A single-effect evaporator is to pr

A single-effect evaporator is to produce a 35% solids tomato concentrate from a 6% solids raw juice entering at 18°C. The pressure in the evaporator is 20 kPa(absolute) and steam is available at 100 kPa gauge. The overall heat-transfer coefficient is 440 J m-2 s-1 °C-1, the boiling temperature of the tomato juice under the conditions in the evaporator is 60°C, and the area of the heat-transfer surface of the evaporator is 12 m2. Estimate the rate of raw juice feed that is required to supply the evaporator.
요약: raw juice 6%고체로부터 토마토35% 증발 ,증발에 요구되는 raw juice 속도?
토마토 비열 어는점 이상 3.98
고형분 수지 0.06f=0.35p
스팀21kpa로 가정했을때 (2539.9-88.14)
100kpa일 때 포화수증기표보고 했는데요.
s(질량속도)= (440)*(12)(온도차)/스팀Hs-hs(2676.1-419.04*10^3)
그 다음부터 어떻게 해야 될지 잘 모르겠어요. 힌트 좀 주세요.

Results (English) 2:

A single-effect evaporator is to produce a 35% solids tomato concentrate from a 6% solids raw juice entering at 18 ° C. The pressure in the evaporator is 20 kPa (absolute) and steam is available at 100 kPa gauge. The overall heat -transfer coefficient is 440 J m-2 s-1 ° C-1, the boiling temperature of the tomato juice under the conditions in the evaporator is 60 ° C, and the area of the heat-transfer surface of the evaporator is 12 m2 . Estimate the rate of feed raw Juice That is required to supply the Evaporator.
요약:? raw Juice 6% 35% 증발 고체 로부터 토마토, 증발 에 요구 되는 raw Juice 속도
토마토 비열 어는점 이상 3.98
고형분 수지 0.06f = 0.35p
스팀 21 kPa 로 가정 했을 때 (2539.9-88.14)
100 kPa 일때 포화 수증기 표 보고 했는데요.
s (질량 속도) = (440) * (12) (온도차) / 스팀 Hs-HS (2676.1-419.04 * 10 ^ 3)
그다음 부터 어떻게 해야 될지 잘 모르겠어요. 힌트 좀 주세요.

Results (English) 3:

A single-effect evaporator is to produce a 35% solids tomato concentrate from a 6% solids raw juice entering at The pressure in the 18 C. evaporator is 20 kPa and steam is (absolute) available at 100 kPa gauge. The overall heat-transfer coefficient is 440 J m-2 S-1 C-1, the boiling temperature of the tomato juice under the conditions in the evaporator is 60 C,And the area of the heat-transfer surface of the evaporator is 12 m2. Estimate the rate of raw juice feed that is required to supply the evaporator.
raw juice 6% days: 35%, days raw juice?
Bi 3.98
to 0.06f=0.35p
21kpa not (2539.9-88.14)
100KPA you
days.S () = (440) * (12) () / Hs-hs (2676.1-419.04*10
3)
years to days. Days.