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Volgvan
3 years ago
9

) A brick of clay is being dried in a batch dryer using constant drying conditions. You have been asked to determine the amount

of time it will take to dry the clay from a total moisture content of 0.6 to a free moisture content of 0.08. You are given the following information: 1. Previously, the drier was able to dry a piece of clay with a total moisture content of 0.64 in 8 hours and 80% of the drying occurred with a constant rate. 2. The equilibrium moisture content is 0.1. 3. The critical free moisture content is 0.15.
Engineering
1 answer:
dimaraw [331]3 years ago
4 0

Answer:

The drying time is calculated as shown

Explanation:

Data:

Let the moisture content be = 0.6

the free moisture content be = 0.08

total moisture of the clay  = 0.64

total drying time for the period = 8 hrs

then if the final dry and wet masses are calculated, it follows that

t = (X0+ Xc)/Rc) + (Xc/Rc)* ln (Xc/X)

 = 31.3 min.

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Answer:

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if (distance % 500 == 0) {

segments = distance / 500;

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segments = distance / 500 + 1;

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3 years ago
technician A says that in any circuit, electrical current takes the path of least resistance. technician B says that while this
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Answer:

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3 years ago
______ are an idication that your vehicle may be developing a cooling system problem.
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Answer:

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3 years ago
A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu
Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,  

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

  • Negligible heat loss through the insulation
  • Steady state system
  • One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}

During steady state

\frac{dT}{dt} = 0 which gives k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0

From which we have \frac{d^{2}T }{dx^{2} }  = -\frac{q'_{G}}{k}

Considering the boundary condition at x =0 where there is no heat loss

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\frac{dT }{dx }  = \frac{q'_{G}}{k} x+ C_{1} from which C₁ is evaluated from the first boundary condition thus

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From the second integration we have

T  = -\frac{q'_{G}}{2k} x^{2} + C_{2}

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Substituting the values we get

T = 167 ° C

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