Question

$100m^3$ of air per minute at ${15}^{\circ }C$ DBT and 80% RH is heated sensibly until its temperature becomes ${22}^{\circ }C$. The saturation pressure at ${22}^{\circ }C$ and ${15}^{\circ }C$ are 0.02645 bar and 0.017 bar. The atmosphere pressure is 1.013 bar. What is the heat added to air per minute?

• A. 14.4 kJ/min
• B. 875.8 kJ/min
• C. 864.08 kJ/min
• D. 1459 kJ/min

Answer 1

The correct option is C 864.08 kJ/min

$\text{Heat added}=m_a(h_2-h_1)$

So, $h_1=1.005\times t_1+{\omega }_1(2500+1.88t_1)$.....(1)

$\varphi =\frac{P_{v1}}{P_{vs}}$

$0.8=\frac{P_{v1}}{0.017}$

$P_{v1}=0.0136bar$

${\omega }_1=0.622\frac{P_{v1}}{P_t-P_{v1}}=\frac{0.622\times 0.0136}{(1.013-0.0136)}=8.4642\times {10}^{-3}kg/kgda$

So, equation, (1)
$h_1=1.005\times 15+8.4642\times {10}^{-3}(2500+1.88\times 15)$

$h_1=36.474kJ/kgda$

For sensible heating specific humidity is same So, $P_v$ is same.

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

$at{22}^{\circ }C=h_2=1.005\times 22+8.4642\times {10}^{-3}(2500+1.88\times 22)$

$h_2=43.6205kJ/kgda$

$P_1{V}_1=maR{T}_1$

$m_a=\frac{(1.013-0.0136)\times 100\times 100}{0.287\times 288}$

$m_a=120.91kg/min$

So, Heat added = $120.91\times (43.6205-36.474)$

$864.083kJ/min$