(a) Explanation The following figure shows the schematic diagram of the well insulated chamber device within the control volume. Figure-(1) At steady state the rate of work within the control volume and the heat transfer rate is equal to zero. W ˙ c v = 0 Q ˙ c v = 0 Write the expression for the mass flow of dry air rate at stage 3 using the mass balance of dry air at steady state. m ˙ a 1 + m ˙ a 2 = m ˙ a 3 (I) Here, the mass flow rate of dry air at stage 1 is m ˙ a 1 and the mass flow rate of dry air at stage 2 is m ˙ a 2 . Write the expression for the mass flow rate of vapour at stage 3 using the mass balance of vapour at steady state. m ˙ v 1 + m ˙ v 2 = m ˙ v 3 (II) Here, the mass flow rate of vapour at state 1 is m ˙ v 1 and the mass flow rate of vapour at state 2 is m ˙ v 2 . Write the expression for the mass flow rate of vapour at state 3 . m ˙ v 3 = ω 3 m ˙ a 3 Here, the humidity ratio is ω 3 at state 3 . Write the expression for the mass flow rate of vapour at state 2 . m ˙ v 2 = ω 2 m ˙ a 2 Here, the humidity ratio is ω 2 at state 2 . Write the expression for the mass flow rate of vapour at state 1 . m ˙ v 1 = ω 1 m ˙ a 1 Here, the humidity ratio is ω 1 at state 1 . Substitute ω 3 m ˙ a 3 for m ˙ v 3 , ω 2 m ˙ a 2 for m ˙ v 2 , ω 1 m ˙ a 1 for m ˙ v 1 in Equation (II). ω 3 m ˙ a 3 = ω 2 m ˙ a 2 + ω 1 m ˙ a 1 ω 3 = ω 2 m ˙ a 2 + ω 1 m ˙ a 1 m ˙ a 3 (III) Substitute m ˙ a 1 + m ˙ a 2 for m ˙ a 3 in Equation (III). ω 3 = ω 2 m ˙ a 2 + ω 1 m ˙ a 1 m ˙ a 1 + m ˙ a 2 ω 3 = ω 1 + ( m ˙ a 2 m ˙ a 1 ) ω 2 1 + ( m ˙ a 2 m ˙ a 1 ) (IV) Write the expression for specific humidity ratio.at state 1 . ω ′ 1 = 0.62 ( p g 1 p 1 − p g 1 ) (V) Here, the partial pressue of saturated water is p g 1 at state 1 and the pressure at state 1 is p 1 . Write the expression for the humidity ratio at state 1 . ω 1 = c p ( T w b 1 − T 1 ) + ω ′ 1 ( h f g ( T w b 1 ) ) h g ( T 1 ) − h f ( T w b 1 ) (VI) Here, wet bul temperature at state 1 is T w b 1 ,temperature at state 1 is T 1 . Specific heat at contant pressure at temperature of wet bulb is c p , specific humidity at state 1 is ω ′ 1 , specific enthalpy of saturated vapour and liquid at temperature of wet bulb at state 1 h f g ( T w b 1 ) , specific enthalpy of saturated liquid at temperature at state 1 is h g ( T 1 ) and the specific enthalpy of saturated liquid at temperature wet bulb at state 1 is h f ( T w b 1 ) . Write the expression for specific humidity ratio at state 2 . ω ′ 2 = 0.62 ( p g 2 p 2 − p g 2 ) (VII) Here, the partial pressue of saturated water at state 2 is p g 2 and the pressire at state 2 is p 2 . Write the expression for the humidity ratio at state 2 . ω 2 = c p ( T w b 2 − T 2 ) + ω ′ 2 ( h f g ( T w b 2 ) ) h g ( T 2 ) − h f ( T w b 2 ) (VIII) Here, wet bul temperature at state 2 is T w b 2 , temperature at state 2 is T 2 . Specific heat at contant pressure at temperature of wet bulb is c p , specific humidity at state 1 is ω ′ 2 , specific enthalpy of saturated vapour and liquid at temperature of wet bulb at state 1 is h f g ( T w b 2 ) , specific enthalpy of saturated liquid at temperature at state 2 is h g ( T 2 ) and the