Question

Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters a cooling and dehumidifying coil with an enthalpy of 85 kJ/kg of dry air and a humidity ratio of 19 grams/kg of dry air. The air leaves the coil with an enthalpy of 43 kJ/kg of dry air and a humidity ratio of 8 grams/kg of dry air. If the condensate water leaves the coil with an enthalpy of 67 kJ/kg, the required cooling capacity of the coil in kW is

• A. 75.0
• B. 123.8
• C. 128.2
• D. 159.0

Answer 1

The correct option is B 123.8

Given data:

Mass flow rate of dry air, (${\dot{m}}_a)=3kg/s$

$h_1=85\text{kJ/kg of dry air}$

${\omega }_1=19\text{gm/kg of dry air}$
= 0.019 kg/kg of dry air

$h_2=43\text{kJ/kg of dry air}$

${\omega }_2=8\text{gm/kg of dray air}$
= 0.008 kg/kg of dry air



Enthalpy of ondensate water leaves the cooling

$coil:h_{fg}=67kJ/kg$

${\omega }_1=\frac{{\dot{m}}_{v1}}{{\dot{m}}_a}$

or ${\dot{m}}_{v1}={\omega }_1{\dot{m}}_a$

$=0.019\times 3kg/s$

= 0.057 kg/s, mass of water vapour at intel

Similarly,

${\dot{m}}_{v2}={\omega }_2{\dot{m}}_a=0.008\times 3$
= 0.024 kg/s,

mass of water vapour at outlet Mass flow rate of condensate,

${\dot{m}}_w$ = mass of water vapour at inlet - mass of water vapour at outlet

${\dot{m}}_w={\dot{m}}_{m1}-{\dot{m}}_{v}2$

=0.057- 0.024

=0.033 kg/s

Cooling capacity of the coil,

$Q={\dot{m}}_a(h_1-h_2)-{\dot{m}}_w{h}_{fg}$

Where $\dot{m}(h_1-h_2)=$ sensible heat removed

${\dot{m}}_w{h}_{fg}=$ Latent heat removed

$\therefore Q=3(85-43)-0.033\times 67$
=123.79 kW

Alternatively

Mass flow rate of condensate,

${\dot{m}}_w={\dot{m}}_a({\omega }_1-{\omega }_2)=3\times (0.019-0.008$

= 0.033 kg/s

Applying energy balance equation, we get

${\dot{m}}_a{h}_1={\dot{m}}_a{h}_2+{\dot{m}}_w{h}_{fg}+Q$

$3\times 85=3\times 43+0.033\times 67+Q$

Q = 123.8 kW