In order to test the feasibility of drying a certain foodstuff, drying data were obtained in a tray dryer with air flow over the
top exposed surface having an area of 0.186 m2. The bone dry sample weight was 3.765 kg dry solid. At equilibrium after a long period, the wet sample weight was 3.955 kg (water plus solid). Hence, 3.955-3.765 or 0.190 kg of equilibrium moisture was present. The following table provides the sample weights versus time during the drying test:
Time (hr)
Weight (kg)
Time (hr)
Weight (kg)
Time (hr)
Weight (kg)
0
4.944
2.2
4.554
7.0
4.019
0.4
4.885
3.0
4.404
9.0
3.978
0.8
4.808
4.2
4.241
12.0
3.955
1.4
4.699
5.0
4.150
A) Calculate the free moisture content, X (kg water per kg dry solid) for each data point and plot X versus t;
B) Using the slope, calculate the drying rate, R in kg water per hour per square meter and plot R versus X;
C) Using this drying rate curve, predict the total time to dry the sample from X = 0.20 to X = 0.04. Use numerical integration in the falling rate period. What is the drying rate in the constant rate period and what is X in the constant rate period?
Time (hr)
Weight (kg)
Time (hr)
Weight (kg)
Time (hr)
Weight (kg)
0
4.944
2.2
4.554
7.0
4.019
0.4
4.885
3.0
4.404
9.0
3.978
0.8
4.808
4.2
4.241
12.0
3.955
1.4
4.699
5.0
4.150
A) Calculate the free moisture content, X (kg water per kg dry solid) for each data point and plot X versus t;
B) Using the slope, calculate the drying rate, R in kg water per hour per square meter and plot R versus X;
C) Using this drying rate curve, predict the total time to dry the sample from X = 0.20 to X = 0.04. Use numerical integration in the falling rate period. What is the drying rate in the constant rate period and what is X in the constant rate period?
1 answer:
Answer / Explanation:
(A) To start, we calculate the free moisture content multiplies by Kg H₂O / Kg Dry solid for each data point and plot X versus Time
Therefore,
X = Wt - Ws - Equilibrium Moisture / Ws
= 4.944 - 3.765 - 0.190 / 3.785
= 0.26268 Kg H₂0 / Kg dry solid
(a) To also calculate the free moisture content X Kg H₂O / Kg Dry solid for each data point and plot X versus Time.
Kindly refer to the plotted graph below of X against or versus Time
(B) To measure the slope, we calculate the drying rate in Kg H₂0 /H.M²
Now, recalling the equation for calculating slope:
we have Slope = ΔX /ΔT
Therefore, R =Ls ΔX / A ΔT
Hence, R = 3.765 ( 0.24701 - 0.26268) / 0.186 ( 0.4 - 0)
= 0.79301
(b) To measure the slope, we calculate the drying rate in Kg H₂0 /H.M²
Kindly refer to the plotted graph below of R against or versus X
(C) Since t ( 0.4) = 4.8 hours
t (0.2 ) = 0.7 hours
total time (t) = 4.1 hours
This can also be represented in the graph below:
(c) USING THIS DRYING RATE CURVE, PREDICT THE TOTAL TIME TO DRY THE SAMPLE FROM X=0.20 TO X=0.04. USE GRAPHICAL INTERGRATION FOR THE FALLING-RATE PERIOD. WHAT IS THE DRYING RATE R IN THE CONSTANT-RATE PERIOD AND X
Moving forward,
Rc = 0.998 Kg H₂0 /hm²
= t = Ls dx/ A R = 3.765 / 0.186 ( 0.20 - 0.12 / 0.998)
= 1.63
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Answer:
Suction and exhaust processes do not affect the performance of Otto cycle.
Explanation:
Step1
Inlet and exhaust flow processes are not including in the Otto cycle because the effect and nature of both the process are same in opposite direction.
Step2
Inlet process or the suction process is the process of suction of working fluid inside the cylinder. The suction process is the constant pressure process. The exhaust process is the process of exhaust out at constant pressure.
Step3
The suction and exhaust process have same work and heat in opposite direction. So, net effect of suction and exhaust processes cancels out. The suction and exhaust processes are shown below in P-V diagram of Otto cycle:
Process 0-1 is suction process and process 1-0 is exhaust process.
