Question

Moist air enters a duct at ${10}^{\circ }C$, 80% relative humidity and mass flow rate of $150m^3/min.$ The mixture is heated (without addition or removal of moisture) as it flows through the duct and leaves at ${30}^{\circ }C$. Assuming the mixture pressure remains constant at 100 kPa, the rate of heat transfer is closest to

For water vapour $P_{sat}$ at ${10}^{\circ }C$ = 1.228 kPa

$P_{sat}$ at ${30}^{\circ }C$ = 4.246 kPa

take$(c_p{)}_{air}=1kJ/kgK$

• A. 3694 kJ/min
• B. 3611 kJ/min
• C. 3623 kJ/min
• D. 3712 kJ/min

Answer 1

The correct option is A 3694 kJ/min

${\varphi }_1=0.8=\frac{P_{V1}}{P_{Sat}}=\frac{P_{V1}}{1.228}\Rightarrow P_{V1}=0.9824kPa$

${\omega }_1=\frac{0.622\times P_{V1}}{100-P_{V1}}=\frac{0.622\times 0.9824}{100-0.9824}=0.00617kg/kgdryair$

$h_1=c_{pa}\times t_1+{\omega }_1[2500+1.88\times t_1]$

$=1\times 10+0.00617\times [2500+1.88\times 10]$

= 25.5 kJ/kg

${\omega }_1={\omega }_2$

$h_2=c_{pa}\times t_2+{\omega }_2[2500+1.88\times t_2]$

$=1\times 30+0.00617\times [2500+1.88\times 30]$

= 45.7 kJ/kg

The mass flow rate of dry air can be calculated by using perfect gas relation:

$\dot{m_a}=\frac{(100-0.9824)\times 150}{0.287\times 283}=182.87kg/min$

The total heat transfer to air in the duct is,

$\dot{Q}=\dot{m}\times (h_2-h_1)$

$=182.87\times (45.7-25.5)\approx 3694kJ/min$