# Module 7: Solved Problems

Module 7: Solved Problems . 1. Deionized water flows through the inner tube of 30-mm diameter in a thin-walled concentric tube heat exchanger of 0.19-...

Module 7: Solved Problems 1. Deionized water flows through the inner tube of 30-mm diameter in a thin-walled concentric tube heat exchanger of 0.19-m length. Hot process water at 95C flows in the annulus formed with the outer tube of 60-mm diameter. The deionized water is to be heated from 40 to 60C at a flow rate of 5 kg/s. The thermo physical properties of the fluids are:

kg/m3) cp(J/kg.K

DEIONIZED WATER

PROCESS WATER

982.3

967.1

4181

4197

0.643

0.673

548

324

3.56

2.02

k(W/m.K N.s/m2 pr

(a) Considering a parallel-flow configuration of the heat exchanger, determine the minimum flow rate required for the hot process water. (b) Determine the overall heat transfer coefficient required for the conditions of part a. (c) Considering a counter flow configuration, determine the minimum flow rate required for the hot process water. What is the effectiveness of the exchanger for this situation?

Schematic: Process water,h

Th,i=95 C

Th,o

T

Concentric tube

T2 mc=5kg/s

ΔT1 Tc,i=40 C

Deionized Tc,0i=60 C water,c

x

PF,L=0.19m D=30mm

0

Assumptions: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy changes. Analysis: (a) from overall energy balances, .

.

q  ( m c ) h (Th , i  Th,o )  ( m c ) h (Tc ,o  Tc , i ) .

For a fixed term Th,i , (m )h will be a minimum when Th,o is a minimum. With the parallel flow configuration, this requires that Th,o=Tc,o=60C. Hence, .

.

m h, min 

( mc ) c (Tc ,o  Tc ,i ) c h (Th,i  Th,o )

5kg / s  4181J / kg . K (60  40)  C  2.85kg / s 4197 J / kg . K (95  60)C

(b)From the rate equation and the log mean temperature relation, q  UATlm , PF

Tlm , PF 

T1 - T2 T  ln 1   T 2 

And since ΔT2=0, ΔTlm=0 so that UA=. Since A=DL is finite, U must be extremely large. Hence, the heating cannot be accomplished with this arrangement. .

(c) With the CF arrangements m h will be a minimum when Tho is a minimum. This requires that Th,o is a minimum. This requires that Th,o is a minimum. This requires that Th,o=Tc,i=40C. Hence, from the overall energy balance, .

m

5kg / s  4181J / kg . K (60  40) K  1.81kg / s 4197 J / kg . K (95  40) K

For this condition, Cmin=Ch which is cooled from Th,i to Tc,i, hence =1 Comments: For the counter flow arrangement, the heat exchanger must be infinitely long.

2. Water with a flow rate of 0.05kg/s enters an automobile radiator at 400K and leaves at 330 K. The water is cooled by air in cross flow which enters at 0.75kg/s and leaves at 300K. If the overall heat transfer coefficient is 200W/m2.K, what is the required heat transfer surface area? Schematic:

Water

Th,i=400K mh=0.05kg/s

Air Tc,i=300K mc=0.75kg/s

Tc,o

Th,o=330K

Assumptions: (1) Negligible heat loss to surroundings and kinetic and potential energy changes, (2) Constant properties. Analysis: The required heat transfer rate is .

q  ( m c ) h (Th ,i  Th,o )  0.05kg / s(4209 J / kg . K )70 K  14,732W

Using the -NTU method,

C min  C h  210.45W / K C max  C c  755.25W / K , hence , C min / C max (Th ,i  Tc ,i )  210.45W / K (100 K )  21,045W and

  q / q max  14,732W / 21,045W  0.700

From figure, NTU1.5, hence A  NTU (C min / U )  1.5  210.45W / K ( 200W / m 2 . K )  1.58m 2

Comments: (1) the air outlet temperature is Tc ,o  Tc ,i  q / C c  300 K  (14,732W / 755.25W / K )  319.5 K

(2) Using the LMTD approach, ΔTlm=51.2 K, R=0.279 and P=0.7. Hence from fig F0.95 and A  q / FUTlm  (14,732W ) /[0.95( 200W / m 2 . K )51.2 K ]  1.51m 2 .

