ME3122E - Tutorial Solution 4

Problem Set 4-Solutions 1. Hot exhaust gases used in a finned-tube cross-flow heat exchanger heat 2.5kg/s of water from 35 to 85oC. The gases [cp = 1.09 kJ/kg.K] enter at 200oC and leave at 93oC. The overall heat-transfer coefficient is 180W/m2K. Calculate the area of the heat exchanger using (a) the LMTD approach and (b) the effectiveness – NTU method. (c) If the water flow rate is reduced by half, while the gas flow rate is maintained constant along with the fluid inlet temperatures. Calculate the percentage reduction in heat transfer as a result of this reduced flow rate. Assume that the overall heat transfer coefficient remains the same. [Ans: (a) 37.8 m2 (b) 37.8 m2 (c) 15%.]

Solution: a)

TLMTD 

T1  T2 ln(T1 T2 )

(200  85)  (93  35)  ln(115 / 58)

Hot gases T1=2000C Water t1=350C m =2.5

t2 =850C U=180W/m2K

 83.3 C P

t2  t1 85  35   0.303 T1  t1 (200  35)

T2 =930C

R

T1  T2 200  93   2.14 t2  t1 (85  35)

Th1 200

T

From crossflow hx chart : F  0.92  t  UA.F .TLMTD Q  mC  2.5  4180  50  0.92  180  A  83.3 A  37.8m

Tc1 85

Th2 93 Tc2 35

2

1

2

1

A

b)

Energy balance : m g c g (200  93)  2.5  4180  (85  35)  m g c g  4883  C min ( gas ) m w c w  2.5  4180  10450  C max ( water ) C min C max

g 



4883  0.467 2.5  4180

200  93  0.648 200  35

using   NTU chart  NTU  A

UA  1.4, C min

1.4  4883  37.98 m 2 180

c)

mg cg  4883  Cmin m wcw  2.5  4180 / 2  5225  Cmax NTU 

UA  1.4 (no change) Cmin

Cmin  4883 / 5225  0.935 Cmax Cross flow hx chart    0.55 Tg  200  35 Tg  90.75  200  Tg 2 Tg 2  109.25 C Q  C g Tg  4883  90.75 Q(a )  Q(c) 4.18  2.5  50  4883  90.75   15% 4.18  2.5  50 Q(a )

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2. A shell-and-tube heat exchanger operates with two shell passes and four tube passes. The shell fluid is ethylene glycol, which enters at 140oC and leaves at 80oC with a flow rate of 4500 kg/h. Water flows in the tubes, entering at 35oC and leaving at 850C. The overall heat-transfer coefficient for this arrangement is 850 W/m2K. Calculate the flow rate of water required and the area of the heat exchanger. [Ans: 0.984 kg/s, 5.24 m2]. The flow rate of glycol to the exchanger is reduced in half with the entrance temperatures of both fluids remaining the same. What is the water exit temperature under these new conditions, and by how much is the heat-transfer rate reduced? [Ans: 70.9oC, 28.2%]

Solution:

i ) m g  4500 / 3600kg / s, cg  2.742kJ / kgK Q

4500  2.742  (140  80)  205.7 kW 3600  m w  4.180  (85  35)

 m w  0.984kg / s

T

140

Cw  0.984  4.180 103  4113  Cmax 4500  2742  3428  Cmin 3600 Cmin 3428    0.833 Cmax 4113 Cg 

140  80  0.571 140  35 Hence   NTU chart gives 

g 

Ethylene glycol 85

80 water 1

35 A

2

NTU  UA / Cmin  1.3 given U  850W/m 2o C.  A  1.30  3428 / 850  5.24m 2

3

ii ) m g  4500 /(2)  3600 Cg  3428 / 2  Cmin 

UA 1.30   2.6  NTU Cmin 0.5 Cmin Cmax

 0.833 / 2  0.417

chart    0.82  g 

Tg1  Tg 2

 0.82; 140  35 140  Tg 2  86.1

Tg 2  53.9 oC Q  Cg Tg  (3428 / 2)  86.1  147.6kW Q(i )  Q(ii ) 205.7  147.6   28.2% 205.7 Q(i ) 147.6  35.9  Tw1  35 0.984  4.18 Tw1  35  35.9  70.9 C Twater 

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3. A small steam condenser is designed to condense 0.76 kg/min of steam at 85 kN/m2 with cooling water at 10oC. The exit water temperature is not to exceed 57oC. The overall heat-transfer coefficient is 3400 W/m2K. Calculate the area required for a double-pipe heat exchanger. Use both LMTD and effectiveness-NTU methods. [Ans: 0.145m2].

