PROBLEM 4.9-4.OUTLET TEMPERATURE AND EFFECTIVENESS OF AN EXCHANGER. HOT OIL AT A FLOW RATE OF 3.00 KG/S (CP=1.92 KJ/KGK) ENTERS AN EXISTING COUNTERFLOW EXCHANGER AT 400 K AND IS COOLED BY WATER ENTERING AT 325 K (UNDER PRESSURE) AND FLOWING AT A RATE OF 0.70 KG/S. THE OVERALL U=350 W/m2K) AND A=12.9 m2. CALCULATE THE HEAT-TRANSFER RATE AND EXIT OIL TEMPERATURE.

GIVEN:m (OIL) = 3.00 KG/S m (H20) =0.70 KG/SCP=1.92 KJ/KGKU=350 W/m2K) AND A=12.9 m2REQUIRED:THO , qSOLUTION: THI= 400 K TCO H= OIL C= H2O TCI= 325 K THO

ASSUME WATER OUTLET TCO = 374 K TCO = = 349 K FROM A.2 FOR H2O AT 349 K, CP=4.196 KJ/KGK

OIL ( mCP ) H = CH = (3.00) (1.92X103 ) = 5760 W/K H20 (mCP ) H = CC = (0.70) (4.196X103 ) = 2937 W/K = CMIN

NTU = = = 1.537 From Fig. 4.10-7a , E = 0.70 q= E CMIN ( THI - TCI ) = ( 0.70) (2937)(400-325) = 154 190 W q= 154 190= (mCP )C ( TCO - TCI ) = 2936 ( TCO 325 ) TCO = 373.2 K

PROBLEM 9.8-1DRYING OF BIOLOGICAL MATERIAL IN TRAY DRYER. A GRANULAR BIOLOGICAL MATERIAL WET WITH WATER IS BEING DRIED IN A PAN 0.305 X 0.305 m AND 38.1 mm DEEP. THE MATERIAL IS 38.1 mm DEEP IN THE PAN, WHICH IS INSULATED ON THE SIDES AND BOTTOM. HEAT TRANSFER IS BY CONVECTION FROM AN AIR STREAM FLOWING PARALLEL TO THE TOP SURFACE AT A VELOCITY OF 3.05 m/s , HAVING A TEMPERATURE OF 65.6 AND HUMDITY H= 0.010 KG H2O/ KG DRY AIR. THE TOP SURFACE RECEIVES RADIATION FROM STEAM-HEATED PIPES WHOSE SURFACE TEMPERATURE TR = 93.3IT IS DESIRED TO KEEP THE SURFACE TEMPERATURE OF THE SOLID BELOW 32.2 SO THAT DECOMPOSITION WILL BE KEPT LOW. CALCULATE THE SURFACE TEMPERATURE AND THE RATE OF DRYING FOR THE CONSTANT- RATE PERIOD.

GIVEN: 0.305 X 0.305 m38.1 mm DEEPH= 0.010 KG H2O/ KG DRY AIRv= 3.05 m/sTR = 93.3 T= 65.6 E = 0.95

REQUIRED:TS , RC

SOLUTION:TS KEPT BELOW 32.2 VH= ( 2.83X 10-3 + 4.56 x10-3 H ) T = ( 2.83X 10-3 + 4.56 x10-3 (0.010) ) (273+65.6) = 0.974 m3/ kg dry air = = 1.037 kg/ m3G= v = 3.05 (1.037) (3600) =11 386 kg/ h m2hc= 0.0204 G 0.8 = 0.0204 (11 386) 0.8 = 35.87 W/ m2KTrial 1:AssumeTs = 32.2 + 273.2 = 305.4 K TR= 93.3 + 273.2 = 366.5 K E = 0.95hr = = 8.42 W/ m2 K@ Uk = 0 ( no conduction, bottom insulated )CS = (1.005 + 1.88 (0.01))103 = 1.024X103 J/kgK s = ( 1 + ) ( T- Ts ) + ( TR Ts )For Ts = 32.2 , Hs = 0.031 s = 2425.3x103 J/KGThus, (2425x103)= ( 1 +0) ( 65.6- Ts ) + ( 93.3 Ts ) Ts = 30.3Trial 2.Assume Ts = 31.7 HS = 0.0305 , s = 2426.5 x103

(2426.5x103)= ( 1 +0) ( 65.6- Ts ) + ( 93.3 Ts ) Ts = 31.3 close enough to assumed Ts

Rc = = Rc =2.583 Kg h2o/hm2

PROBLEM 8.4-7EFFECT OF FEED TEMPERATURE ON EVAPORATING AN NaOH SOLUTION. A SINGLE-EFFECT EVAPORATOR IS CONCENTRATING A FEED OF 9072 Kg/H OF A 10% WT. SOLUTION OF NaOH IN WATER TO A PRODUCT OF 50% SOLIDS. THE PRESSURE OF THE SATURATED STEAM USED IS 42 KPA (GAGE) AND THE PRESSURE IN THE VAPOR SPACE OF THE EVAPORATOR IS 20 KPA (ABS). THE U=1988 W/m2K. CALCULATE THE STEAM USED, THE STEAM ECONOMY IN KG VAPORIZED/KG STEAM, AND THE AREA FOR THE FOLLOWING FEED CONDITIONS.(A) FEED TEMPERATURE OF 288.8 K (B) FEED TEMPERATURE OF 322.1 K

PROBLEM 9.7-2In order to test the feasibility of drying a certain foodstuff, drying data were obtained in a tray dryer with air flow over the to exposed surface having an area of 0.186 M2 . The bone-dry sample weight was 3.765 kg dry solid. At equilbrium after a long period, the wet sample weight was 3.955 kg H2O + solid. Hence, 3.955-3.765, or 0.190 kg of equilibrium moisture was present.The following sample weights versus time were obtained in the drying test.

GIVEN: A= 0.186 m2WS= LS=3.765 kg dry solidEquilibrium moisture= 0.190 kg h20

reQUIRED:A. calculate the free moisture content x kg h20/kg dry solid for each data point and plot x versus time.B. Measure the slopes, calculate the drying rates R in kg h20/hm2C. Using this drying rate curve, predict the total time to dry the sample from x=0.20 to x=0.04. use graphical INTERGRATION FOR THE FALLING-RATE PERIOD. WHAT IS THE DRYING RATE RC IN THE CONSTANT-RATE PERIOD AND xC ?

SOLUTION: A.

B.

C.

Rc= 0.998 kg H20/ h m2

xc= 0.12 kg H20/kg Dry solid

A3A2

A1