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MAE4703 Fall 2015 – Midterm Exam solution COVER PAGE
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Instructions. You must use the blank spaces in these pages for your exam solution. You cannot add any additional pages as part of your solution. If needed, you can use the psychrometric charts placed as appendix and indicate the question in your solution. In that case, make sure to clearly show your psychrometric process on the psychrometric chart and indicate the state points. In your solution and in the charts, use only black or blue pen (do not use pencil, red pen, nor green pen).
In each problem, explain your answer, shows the equations, and discuss the figures, results and tables as needed. If the problem is given in SI units, your final answer(s) should be in SI units. If the problem is given in English IP units, your final answer(s) should be in IP units.
MAE4703 Fall 2015 – Midterm Exam solution Page 2 of 7
Question M-1 ( 4 Points ): Use psychrometrics to estimate water condensate ( 2 points ) Humid air at 101 kPa, 35 °C, 80% relative humidity is conditioned to 101 kPa, 25 °C, 50% relative humidity. How much condensate is formed during this process? ( 2 points ) In your hardcopy solution, show and explain briefly your answer.
A) about 0.008 kg/kg-da B) about 0.010 kg/kg-da C) about 0.019 kg/kg-da D) about 0.029 kg/kg-da E) none of the above
Solution: Answer C is the correct answer. Using the psychometric chart, 1 = 0.029 kg-wv/kg-da and 2 = 0.0100 kg-wv/kg-da. According to the water mass balance, mcondensate = 1 - 2 = 0.019 kg/kg-da.
Question M-2 ( 18 points ). Mixing of air streams in a multi-zone air handler system A multi-zone air handler system similar to the one shown in Figure 2-12 of the textbook operates with 4, cfm of air entering at 100F db and 40 percent relative humidity. One third of the air flows through the inactive heating coil while the remaining two thirds of the air flows over the cooling coil and leaves at 61F db and 59F wb. Then, the two air streams mix before entering the space. Determine the ( 4 points ) dry bulb temperature, ( 2 points ) relative humidity, ( 2 points ) absolute moisture content, and ( 2 points ) enthalpy of the mixed air supplied to the space. ( 4 points ) Calculate the volume flow rate of the supplied air to the space. ( 4 points ) Show the process on a psychrometric chart. Assume standard sea level pressure. Solution This is a classical adiabatic mixing problem involving the air leaving the heating and cooling coils that may be solved using a psychrometric chart or a computer program. Using Chart 1 (IP):
Point 1 is at 100F db and 40 % R.H
Point 2 is at 61F and 59F wb (about 90 percent relative humidity)
Divide the straight line between 1 and 2 in three identical segments. Locate the mixed state 3 one segment away from point 2 (which has two third of the flow – larger flow between the two). The point 3 state is at 73 F db 1 F 66 F wb, about 65 to 75% R.H. , moisture content of 0.011 to 0.013 lbm_moisture/lbm_dry,air , and enthalpy of about 30.3 to 31.3 Btu/lbm.
To determine the leaving volume flow rate, the specific volumes of the entering and mixed air are required. The specific volumes may be estimated from Chart 1a as v 1 = 14.5 ft^3 /lbma and v 3 = 13.7 ft^3 /lbma. Then it results as follows:
𝑉̇ 3 = 𝑉̇ 1
1
2
3
MAE4703 Fall 2015 – Midterm Exam solution Page 4 of 7
Question M-4 ( 14 Points ). Determine indoor design (or maximum) relative humidity Select an indoor design relative humidity for the structures located in the Denver, CO. Assume an indoor design dry bulb temperature of 72F. Windows in the building are double glass, aluminum frame with thermal break. Other external surfaces are well insulated. (Hint: The design indoor relative humidity is such that condensation on inside of glass surface is avoided. Find the glass surface temperature first, which is the maximum allowed dew-point temperature of the indoor air.)
Solution Design relative humidity is determined by possible condensation on inside of glass. Find glass surface temperature (which is the maximum dew-point temperature of the inside air allowed).
Q/A= (ti – to)/Rtotal = (t 1 -t 0 )/Rc
Q/A= U (ti – to) = C 1 (t 1 -t 0 ) ti = 72°F; t 1 = glass surface temperature
U = 0.65 Btu/(hr-ft2-F) [from Table 5-5a]
Rc = Rtotal -Rh C U hi
1
^ ; hi = 1.46 Btu/(hr-ft2-F)
Then C 1 = 1.172 Btu/(hr-ft2-F).
1
1 (^1) C
U t t C U t i^ o
To avoid condensation at all times, use the outdoor design conditions for heating at 99.6% in the Appendix B
City
Indoor Tdb, °F
Outdoor Tdb, °F t 1 = tdp, °F Design or Max.^ RH ~ % Denver, CO 72 - (^3) 38.6 ~ 28 to 30%
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