Solution Ch8

8/18/2019 Solution Ch8

1/15

PROBLEM 8.6

KNOWN: Water, engine oil and NaK flowing in a 20 mm diameter tube, temperature of the

fluids.

FIND: (a) The mean velocity as well as hydrodynamic and thermal entrance lengths, for a flow

rate of 0.01 kg/s and mean temperature of 366 K, (b) The mass flow rate as well as hydrodynamicand thermal entrance lengths for water and oil at a mean velocity of 0.02 m/s at mean

temperatures of 300 and 400 K.

SCHEMATIC:

ASSUMPTIONS: (1) Constant properties.

PROPERTIES:

Liquid T(K) Table ρ(kg/m3) μ(N⋅s/m2) ν(m2/s) PrWater 300 A.6 997 855 × 10-6 - 5.83

366 A.6 963 303 × 10-6 - 1.89400 A.6 937 217 × 10-6 - 1.34

Oil 300 A.5 884 48.6 × 10-2 - 6400366 A.5 844 2.12 × 10-2 - 338400 A.5 825 0.874 × 10-2 - 152

NaK 366 A.7 849 - 5.797 × 10-7 0.019

ANALYSIS: (a) The mean velocity is given by

2 2m cu = m ρA = 0.01 kg/s/ (0.020m) /4 = 31.8 kg/s m /⎡ ⎤ρπ ⋅ ρ⎣ ⎦

& (1)

The Reynolds number is

D

4m 4 × 0.01 kg/s 0.636 kg/s mRe = = =

πDμ π(0.020 m)μ μ

⋅& (2)

The hydrodynamic entrance length is

fd,h D

0.636 kg/s mx = 0.05Re D = 0.05 × × (0.020 m)

μ

⋅

-6636 × 10 kg/s m=

μ

⋅ (3)

Continued…

D = 0.020 m

Water,

NaK, or

Engine oil

D = 0.020 m

Water,

NaK, or

Engine oil


2/15

PROBLEM 8.6 (Cont.)

The thermal entrance length is

fd,t D fd,hx = 0.05Re DPr = x Pr

-6636 × 10 kg/s m

= Pr μ

⋅ (4)

Solving Equations (1), (3) and (4) yields


3/15

PROBLEM 8.17

KNOWN: Surface heat flux for air flow through a rectangular channel.

FIND: (a) Differential equation describing variation in air mean temperature, (b) Air outlet

temperature for prescribed conditions.

SCHEMATIC:

ASSUMPTIONS: (1) Ideal gas with negligible viscous dissipation and pressure variation,

(2) No heat loss through bottom of channel, (3) Uniform heat flux at top of channel.

PROPERTIES: Table A-4, Air (T ≈ 50°C, 1 atm): c p = 1008 J/kg⋅K.

ANALYSIS: (a) For the differential control volume about the air,

in outE =E& &

( ) ( ) p m o p m mm c T q w dx m c T d T′′+ ⋅ = +& &

om

p

q wd T

dx m c

′′ ⋅=

&

Separating and integrating between the limits of x = 0 and x, find

( ) ( )o

m m,i p

q w xT x T

m c

′′ ⋅= +

&

( )om,o m,i

p

q w LT T .

m c

′′ ⋅= +

&


4/15

PROBLEM 8.25

KNOWN: Inlet temperature and flowrate of oil flowing through a tube of prescribed surface temperature

and geometry.

FIND: (a) Oil outlet temperature and total heat transfer rate, and (b) Effect of flowrate.

SCHEMATIC:

ASSUMPTIONS: (1) Negligible temperature drop across tube wall, (2) Incompressible liquid with

negligible viscous dissipation.

PROPERTIES: Table A.5, Engine oil (assume Tm,o = 140°C, hencem

T = 80°C = 353 K): ρ = 852

kg/m3, ν = 37.5 × 10-6 m2/s, k = 138 × 10-3 W/m⋅K, Pr = 490, μ = ρ⋅ν = 0.032 kg/m⋅s, c p = 2131 J/kg⋅K.

ANALYSIS: (a) For constant surface temperature the oil outlet temperature may be obtained from Eq.

8.41b. Hence

( )m,o s s m,i p

DLT T T T exp h

mc

π ⎛ ⎞= − − −⎜ ⎟

⎜ ⎟⎝ ⎠&

To determine h , first calculate ReD from Eq. 8.6,

( )

( )( )D

4 0.5 kg s4mRe 398

D 0.05m 0.032kg m sπ μ π = = =

⋅

&

Hence the flow is laminar. Moreover, from Eq. 8.23 the thermal entry length is

( )( )( )fd,t Dx 0.05D Re Pr 0.05 0.05m 398 490 486m≈ = = .

