CHEMICAL ENGINEERING CALCULATIONS
MATERIAL BALANCE WITHOUT CHEMICAL REACTIONS PROBLEMS
1. A moist paper containing 20% water by weight goes into a drier in a continuous process. The
paper leaves the drier containing 2% water by weight. Calculate the weight of water removed
from the paper per 100 kg of the original moist paper.
Given:
Required: kg water removed
Solution:
𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵: 100 𝑘𝑘𝑘𝑘 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝐿𝐿𝐿𝐿𝐿𝐿: 𝑊𝑊 = 𝑘𝑘𝑘𝑘 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
𝑃𝑃 = 𝑘𝑘𝑘𝑘 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑂𝑂𝑂𝑂𝑂𝑂: 100 = 𝑊𝑊 + 𝑃𝑃
100(1 − 0.2) = 𝑃𝑃(1 − 0.02)
80 = 𝑃𝑃(0.98)
𝑃𝑃 = 81.63 𝑘𝑘𝑘𝑘
100 = 𝑊𝑊 + 81.63
𝑊𝑊 = (100 − 81.63)𝑘𝑘𝑘𝑘
𝑊𝑊 = 18.37 𝑘𝑘𝑘𝑘
Answer: 18.37 kg of water removed
2. Fifty pounds (lb) of NaCl solution (40% NaCl and 60% H2O), 100 lb of a sugar solution (20%
sugar and 80% H2O), and 40 lb of a water solution (10% NaCl, 5% sugar and 85% H2O) are
mixed together and heated. Some of the water is lost by evaporation. If the resulting mixture
contains 15% NaCl, what percent of the total water charged were lost through evaporation? (All
%’s are weight percent)
Given:
Required: percent of the total water lost through evaporation
Solution:
𝐿𝐿𝐿𝐿𝐿𝐿: 𝑊𝑊 = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑏𝑏𝑏𝑏 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
𝑃𝑃 = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑤𝑤𝑤𝑤𝑤𝑤ℎ 15% 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁
𝑂𝑂𝑂𝑂𝑂𝑂: (50 + 100 + 40)𝑙𝑙𝑙𝑙 = 𝑃𝑃 + 𝑊𝑊
, 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏: 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑖𝑖𝑖𝑖 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑖𝑖𝑖𝑖 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
(50 𝑙𝑙𝑏𝑏)(0.4) + (40 𝑙𝑙𝑙𝑙)(0.1) = 𝑃𝑃(0.15)
24 𝑙𝑙𝑙𝑙 = 𝑃𝑃(0.15)
𝑃𝑃 = 160 𝑙𝑙𝑙𝑙
160 𝑙𝑙𝑙𝑙 + 𝑊𝑊 = 190 𝑙𝑙𝑙𝑙
𝑊𝑊 = (190 − 160) 𝑙𝑙𝑙𝑙
𝑊𝑊 = 30 𝑙𝑙𝑙𝑙
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑖𝑖𝑖𝑖 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = (50 𝑙𝑙𝑙𝑙)(0.6) + (100 𝑙𝑙𝑙𝑙)(0.8) + (40 𝑙𝑙𝑙𝑙)(0.85)
= 144 𝑙𝑙𝑙𝑙 𝐻𝐻2 𝑂𝑂 𝑖𝑖𝑖𝑖 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑏𝑏𝑏𝑏 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
% 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑏𝑏𝑏𝑏 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 𝑥𝑥 100%
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑖𝑖𝑖𝑖 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
30 𝑙𝑙𝑙𝑙
% 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑏𝑏𝑏𝑏 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 𝑥𝑥 100%
144 𝑙𝑙𝑙𝑙
% 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑟𝑟 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑏𝑏𝑏𝑏 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 20.83%
Answer: 20.83% water lost by evaporation
3. A labeled flowchart of a continuous steady-state two-unit process is shown below. Each stream
contains two components, A and B, in different proportions. Three streams whose flowrates
and/or compositions are not known are labeled x, y, and z. Calculate the known flowrates and
compositions of streams x, y and z.
Given:
Required: flowrates and compositions of streams x, y and z.
Solution:
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘
𝑂𝑂𝑂𝑂𝑂𝑂: 100 = 40 + 𝑥𝑥
ℎ𝑟𝑟 ℎ𝑟𝑟
𝑘𝑘𝑘𝑘
𝑥𝑥 = 60
ℎ𝑟𝑟
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝐴𝐴 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵
𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝐴𝐴 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝐴𝐴
�100 � �0.50 � = �40 � �0.90 � + 𝑥𝑥 ∙ 𝑥𝑥𝐴𝐴
ℎ𝑟𝑟 𝑘𝑘𝑘𝑘 ℎ𝑟𝑟 𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝐴𝐴 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝐴𝐴 𝑘𝑘𝑘𝑘
�100 � �0.50 � − �40 � �0.90 � = �60 � ∙ 𝑥𝑥
MATERIAL BALANCE WITHOUT CHEMICAL REACTIONS PROBLEMS
1. A moist paper containing 20% water by weight goes into a drier in a continuous process. The
paper leaves the drier containing 2% water by weight. Calculate the weight of water removed
from the paper per 100 kg of the original moist paper.