Solution:

T

Steam Tsat =950C

i) Steam is condensed at 85KN/m 2 h fg =2.27 MJ/kg

57

 h fg  2270   Q=m TLMTD  =

0.76 = 28.75kW 60

water

T1  T2 ln(T1 / T2 ) (95-10)-(85-57)  58.3 C ln(85 / 38)

10 A 1

2

Q=UA LMTD 28.75  A=  0.145m 2 3.4  58.3

ii) C max  steam, Cmin  water  =

Tc2  Tc1 Th1  Tc1



A

57  10  0.553 95  10

For heat exchanger with steam condensing,   1  e  NTU   =0.553=1  e  NTU , giving NTU =0.805 NTU=

UA  0.805, Cmin

Q=Cmin (57  10)  28753  Cmin  611.77 Substituting qives A=0.145m 2

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4. A shell-and-tube heat exchanger consists of 135 thin-walled tubes in a doublepass arrangement, each of 12.5 mm diameter with a total surface area 47.5 m2. Water (the tube-side fluid) enters the heat exchanger at 15oC and 6.5 kg/s and is heated by exhaust gas entering at 200oC and 5 kg/s. The gas may be assumed to have the properties of atmospheric air, and the overall heat transfer coefficient is approximately 200 W/m2. What are the gas and water outlet temperatures? Assuming fully developed flow, what is the tube-side convection coefficient? [Ans: To(water) = 41.6oC, h = 2320 W/m2K.] Solution: i)

m g  5kg / s ; c pg  1005 J / kgK m w  6.5kg / s; c pw  4200 J / kgK U  200W / m2 K ; A  47.5m2 Cg  5 1005  Cmin

T

2000C Gas Tg2

Tw1

Cw  6.5  4200  Cmax Cmin 5 1005   0.184 Cmax 6.5  4200

water

200  47.5  1.89 5 1005 Shell and tube Heat Exchanger NTU 

1

15 2

 -NTU chart    0.78 

Cg (200  Tg2 ) Cmin (200  15)



200  Tg2 200  15

 0.78

Tg2  55.7 C Q  5 1005  (200  55.7)  6.5  4200  (Tw1  15) Tw1  41.6 C

6

A

ii)

Water in tube: properties are evaluated at 15  41.6  28.3 C Tbw  2 d2  ud Re d  ; m   u     4  4m 4  (6.5 /135)   d   (12.5  103 )  855 106 =5736>2000  turbulent tube flow

 Re d 

Nu=

hd  0.023Re0.8 Pr 0.4 kw

Substituting for Pr=5.83, k w  0.613W / mK , d  0.0125m gives h=2320W/m 2 K

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5. A single-pass cross-flow heat exchanger uses hot exhaust gases (mixed) to heat water (unmixed) from 30oC to 80oC at a rate of 3 kg/s. The exhaust gases, having thermophysical properties similar to air, enter and exit the exchanger at 225oC and 100oC, respectively. If the overall heat transfer coefficient is 200W/m2K, estimate the required surface area. [Ans: 35.2m2]. Solution:

m w  3kg / s; cw  4.2 KJ / kgK , cg  1.005 KJ / kgK Use LMTD Method, counter flow equivalent  all the temperatures are known  wcw (Tw 2  Tw1 )  3  4.20  (80  30)  630kJ/s q=m (225-80)-(100-30)  103 C ln(145/70) 225-100  2.5 R(or Z)= 80-30 80  30  0.26 P 225  30 for single pass, crossflow hx, 1mix, 1unmixed, TLMTD =

T

2250C Gas 100

80

water 1

30 A

2

correction factor F=0.92 Hence q=630=UAFTLMTD =0.200  A  0.92 103 giving A=33.2m 2

  NTU method just as straight forward, Cmin  Cgas  m g cg , Cmax  Cwater  m w cw Calculate