specific enthalpy of saturated liquid at temperature wet bulb at state 2 is h f ( T w b 2 ) . Write the expression for the pressure at vapour state 1 . p v 1 = p 1 ω 1 0.622 + ω 1 (IX) Here, the pressure at state 1 is p 1 and the humidity ratio at state 1 is ω 1 . Write the expression for the pressure at dry air state 1 . p a 1 = p − p v 1 (X) Here, the pressure at saturated vapour state 1 is p v 1 . Write the expression for the pressure at vapour state 2 . p v 2 = p 2 ω 2 0.622 + ω 2 (XI) Here, the pressure at state 2 is p 2 and the humidity ratio at state 2 is ω 2 . Write the expression for the pressure at dry air state 2 . p a 2 = p − p v 2 (XII) Here, the pressure at saturated vapour state 2 is p v 2 . Write the expression for mass flow rate of dry air at state 1 to state 2 . m ˙ a 2 m ˙ a 1 = 2 ( p a 2 p a 1 ) T 1 T 2 (XIII) Write the expression for the pressure vapour at state 3 . p v 3 = p ω 3 0.622 + ω 3 (XIV) Write the expression for the relative humidity at state 3 . ϕ 3 = p v 3 p g ( T 3 ) (XV) Here, pressure at vapour state 3 is p v 3 and the pressure at vapour state 3 is p g ( T 3 ) . Write the expression for the energy rate balance at steady state. ( h a 3 + ω 3 h g 3 ) = ( h a 1 + ω 1 h g 1 ) + m ˙ a 2 m ˙ a 1 ( h a 2 + h g 2 ) 1 + ( m ˙ a 2 m ˙ a 1 ) (XVI) Here, specific enthalpy of saturated vapour as state 1 is h v 1 , specific enthalpy of saturated vapour at state 2 is h v 2 , soicific enthalpy fo dry air at state 1 is h a 1 , specific enthalpy of dry air at state 2 is h a 2 , specific enthalpy of saturated liquid at state 3 is h v 3 . Conclusion: Refer to Table A-2, “Properties of saturated water (liquid vapour): temperature table” to obtain the value of p g 1 as 0.04008 bar at 29 ° C . Substitute 0.4008 bar for p g 1 and 1.01325 bar for p 1 in Equation (V). ω ′ 1 = 0.622 ( 0.04008 bar 1.01325 bar − 0.04008 bar ) = 0.622 ( 0.0411 ) = 0.0256 Refer to Table A-2, “Properties of saturated water (liquid vapour): temperature table" to obtain the value of h f g ( T 1 ) as 2432.8 kJ / kg , h f ( T w b 1 ) as 121.61 kJ / kg and h g ( T 1 ) as 2554.5 kJ / kg at 29 ° C . Refer to Table A-20, “Ideal gas specific heats of some common gases” to obtain the value of c p as 1.005 kJ / kg ⋅ K with respect to air at the temperature 300 K . Substitute 1.005 kJ / kg ⋅ K for c p , 29 ° C for T w b 1 , 33 ° C for T 1 , 0.0256 for ω ′ 1 , 2432.8 kJ / kg for h f g ( T w b 1 ) , 2554.5 kJ / kg for h g ( T 1 ) and 121.61 kJ / kg for h f ( T w b 1 ) in Equation (VI). ω 1 = ( 1 (b) To determine The temperature in ° C .

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118412930

Author: Michael J. Moran, Howard N. Shapiro, Daisie D. Boettner, Margaret B. Bailey

Publisher: WILEY

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Chapter 12.9, Problem 87P

(a)

To determine

The relative humidity.

(b)

To determine

The temperature in °C.

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Chapter 12.9, Problem 84P

Chapter 12.9, Problem 88P

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Fundamentals of Engineering Thermodynamics

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