivity K = 65.2 W/(m·K)
Therefore;
The inside temperature, = 247.99 °C ≈ 248 °C.
Answer:
S = 0.5 km
velocity of motorist = 42.857 km/h
Explanation:
given data
speed = 70 km/h
accelerates uniformly = 90 km/h
time = 8 s
overtakes motorist = 42 s
solution
we know initial velocity u1 of police = 0
final velocity u2 = 90 km/h = 25 mps
we apply here equation of motion
u2 = u1 + at
so acceleration a will be
a =
a = 3.125 m/s²
so
distance will be
S1 = 0.5 × a × t²
S1 = 100 m = 0.1 km
and
S2 = u2 × t
S2 = 25 × 16
S2 = 400 m = 0.4 km
so total distance travel by police
S = S1 + S2
S = 0.1 + 0.4
S = 0.5 km
and
when motorist travel with uniform velocity
than total time = 42 s
so velocity of motorist will be
velocity of motorist =
velocity of motorist =
velocity of motorist = 42.857 km/h
Answer: see attached file for the answer
