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Heat Transfer - Engineering

Q1:

In a laboratory test run, the rate of drying was found to be 0.5 x 10-3 kg/m2.s, when the moisture content reduced from 0.4 to 0.1 on dry basis. The critical moisture content of the material is 0.08 on a dry basis. A tray dryer is used to dry 100 kg (dry basis) of the same material under identical conditions. The surface area of the material is 0.04 m2/kg of dry solid. The time required (in seconds) to reduce the moisture content of the solids from 0.3 to 0.2 (dry basis) is

A 2000

B 4000

C 5000

D 6000

ANS:C - 5000

To find the time required to reduce the moisture content of the solids from 0.3 to 0.2 (dry basis), we can use the following equation: 𝑑𝑚𝑑𝑡=𝐴⋅(𝑋1−𝑋2)𝑡dtdm​=tA⋅(X1​−X2​)​ Where:

  • 𝑑𝑚𝑑𝑡dtdm​ is the rate of drying (kg/m²·s),
  • 𝐴A is the surface area of the material (m²/kg dry solid),
  • 𝑋1X1​ is the initial moisture content (dry basis),
  • 𝑋2X2​ is the final moisture content (dry basis),
  • 𝑡t is the time required (s).
Given:
  • 𝑑𝑚𝑑𝑡=0.5×10−3dtdm​=0.5×10−3 kg/m²·s,
  • 𝑋1=0.3X1​=0.3,
  • 𝑋2=0.2X2​=0.2,
  • 𝐴=0.04A=0.04 m²/kg dry solid.
We need to rearrange the equation to solve for 𝑡t: 𝑡=𝐴⋅(𝑋1−𝑋2)𝑑𝑚𝑑𝑡t=dtdm​A⋅(X1​−X2​)​ Substituting the given values: 𝑡=0.04×(0.3−0.2)0.5×10−3t=0.5×10−30.04×(0.3−0.2)​ 𝑡=0.04×0.10.5×10−3t=0.5×10−30.04×0.1​ 𝑡=0.0040.5×10−3t=0.5×10−30.004​ 𝑡=0.0040.0005t=0.00050.004​ 𝑡=8 secondst=8seconds Therefore, the time required to reduce the moisture content of the solids from 0.3 to 0.2 (dry basis) is 8 seconds.