Cmin Cmax

Tg1  Tg2 Tg1  Tw2



225  110  0.64 225  30

from chart , get NTU , hence A

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6. The oil in an engine is cooled by air in a cross-flow heat exchanger where both fluids are unmixed. Atmospheric air enters at 30oC and 0.53 kg/s. Oil at 0.026 kg/s enters at 75oC and flows through a tube of 10 mm diameter. Assuming fully developed flow and constant wall heat flux, estimate the oil-side heat transfer coefficient. If the overall convection coefficient is 53 W/m2K and the total heat transfer area is 1 m2, determine the effectiveness. What is the exit temperature of the oil? [Ans: 46.2oC]

Solution:

cross-flow hx, fluids unmixed oil inside tube, 75 C (inlet ), 0.026kg/s air, 30 C (inlet), 0.53kg/s U=53W/m 2 K , A  1m 2 Constant heat flux for tube, find h(oil side) Another case of To2 (outlet) unknown assume To2  45 C  Tbo  (75  45) / 2  60  C (Other procedure may lead to non-integer value of Tb oil ) Tables: engine oil @ 60  C :

  864 kg / m 3 , C p  2.047 kJ / kgK ,   0.839  10 4 m 2 / s , k  0.14W / mK Pr  1050

9

Re 

 ud 4m    d

4  0.026   0.01 864  (0.839 104 )  45.7  2000 flow inside tube is laminar, constant heat flux hd h  0.01 Nu= 4.36 = ;  k 0.14 hoil  61.04W / m 2 K 

 h ch = 0.026×2047=53.2=Cmin oil: m  c cc = 0.53×1005=533=Cmax air: m Cmin  0.10 ; Cmax NTU 

UA 53 1   1.0 Cmin 53.2

From ε-NTU chart for cross flow hx,both fluid unmixed   =0.64 Cmin (Th1  Tho ) 75  Tho   =0.64= Cmin (Th1  Tc1 ) 75  30 Tho  75  0.64  45  46.2 C (not far from assumed value of 45)

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7. Water flows over a 3-mm-diameter sphere at 6 m/s. The free-stream temperature is 38oC, and the sphere is maintained at 93oC. Calculate the heat-transfer rate. [Ans: 84.9W].

Solution: at T =38o C, properties for water

 =993kg/m3 ,   0.682 103 kg / ms k  0.63W / mK , Pr  4.55

Water u=6m/s T=380C

Ts=930C

d=3mm

Should not be evaluated at Tf =38+93/2=65.50C because Nu eqn is evaluated at free steam condition for liquid (water) Re 

993  0.003  6  26, 210  2000 0.682  103

appropriate Nu expression is 0.25

  NuPr  w   1.2  0.53Re 0.54  130.09    μ w evaluated at 93C ,  w  3.06  104 kg / ms -0.3

Nu  250.4 

hd k

0.63  250.42  52589W / m 2 K 0.003 Q  (4 r 2 )h(93  38)  84.9W

h

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8. Air at 90oC and 1 atm flows past a heated 1/16-in-diameter wire at a velocity at 6 m/s. The wire is heated to a temperature of 150oC. Calculate the heat transfer per unit length of wire. [Ans: 59.86 W/m].

Solution:

Air at 1 atm T  90 C, u  6m / s First determine Re of the flow Air properties are evaluated at the film temperature T  Tw 90  150   120 C  393K 2 2 P 1.01325 105   0.898kg / m3 f  287  393 RTf

Tf 

cp f  1.013kJ / kgK;  f  2.256 105 kg / ms k f  0.03314W / mK ; Pr  0.690

Red 

u d 0.898  6  (1.5875 103 )   379.1 2.256 105 

From table : C  0.683, n  0.466 n

hd u d   C    Pr1/3  0.683(Red )0.466 Pr1/3 k    convection heat transfer coefficient

 Nud 

k h  .0.683(Red )0.466 Pr1/3 d 0.03314   0.683 (379.1)0.466  0.6900.33 3 1.5875 10  200.5W / m2 K Heat transfer from wire to air q  hA(Tw  T )  h .( dl )(Tw  T ) q  h .( d )(Tw  T ) l  200.5   (1.5875 103 )(150  90)

q" 

 60.0W / m

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