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A forced circulation evaporator is to concentrate 50,000 kg/h of 40% KOH to
70% KOH using steam at 3 bar pressure. The feed temperature and the
condensing temperature are both at 45oC. The density of the feed solution is 1.5
gm/cm3. If the overall heat transfer coefficient is 2 kW/m2 oC, calculate the
following,
i. the steam requirement,
ii. the heat transfer area required.
? Related: Evaporators (Part - 1)
Most Upvoted Answer
A forced circulation evaporator is to concentrate 50,000 kg/h of 40% K...
Forced Circulation Evaporator for Concentration of KOH Solution
Given:
- Feed rate, F = 50,000 kg/h
- Feed concentration, X1 = 40% KOH
- Product concentration, X2 = 70% KOH
- Feed temperature, T1 = 45°C
- Condensing temperature, Tc = 45°C
- Steam pressure, P = 3 bar
- Density of feed solution, ρ = 1.5 g/cm3
- Overall heat transfer coefficient, U = 2 kW/m2°C
Assumptions:
- Negligible boiling point rise
- Negligible sub-cooling of product
- Negligible pressure drop in the evaporator
Calculation:
1. Calculation of steam requirement
The heat balance equation for a forced circulation evaporator can be written as:
Q = F*Cp*(T2 - T1) + Qs
where,
- Q = Heat transferred per unit time (kW)
- Cp = Specific heat of the feed solution (kJ/kg°C)
- T2 = Temperature of product (°C)
- Qs = Heat transferred by steam per unit time (kW)
The specific heat of the feed solution can be calculated as:
Cp = X1*Cp1 + (1 - X1)*Cp2
where,
- Cp1 = Specific heat of pure KOH solution (4.18 kJ/kg°C)
- Cp2 = Specific heat of water (4.18 kJ/kg°C)
Substituting the given values in the above equations, we get:
Q = 50,000*(0.4*4.18 + 0.6*4.18)*(0.7*100 - 0.4*100) + Qs
Q = 8,288 kW
The heat transferred by steam can be calculated as:
Qs = U*A*(Tc - T2)
where,
- A = Heat transfer area (m2)
Substituting the given values in the above equation, we get:
Qs = 2*A*(45 - 100)
Qs = -110*A
Equating Q and Qs, we get:
8,288 = -110*A
A = -75.35 m2
The amount of steam required can be calculated as:
ms = Qs/(hfg*η)
where,
- hfg = Lat
Given:
- Feed rate, F = 50,000 kg/h
- Feed concentration, X1 = 40% KOH
- Product concentration, X2 = 70% KOH
- Feed temperature, T1 = 45°C
- Condensing temperature, Tc = 45°C
- Steam pressure, P = 3 bar
- Density of feed solution, ρ = 1.5 g/cm3
- Overall heat transfer coefficient, U = 2 kW/m2°C
Assumptions:
- Negligible boiling point rise
- Negligible sub-cooling of product
- Negligible pressure drop in the evaporator
Calculation:
1. Calculation of steam requirement
The heat balance equation for a forced circulation evaporator can be written as:
Q = F*Cp*(T2 - T1) + Qs
where,
- Q = Heat transferred per unit time (kW)
- Cp = Specific heat of the feed solution (kJ/kg°C)
- T2 = Temperature of product (°C)
- Qs = Heat transferred by steam per unit time (kW)
The specific heat of the feed solution can be calculated as:
Cp = X1*Cp1 + (1 - X1)*Cp2
where,
- Cp1 = Specific heat of pure KOH solution (4.18 kJ/kg°C)
- Cp2 = Specific heat of water (4.18 kJ/kg°C)
Substituting the given values in the above equations, we get:
Q = 50,000*(0.4*4.18 + 0.6*4.18)*(0.7*100 - 0.4*100) + Qs
Q = 8,288 kW
The heat transferred by steam can be calculated as:
Qs = U*A*(Tc - T2)
where,
- A = Heat transfer area (m2)
Substituting the given values in the above equation, we get:
Qs = 2*A*(45 - 100)
Qs = -110*A
Equating Q and Qs, we get:
8,288 = -110*A
A = -75.35 m2
The amount of steam required can be calculated as:
ms = Qs/(hfg*η)
where,
- hfg = Lat
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A forced circulation evaporator is to concentrate 50,000 kg/h of 40% KOH to
70% KOH using steam at 3 bar pressure. The feed temperature and the
condensing temperature are both at 45oC. The density of the feed solution is 1.5
gm/cm3. If the overall heat transfer coefficient is 2 kW/m2 oC, calculate the
following,
i. the steam requirement,
ii. the heat transfer area required. Related: Evaporators (Part - 1)?
