Air at 38°C and 97% relative humidity is to be cooled to 18°C and fed into a plant area at a rate of 510 m/min. a. Calculate the rate (kg/min) at which water condenses. b. Calculate the cooling requirement in tons (1 ton of cooling 12,000 Btu/h), assuming that the enthalpy of water vapor is that of saturated steam at the same temperature and the enthalpy of dry air is given by the expression

a) flow(m_v) = 36.72 kg/min

b) m_L = 437.3 tons

Explanation:

Given:

- T_1 = 38 C

- T_2 = 18 C

- w_r = 97 %

- flow(V) = 510 m^3 / min

Find:

a) Calculate the rate (kg/min) at which water condenses

b) Calculate the cooling requirement in tons (1 ton of cooling 12,000 Btu/h), assuming that the enthalpy of water vapor is that of saturated steam at the same temperature and the enthalpy of dry air is given by the expression

Solution:

- We will use ideal Gas Law to tackle our questions, The law states:

                                 P*flow(V) = flow(m)*R*T

- Where, P : pressure , flow(V) : Volume Flow Rate, flow(m) : mass flow rate  R : Gas constant , T : absolute Temperature.

- Use the ideal gas law to calculate flow(m):

                                flow(m) = P_1*flow(V) / R*T_1

- Use psychometric chart and evaluate P_1:

                                Inputs: T_1 = 38 C and 97 relative humidity

                                Output: P_1 = 0.06626 bar

- Hence, the relationship becomes:

                                 flow(m) = 6626*510 / 287*311

                                 flow(m) = 37.85 kg/min      

- Using the relative humidity we can compute the flow(m_v) at which the water condenses:

                                 flow(m_v) = w_r*flow(m)

                                 flow(m_v) = 0.97*37.85

                                 flow(m_v) = 36.72 kg/min

- The total enthalpy at a state is given by:

                                 H = H_a + H_v

Where, H is the total enthalpy , H_a : dry air enthalpy , H_v : wet air enthalpy

                                H = flow(m)*0.0291*(38 - 14 ) + 2500*flow(m_v)

                                H = 37.85*0.0291*(38 - 14 ) + 2500*36.72

                                H = 91826.43444 KJ / min

- Convert to Btu/h:

                               H = 91826.43444 KJ / min * (ton*min / 210 KJ)

                              H = 437.3 tons

- The cooling requirement in tons is equal to the total enthalpy of vapor and dry air combined. m_L = 437.3 tons