Air is cooled and dehumidified as it flows over the coils of refrigeration system at 100 kPa from 30 ºC and relative humidityof
84%to 15 ºC and relative humidity of 100%. The mass flow rate of dry air is 0.4 kg/s.
Using the formulaand not the Psychrometric chart, determine:
a) the mass flow rate of water,
b)the heat removal from the air.
Using the formulaand not the Psychrometric chart, determine:
a) the mass flow rate of water,
b)the heat removal from the air.
1 answer:
Answer:
(a) The mass glow rate of water is 0.0032kg/s
(b) The heat removal from air is 14.17KJ/sec
Explanation:
In this question, we are asked to use the formula to calculate the mass flow rate and the heat removal from water based on the data in the question.
To answer this question, we take values of absolute humidity and enthalpy values from table from corresponding temperature
Please check attachment for complete solution and step by step explanation
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Answer:
0.00650 Ib s /ft^2
Explanation:
diameter ( D ) = 0.71 inches = 0.0591 ft
velocity = 0.90 ft/s ( V )
fluid specific gravity = 0.96 (62.4 ) ( x )
change in pressure ( P ) = 0 because pressure was constant
viscosity = (change in p - X sin∅ ) / 32 V
= ( 0 - 0.96( 62.4) sin -90 ) * 0.0591 ^2 / 32 * 0.90
= - 59.904 sin (-90) * 0.0035 / 28.8
= 0.1874 / 28.8
viscosity = 0.00650 Ib s /ft^2
Answer:
The force induced on the aircraft is 2.60 N
Solution:
As per the question:
Power transmitted,
Now, the force, F is given by:
(1)
where
v = velocity
Now,
For a geo-stationary satellite, the centripetal force, is provided by the gravitational force,
:
Thus from the above, velocity comes out to be:
where
R =
R =
where
G = Gravitational constant
T = Time period of rotation of Earth
R is calculated as 42166 km
Now, from eqn (1):
F = 2.60 N