3. Saturated steam leaves a steam turbine at a flow rate of 1.5kg/s and a pressure of 0.51 bars. The vapor is to be completely condensed to saturated liquid in a shell-and –tube heat exchanger which uses water as the coolant. The water enters the thin-walled tubes at 17C and leaves at 57C. If the overall heat transfer coefficient of 200W/m2.K, determine the required heat exchanger surface area and the water flow rate. After extended operation, fouling causes the overall heat transfer coefficient to decrease to 100W/m2.K. For the same water inlet temperature and flow rate, what is the new vapor flow rate required for complete condensation? Schematic:

Assumptions: (1) Negligible heat loss to surroundings, (2) Negligible wall conduction resistance. Properties: Table for sat.Water: 

(T c  310 K ) : c p,c  4178J / kg . K ; (p  0.51 bars) : Tsa t  355K, h fg  2304kJ/kg.

Analysis: (a) The required heat transfer rate is .

q  m h h fg  1.5kg / s( 2.304  10 6 J / kg )  3.46  10 6 W

And the corresponding heat capacity rate for the water is

C c  C min  q /(Tc ,o  Tc ,i )  3.48  10 6 W / 40 K  86,400W / K hence ,   q /(C min [Th,i  Tc ,i ])  3.46  10 6 W / 86,400W / K (65 K )  0.62 since C min /C max  0, NTU  -ln(1 -  )   ln(1  0.62)  0.97

And A  NTU (C min / U )  0.97(86,400W / K / 2000W / m 2. K )  41.9m 2 .

m c  C c / c p ,c  86,400W / K / 4178 J / kg . K  20.7 kg / s

(b) using the final overall heat transfer coefficient, find Since C min /C max  0,   1  exp( NTU )  1  exp(0.485)  0.384 hence, q  C min (Th,i  Tc ,i )  0.384(886,400W / K )65 K  2.16106W .

m h  q / h fg  2.16  10 6 W / 2.304  10 6 J / kg  0.936kg / s

.

Comments: The significant reduction (38%) in m h represents a significant loss in turbine power. Periodic cleaning of condenser surfaces should be employed to minimize the adverse effects of fouling.

4. Water at 225 kg/h is to be heated from 35 to 95C by means of a concentric tube heat exchanger. Oil at 225kg/h and 210C, with a specific heat of 2095 J/kg.K, is to be used as the hot fluid. If the overall heat transfer coefficient based on the outer diameter of the inner tube if 550W/m2.K, determine the length of the exchanger if the outer diameter is 100mm. Schematic:

Assumptions: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy changes, (3) Constant properties. Properties: Table for Water: _

(Tc  ( 35  95)  C / 2  338 K ) : c p ,c  4188J / kg . K

Analysis: From rate equation with Ao=DoL, L=q/UoDoΔT  m The heat rate, q, can be evaluated from an energy balance on the cold fluid, .

q  m c c c (Tc , 0  Tc ,i ) 

225kg / h  4188 J / kg . K (95  35) K  15,705W 3600 s / h

In order to evaluate ΔT  m, we need to know whether the exchanger is operating in CF or PF. From an energy balance on the hot fluid, find .

Th,o  Th ,i  q / m h c h  210  C  15,705W /

225kg / h J  2095  90.1C 3600 s / h kg . K

Since Th,o
T1  T2 ( 210  95)  (90.1  35)  C  81.5C n( T1 / T2 ) n(115 / 55.1)

Substituting numerical values, the HXer length is L  15,705W / 550W / m 2 . K (0.10m )  81.4 K  1.12m

Comments: The –NTU method could also be used. It would be necessary to perform the hot fluid energy balance to determining CF operation existed. The capacity rate is Cmin/Cmax=0.50. From eq. for effectiveness, and from with q evaluated from an energy balance on the hot fluid, 

Th,i  Th,o Th ,i  Tc ,i

210  90.1  0.69 210  35

T Th,i=210 C Th,o=90.1 C

ΔT1 ΔT2 Tc,o=95 C

Tc,i=35 C 1

2

x

From fig, find NTU1.5 giving L  NTU .C min / U oDo  1.5  130.94

W W 550 2 . (0.10m )  1.14m K m .K

Note the good agreement by both methods.

