Answer:
a) flow(m_v) = 36.72 kg/min
b) m_L = 437.3 tons
Step-by-step 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