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Correct question is;
An air-conditioning system operates at a total pressure of 1 atm and consists of a heating section and a humidifier that supplies wet steam (saturated water vapor) at 100°C. Air enters the heating section at 10°C and 70 percent relative humidity at a rate of 35 m3/min, and it leaves the humidifying section at 20°C and 60 percent relative humidity. Determine (a) the temperature and relative humidity of air when it leaves the heating section, (b) the rate of heat transfer in the heating section, and (c) the rate at which water is added to the air in the humidifying section.
Answer:
A) Temperature at state 2: T2 = 19.5°C and Relative Humidity at state 2: Φ2 = 37.8%
B) Q' = 420.01 KJ/min
C) m'w = 0.1472 Kg/min
Explanation:
This question depicts a steady state process and as such the mass flow rate of dry air will remain constant during the entire process.
A) Now, from the psychometric chart attached, at temperatures of T1 = 10°C & T3 = 20°C, Relative humidities; Φ2 = 70% & Φ2 = 60% and at pressure of 1 atm, we have;
Enthalpy at state 1;h1 = 23.5 Kj/kg dry air
Absolute humidity at state 1;ω1 = ω2 = 0.0053 kg of water per kg dry air
Enthalpy at state 3;h3 = 42.3 KJ/Kg dry air
Absolute humidity at state 3; ω3 = 0.0087 kg of water per Kg dry air
Specific volume at state 1;υ1 = 0.809 m³/kg
The formula for energy balance for the humidifying is given as;
h3 = h2 + hg(ω3 - ω2)
Where hg is the enthalpy of wet steam.
From second table attached, hg at 100°C is 2675.6 KJ/kg
Thus;
Making h2 the subject, we have;
h2 = h3 + hg(ω2 - ω3)
Plugging in the relevant values we have;
h2 = 42.3 + 2675.6(0.0053 - 0.0087)
h2 = 33.2 KJ/kg
Still using the psychrometric chart attached at ω2 = 0.0053 and h2 = 33.2 KJ/kg, we have;
Temperature at state 2: T2 = 19.5°C and Relative Humidity at state 2: Φ2 = 37.8%
B) to determine the rate of heat transfer, let's first find the mass flow rate first;
m' = V1'/υ1
Thus, m' = 35/0.809
m' = 43.3 kg/min
Thus, rate of heat transfer is given by;
Q' = m'(h2 - h1)
Plugging in the relevant values, gives;
Q' = 43.3(33.2 - 23.5)
Q' = 420.01 KJ/min
C) the rate at which water is added to the air in the humidifying section is given by;
m'w = m'( ω3 - ω2)
m'w = 43.3(0.0087 - 0.0053)
m'w = 0.1472 Kg/min
Answer:
22.90 × 10⁸ kg
Explanation:
Given:
Diameter, d = 0.02 m
ωₙ = 0.95 rad/sec
Time period, T = 0.35 sec
Now, we know
T=
where, L is the length of the steel cable
g is the acceleration due to gravity
0.35=
or
L = 0.0304 m
Now,
The stiffness, K is given as:
K =
Where, A is the area
E is the elastic modulus of the steel = 2 × 10¹¹ N/m²
or
K =
or
K = 20.66 × 10⁸ N
Also,
Natural frequency, ωₙ =
or
mass, m =
or
mass, m =
mass, m = 22.90 × 10⁸ kg