Question Description
A forced circulation evaporator is to concentrate 50,000 kg/h of 40% KOH to 70% KOH using steam at 3 bar pressure. The feed temperature and the condensing temperature are both at 45oC. The density of the feed solution is 1.5 gm/cm3. If the overall heat transfer coefficient is 2 kW/m2 oC, calculate the following, i. the steam requirement, ii. the heat transfer area required. Related: Evaporators (Part - 1)? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about A forced circulation evaporator is to concentrate 50,000 kg/h of 40% KOH to 70% KOH using steam at 3 bar pressure. The feed temperature and the condensing temperature are both at 45oC. The density of the feed solution is 1.5 gm/cm3. If the overall heat transfer coefficient is 2 kW/m2 oC, calculate the following, i. the steam requirement, ii. the heat transfer area required. Related: Evaporators (Part - 1)? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A forced circulation evaporator is to concentrate 50,000 kg/h of 40% KOH to 70% KOH using steam at 3 bar pressure. The feed temperature and the condensing temperature are both at 45oC. The density of the feed solution is 1.5 gm/cm3. If the overall heat transfer coefficient is 2 kW/m2 oC, calculate the following, i. the steam requirement, ii. the heat transfer area required. Related: Evaporators (Part - 1)?.
A forced circulation evaporator is to concentrate 50,000 kg/h of 40% KOH to 70% KOH using steam at 3 bar pressure. The feed temperature and the condensing temperature are both at 45oC. The density of the feed solution is 1.5 gm/cm3. If the overall heat transfer coefficient is 2 kW/m2 oC, calculate the following, i. the steam requirement, ii. the heat transfer area required. Related: Evaporators (Part - 1)? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about A forced circulation evaporator is to concentrate 50,000 kg/h of 40% KOH to 70% KOH using steam at 3 bar pressure. The feed temperature and the condensing temperature are both at 45oC. The density of the feed solution is 1.5 gm/cm3. If the overall heat transfer coefficient is 2 kW/m2 oC, calculate the following, i. the steam requirement, ii. the heat transfer area required. Related: Evaporators (Part - 1)? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A forced circulation evaporator is to concentrate 50,000 kg/h of 40% KOH to 70% KOH using steam at 3 bar pressure. The feed temperature and the condensing temperature are both at 45oC. The density of the feed solution is 1.5 gm/cm3. If the overall heat transfer coefficient is 2 kW/m2 oC, calculate the following, i. the steam requirement, ii. the heat transfer area required. Related: Evaporators (Part - 1)?.
Solutions for A forced circulation evaporator is to concentrate 50,000 kg/h of 40% KOH to
70% KOH using steam at 3 bar pressure. The feed temperature and the
condensing temperature are both at 45oC. The density of the feed solution is 1.5
gm/cm3. If the overall heat transfer coefficient is 2 kW/m2 oC, calculate the
following,
i. the steam requirement,
ii. the heat transfer area required. Related: Evaporators (Part - 1)? in English & in Hindi are available as part of our courses for Chemical Engineering.
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Here you can find the meaning of A forced circulation evaporator is to concentrate 50,000 kg/h of 40% KOH to
70% KOH using steam at 3 bar pressure. The feed temperature and the
condensing temperature are both at 45oC. The density of the feed solution is 1.5
gm/cm3. If the overall heat transfer coefficient is 2 kW/m2 oC, calculate the
following,
i. the steam requirement,
ii. the heat transfer area required. Related: Evaporators (Part - 1)? defined & explained in the simplest way possible. Besides giving the explanation of
A forced circulation evaporator is to concentrate 50,000 kg/h of 40% KOH to
70% KOH using steam at 3 bar pressure. The feed temperature and the
condensing temperature are both at 45oC. The density of the feed solution is 1.5
gm/cm3. If the overall heat transfer coefficient is 2 kW/m2 oC, calculate the
following,
i. the steam requirement,
ii. the heat transfer area required. Related: Evaporators (Part - 1)?, a detailed solution for A forced circulation evaporator is to concentrate 50,000 kg/h of 40% KOH to
70% KOH using steam at 3 bar pressure. The feed temperature and the
condensing temperature are both at 45oC. The density of the feed solution is 1.5
gm/cm3. If the overall heat transfer coefficient is 2 kW/m2 oC, calculate the
following,
i. the steam requirement,
ii. the heat transfer area required. Related: Evaporators (Part - 1)? has been provided alongside types of A forced circulation evaporator is to concentrate 50,000 kg/h of 40% KOH to
70% KOH using steam at 3 bar pressure. The feed temperature and the
condensing temperature are both at 45oC. The density of the feed solution is 1.5
gm/cm3. If the overall heat transfer coefficient is 2 kW/m2 oC, calculate the
following,
i. the steam requirement,
ii. the heat transfer area required. Related: Evaporators (Part - 1)? theory, EduRev gives you an
ample number of questions to practice A forced circulation evaporator is to concentrate 50,000 kg/h of 40% KOH to
70% KOH using steam at 3 bar pressure. The feed temperature and the
condensing temperature are both at 45oC. The density of the feed solution is 1.5
gm/cm3. If the overall heat transfer coefficient is 2 kW/m2 oC, calculate the
following,
i. the steam requirement,
ii. the heat transfer area required. Related: Evaporators (Part - 1)? tests, examples and also practice Chemical Engineering tests.
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