Question

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 1008C. Air enters the heating section at 108C and 70 percent relative humidity at a rate of 35 m3/min, and it leaves the humidifying section at 208C 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 1

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

Step-by-step 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 2

T₂ = 19.5 °C

∅₂ = 37.8%

Q = 420.01 kJ/min

m
`_(w)` = 0.015 kg/min

Step-by-step explanation:

Given:

total pressure = p = 1 atm

Temperatures:

T₁ = 10°C

T₃ = 20°C

Relative Humidity:

∅₁ = 70%

∅₃ = 60%

Volume = V₁ = 35 m³/min

Solution:

a) Use psychometric chart to determine the enthalpies by using the given values of temperatures and relative humidity:

h₁ = 23.5 kJ/kg

w₁ = 0.0053

w₁ = w₂

h₃ = 42.3 kJ/kg

w₃ = 0.0087

v₁ = 0.809 m³/kg

When the airs flows through the heating section, the amount of moisture in it remains constant. So

w₁ = w₂

When the airs flows through the humidifying section, the amount of moisture in it increases. So

w₃ > w₂

Compute enthalpy h₂

h₂ = h₃ - ( w₃ - w₂) hs

where hs is the enthalpy of wet steam at 100°C from steam table

hs = 2676 kJ/kg

h₂ = 42.3 - ( 0.0087 - 0.0053 ) 2676

= 42.3 - (0.0034) 2676

= 42.3 - 9.0984

= 33.2016

h₂ = 33.2 kJ/kg

Compute T₂ and ∅₂

Using h₂ = 33.2 kJ/kg and w₂ = 0.0053 and psychometric chart:

T₂ = 19.5 °C

∅₂ = 37.8%

Compute mass flow rate:

mass flow rate = m = V₁ /v₁

= 35/0.809

= 43.26 kg/min

m = 43.3 kg/min

b) Compute heat transfer in the heating section:

Q = m x (h₂ - h₁)

= 43.3 kg/min ( 33.2 kJ/kg - 23.5 kJ/kg )

= 43.3 ( 9.7 )

Q = 420.01 kJ/min

c) Compute rate at which water is added to the air in the humidifying section

Let m
`_(w)` be the mass flow rate equation of water in humidifying section is:

m
`_(w)` = m(w₃ - w₂)

= 43.3 ( 0.0087 - 0.0053 )

= 43.3(0.0034)

= 0.14722

= 0.147 kg/min

m
`_(w)` = 0.015 kg/min