The dry bulb temperature for sample of moisture is 37oC- while its relative humidity is 50Saturation pressure of water vapour at 37oC is 6-2795 kPa-

Φ=P_vP_vs P_v=0.5× P_vs=0.5× 6.2795 =3.13975 kPa Humidity ratio, ω =0.622P_vP_t P_v =0.622×313975101.325 3.13975=0.01989 kg/kgda Enthalpy of moist air, h=1 ...

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Standard XII

Chemistry

Internal Energy

The dry bulb ...

Question

The dry bulb temperature for sample of moisture is $37^{o} C$ , while its relative humidity is $50$ % and the barometric pressure is $101.325 k P a$ . What is the enthalpy of moist air?
(Saturation pressure of water vapour at $37^{o} C$ is $6.2795 k P a)$ .

88.31 k J / k g d a

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86.9 k J / k g d a

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37.185 k J / k g d a

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50.1 k J / k g d a

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Solution

The correct option is A $88.31 k J / k g d a$
$Φ = \frac{P_{v}}{P_{v s}}$

$P_{v} = 0.5 \times P_{v s} = 0.5 \times 6.2795$

$= 3.13975 k P a$

Humidity ratio, $ω = 0.622 \frac{P_{v}}{P_{t} - P_{v}}$

$= 0.622 \times \frac{313975}{101.325 - 3.13975} = 0.01989 k g / k g d a$

Enthalpy of moist air, $h = 1.005 \times t + ω (2500 + 1.88 t)$

$= 1.005 \times 37 + 0.01989 (2500 + 1.88 \times 37)$

$= 88.31 k J / k g d a$

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The dry bulb temperature for sample of moisture is 37oC, while its relative humidity is 50% and the barometric pressure is 101.325kPa. What is the enthalpy of moist air? (Saturation pressure of water vapour at 37oC is 6.2795kPa).

The correct option is A 88.31kJ/kgdaΦ=PvPvs Pv=0.5×Pvs=0.5×6.2795 =3.13975kPa Humidity ratio, ω=0.622PvPt−Pv =0.622×313975101.325−3.13975=0.01989kg/kgda Enthalpy of moist air, h=1.005×t+ω(2500+1.88 t) =1.005×37+0.01989(2500+1.88×37) =88.31kJ/kgda

The dry bulb temperature for sample of moisture is $37^{o} C$ , while its relative humidity is $50$ % and the barometric pressure is $101.325 k P a$ . What is the enthalpy of moist air?
(Saturation pressure of water vapour at $37^{o} C$ is $6.2795 k P a)$ .