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

energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this