Chemical Engineering - Heat Transfer - Discussion
Discussion Forum : Heat Transfer - Section 7 (Q.No. 42)
42.
1000 Kg of liquid at 30°C in a well stirred vessel has to be heated to 120°C, using immersed coils carrying condensing steam at 150°C. The area of the steam coils is 1.2 m2 and the overall heat transfer co-efficient to the liquid is 1500 W/m2.°C. Assuming negligible heat loss to the surrounding and specific heat capacity of the liquid to be 4 kJ/kg.°C, the time taken for the liquid to reach desired temperature will be
Discussion:
5 comments Page 1 of 1.
Reece said:
2 years ago
First, solve for the logarithmic mean temperature difference ΔTₗₘ:
ΔTₗₘ = [(150°C - 120°C)-(150°C - 30°C)]/ln((150°C - 120°C)/(150°C - 30°C))
ΔTₗₘ = 64.92°C = 64.92 K.
From here, consider the energy balance around the vessel:
mCₚΔT = thAΔTₗₘ
Plugging in the known values and applying appropriate conversions,
(1,000 kg)(4 kJ/kg/°C)(120°C - 30°C)(1,000 J/1 kJ) = t(1,500 W/m²/K)(1.2 m²)(64.92 K)(60 s/1 min).
Solving for the time required for the liquid to reach the desired temperature t:
t = 51.34 min ≈ 51 min.
ΔTₗₘ = [(150°C - 120°C)-(150°C - 30°C)]/ln((150°C - 120°C)/(150°C - 30°C))
ΔTₗₘ = 64.92°C = 64.92 K.
From here, consider the energy balance around the vessel:
mCₚΔT = thAΔTₗₘ
Plugging in the known values and applying appropriate conversions,
(1,000 kg)(4 kJ/kg/°C)(120°C - 30°C)(1,000 J/1 kJ) = t(1,500 W/m²/K)(1.2 m²)(64.92 K)(60 s/1 min).
Solving for the time required for the liquid to reach the desired temperature t:
t = 51.34 min ≈ 51 min.
(3)
JayN said:
9 years ago
Use the following formula to get the solution.
t = (( mCp/ha ) ln ((Tc -t)/(Tc -T)))/60.
t = (( mCp/ha ) ln ((Tc -t)/(Tc -T)))/60.
Prakash Dhanaraj said:
7 years ago
t= (1000*4000/1500*4 ) * ln(120/30).
= 15.5 mins.
So, option A is the right answer.
= 15.5 mins.
So, option A is the right answer.
@nkit said:
6 years ago
51 min is right
Given HTA 1.2.
Given HTA 1.2.
Saniya said:
1 decade ago
Explain please.
(1)
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