Given:
Required: kg water removed
Solution:
𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵: 100 𝑘𝑘𝑘𝑘 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝐿𝐿𝐿𝐿𝐿𝐿: 𝑊𝑊 = 𝑘𝑘𝑘𝑘 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
𝑃𝑃 = 𝑘𝑘𝑘𝑘 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑂𝑂𝑂𝑂𝑂𝑂: 100 = 𝑊𝑊 + 𝑃𝑃
100(1 − 0.2) = 𝑃𝑃(1 − 0.02)
80 = 𝑃𝑃(0.98)
𝑃𝑃 = 81.63 𝑘𝑘𝑘𝑘
100 = 𝑊𝑊 + 81.63
𝑊𝑊 = (100 − 81.63)𝑘𝑘𝑘𝑘
𝑊𝑊 = 18.37 𝑘𝑘𝑘𝑘
Answer: 18.37 kg of water removed
2. Fifty pounds (lb) of NaCl solution (40% NaCl and 60% H2O), 100 lb of a sugar solution (20%
sugar and 80% H2O), and 40 lb of a water solution (10% NaCl, 5% sugar and 85% H2O) are
mixed together and heated. Some of the water is lost by evaporation. If the resulting mixture
contains 15% NaCl, what percent of the total water charged were lost through evaporation? (All
%’s are weight percent)
Given:
Required: percent of the total water lost through evaporation
Solution:
𝐿𝐿𝐿𝐿𝐿𝐿: 𝑊𝑊 = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑏𝑏𝑏𝑏 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
𝑃𝑃 = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑤𝑤𝑤𝑤𝑤𝑤ℎ 15% 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁
𝑂𝑂𝑂𝑂𝑂𝑂: (50 + 100 + 40)𝑙𝑙𝑙𝑙 = 𝑃𝑃 + 𝑊𝑊
, 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏: 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑖𝑖𝑖𝑖 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑖𝑖𝑖𝑖 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
(50 𝑙𝑙𝑏𝑏)(0.4) + (40 𝑙𝑙𝑙𝑙)(0.1) = 𝑃𝑃(0.15)
24 𝑙𝑙𝑙𝑙 = 𝑃𝑃(0.15)
𝑃𝑃 = 160 𝑙𝑙𝑙𝑙
160 𝑙𝑙𝑙𝑙 + 𝑊𝑊 = 190 𝑙𝑙𝑙𝑙
𝑊𝑊 = (190 − 160) 𝑙𝑙𝑙𝑙
𝑊𝑊 = 30 𝑙𝑙𝑙𝑙
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑖𝑖𝑖𝑖 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = (50 𝑙𝑙𝑙𝑙)(0.6) + (100 𝑙𝑙𝑙𝑙)(0.8) + (40 𝑙𝑙𝑙𝑙)(0.85)
= 144 𝑙𝑙𝑙𝑙 𝐻𝐻2 𝑂𝑂 𝑖𝑖𝑖𝑖 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑏𝑏𝑏𝑏 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
% 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑏𝑏𝑏𝑏 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 𝑥𝑥 100%
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑖𝑖𝑖𝑖 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
30 𝑙𝑙𝑙𝑙
% 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑏𝑏𝑏𝑏 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 𝑥𝑥 100%
144 𝑙𝑙𝑙𝑙
% 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑟𝑟 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑏𝑏𝑏𝑏 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 20.83%
Answer: 20.83% water lost by evaporation
3. A labeled flowchart of a continuous steady-state two-unit process is shown below. Each stream
contains two components, A and B, in different proportions. Three streams whose flowrates
and/or compositions are not known are labeled x, y, and z. Calculate the known flowrates and
compositions of streams x, y and z.
Given:
Required: flowrates and compositions of streams x, y and z.
Solution:
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘
𝑂𝑂𝑂𝑂𝑂𝑂: 100 = 40 + 𝑥𝑥
ℎ𝑟𝑟 ℎ𝑟𝑟
𝑘𝑘𝑘𝑘
𝑥𝑥 = 60
ℎ𝑟𝑟
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝐴𝐴 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵
𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝐴𝐴 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝐴𝐴
�100 � �0.50 � = �40 � �0.90 � + 𝑥𝑥 ∙ 𝑥𝑥𝐴𝐴
ℎ𝑟𝑟 𝑘𝑘𝑘𝑘 ℎ𝑟𝑟 𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝐴𝐴 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝐴𝐴 𝑘𝑘𝑘𝑘
�100 � �0.50 � − �40 � �0.90 � = �60 � ∙ 𝑥𝑥