5. Consider a very long, concentric tube heat exchanger having hot and cold water inlet temperatures of 85 and 15C. The flow rate of the hot water is twice that of the cold water. Assuming equivalent hot and cold water specifies heats; determine the hot water outlet temperature for the following modes of operation (a) Counter flow, (b) Parallel flow. Schematic:

Tc,i=15 C Th,i=85 C

Ch=2Cc Cc

Assumptions: (1) equivalent hot and cold water specific heats, (2) Negligible Kinetic and potential energy changes, (3) No eat loss to surroundings.

Analysis: the heat rate for a concentric tube Heat exchanger with very large surface area Operating in the counter flow mode is q  q max  C min (Th,i  Tc ,i )

Combining the above relation and rearranging, find Th,o  

C C min (Th ,i  Tc ,i )  Th,i   c (Th,i  Tc ,i )  Th,i Ch Ch

Substituting numerical values

1 Th,o   (85  15)C  85C  50C 2

For parallel flow operation, the hot and cold outlet temperatures will be equal; that is Tc,o=Th,o. Hence

C c (Tc ,o  Tc ,i )  C h (Th,i  Th,o )

Setting Tc,o=Th,o and rearranging    C C  Th,o  Th,i  c Tc ,i  / 1  c  Ch    Ch  1    1 Th,o  85   15 C / 1    61.7  C 2 2   

Comments: Note that while  =1 for CF operation, for PF operation find = q/qmax=0.67.

6. A concentric tube heat exchanger uses water, which is available at 15°C, to cool ethylene glycol from 100 to 60°C. The water and glycol flow rates are each 0.5 kg/s. Determine the maximum possible heat transfer rate and effectiveness of the exchanger. Determine which is preferred, a parallel –flow or counter flow mode of operation? Known: Inlet temperatures and flow rate for a concentric tube heat exchanger. Find: (a) Maximum possible heat transfer rate and effectiveness, (b) Proffered mode of operation. Schematic: U,A

Ethylene glycol Th,i=100 C Tho=60 C mhi=0.5kg/s

Tc,i=15 C

Water

mc=0.5kg/s

Assumptions: (1) Steady-state operation, (2) Negligible KE and PE changes, (3) Negligible heat loss to surroundings, (4) Fixed overall heat transfer and coefficient. _

Properties: Table: Ethylene glycol ( T in  80C ); cp=2650J/kg.K; _

Water (Tm  30C ) : c p  4178J / kg . K

Analysis: (a) Using the -NTU method, find .

C min  C h  m h c p ,h  (0.5kg / s )( 2650J / kg . K )  1325W / K q ma x  C min (Th. i  Tc ,i )  (1325W / K )(100  15)C  1.13  10 5 W

.

q  m h c p ,h (Th. i  Tc ,i )  0.5kg / s( 2650 J / kg . K )(100  60)C  0.53  10 5 W

  q / q max  0.53  10 5 / 1.13  10 5  0.47

(b) Tc ,o  Tc ,i 

q .

m c c p ,c

 15C 

0.53  10 5  40.4C 0.5kg / s  4178 J / kg . K

Since Tc,o
.

However, with (Cmin/Cmax) = ( m c p ,h / m c c p , c ) =0.63, h

From fig (NTU)PF0.95, (NTU)CF0.75 Hence (ACF/APF)= (NTU) CF/ (NTU) PF (0.75/0.95)=0.79 Because of the reduced size requirement, hence capital investment, the counter flow mode of operation is preferred.