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Heat Transfer - Engineering

Q1:

200 kg of solids (on dry basis) is subjected to a drying process for a period of 5000 seconds. The drying occurs in the constant rate period with the drying rate as, Nc = 0.5 x 10-3 kg/m2.s. The initial moisture content of the solid is 0.2 kg moisture/kg dry solid. The interfacial area available for drying is 4 m2/1000 kg of dry solid. The moisture content at the end of the drying period is (in kg moisture/kg dry solid)

A 0.5

B 0.05

C 0.1

D 0.15

ANS:C - 0.1

To find the moisture content at the end of the drying period, we can use the equation for drying in the constant rate period: dtdm​=Nc​A Where:

  • dtdm​ is the rate of moisture removal (kg/s)
  • Nc​ is the drying rate (kg/m².s)
  • A is the interfacial area available for drying (m²)
  • m is the mass of moisture in the solid (kg)
  • t is the drying time (s)
Given:
  • =0.5×10−3Nc​=0.5×10−3 kg/m².s
  • =41000×200=0.8A=10004​×200=0.8 m² (since 1 m² is available for 1000 kg of dry solid, and we have 200 kg of dry solid)
  • =5000t=5000 s
We can rearrange the equation to solve for dm: =dm=Nc​A×dt Substituting the given values: =(0.5×10−3 kg/m².s)×(0.8 m²)×(5000 s)dm=(0.5×10−3kg/m².s)×(0.8m²)×(5000s) 2 kgdm=2kg Now, let's find the total mass of moisture removed: initial=0.2 kg moisture/kg dry solid×200 kg dry solid=40 kg moistureminitial​=0.2kg moisture/kg dry solid×200kg dry solid=40kg moisture final=initial−=40 kg−2 kg=38 kg moisturemfinal​=minitial​−dm=40kg−2kg=38kg moisture Now, we need to find the moisture content at the end of the drying period: Moisture content=finalMass of dry solidMoisture content=Mass of dry solidmfinal​​ Moisture content=38 kg200 kg=0.19 kg moisture/kg dry solidMoisture content=200kg38kg​=0.19kg moisture/kg dry solid So, the moisture content at the end of the drying period is approximately 0.19 kg moisture/kg dry solid0.19kg moisture/kg dry solid.