Since L = 25 m the flow is far from being thermally fully developed. Since Pr > 5, h may be determined

from Eq. 8.57

DD 2/3

D

0.0668Gz Nu 3.66

1 0.04Gz= +

+.

With GzD = (D/L)ReDPr = (0.05/25)398 × 490 = 390, it follows that

D26

Nu 3.66 11.95.1 2.14= + =+

Hence,2

Dk 0.138 W m K

h Nu 11.95 33W m K D 0.05m

⋅= = = ⋅ and it follows that

Continued...


5/15

PROBLEM 8.25 (Cont.)

( ) ( ) ( ) 2m,o0.05 m 25 m

T 150 C 150 C 20 C exp 33W m K 0.5kg s 2131J kg K

π ⎡ ⎤= − − − × ⋅⎢ ⎥× ⋅⎣ ⎦

o o o

Tm,o = 35°C.


6/15

PROBLEM 8.30

KNOWN: Diameter and length of tube, air flow rate, air temperature and pressure at the tube inlet.

Surface temperature at the tube exit.

FIND: (a) The heat transfer rate of the problem. (b) Conditions at the tube exit for reduced tube

length. (c) Conditions at the tube exit for increased air flow rate.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible viscous

dissipation.

PROPERTIES: Table A.4, Air ( mT ≈ 400 K, p = 1 atm): μ = 230.1×10-7 N⋅s/m2, Pr = 0.690, k =

0.0338 W/m⋅K, c p = 1014 J/kg⋅K.

ANALYSIS: (a) We begin by calculating the Reynolds number

6

7 2

4 4 135 10 kg/s1494

0.005m 230.1 10 N s/m D

m Re

Dπ μ π

−

−

× ×= = =

× × × ⋅

&

Therefore, the flow is laminar. The hydrodynamic and thermal entrance lengths are

,0.05 0.05 0.005 m 1494 0.37 m

= = 0.37 m 0.690 = 0.26 m fd h D

fd,t fd,h

x DRe

x x Pr

= = × × =

×

Therefore, the flow is fully-developed at the tube exit. For fully-developed laminar flow with constant

heat flux conditions, the Nusselt number is Nu D = 4.36. Therefore, the local heat transfer coefficient at

the tube exit is

24.36 / 4.36 0.0338 W/m K /0.005m 29.47 W/m K h k D= = × ⋅ = ⋅

Two independent expressions for the heat flux may be written based upon application of Newton’s law

of cooling at the tube exit and an overall energy balance.

, ,, ,

( )" ( ) ;

p m o m is o m o

mc T T q h T T q" =

DLπ

−= −

& (1, 2)

Equating Eqs. (1) and (2) yields

Continued…

Air

m = 135 × 10-6 kg/s·

Tm,i = 100°C

Ts,o = 160°C

Tm,o

L = 2 m

D = 5 mm

Constant wall heat flux

Air Air

m = 135 × 10-6 kg/s·

Tm,i = 100°C

Ts,o = 160°C

Tm,o

L = 2 m

D = 5 mm

Constant wall heat flux


7/15


, , ,

62

62

135 10 kg/s 1014 J/kg K 29.47W/m K 160 C 100 C0.005 m 2 m

135 10 kg/s 1014 J/kg K

24.97W/m K 0.005 m 2 m

p pm o s o m i

mc mcT hT T h

DL DLπ π

π

π

−

−

⎡ ⎤ ⎡ ⎤= + +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦

⎡ ⎤× × ⋅⋅ × ° + × °⎢ ⎥× ×⎣ ⎦=

⎡ ⎤× × ⋅+ ⋅⎢ ⎥

× ×⎣ ⎦

& &

= 152.3°C

Hence, the heat rate is

6, ,( ) 135 10 kg/s 1014J/kg K 52.3 C = 7.16 W p m o m iq mc T T

−= − = × × ⋅ × °&


8/15

PROBLEM 8.37

KNOWN: Inner and outer diameter of a steel pipe insulated on the outside and experiencing

uniform heat generation. Flow rate and inlet temperature of water flowing through the pipe.

FIND: (a) Pipe length required to achieve desired outlet temperature, (b) Location and value

of maximum pipe temperature.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Incompressible

liquid with negligible viscous dissipation, (4) One-dimensional radial conduction in pipe wall,

(5) Outer surface is adiabatic.PROPERTIES: Table A-1, Stainless steel 316 (T ≈ 400K): k = 15 W/m⋅K; Table A-6 ,

Water ( )mT 303K := c p = 4178 J/kg⋅K, k = 0.617 W/m⋅K, μ = 803 × 10-6

N⋅s/m2, Pr = 5.45.

ANALYSIS: (a) Performing an energy balance for a control volume about the inner tube, it

follows that

( ) ( ) ( )2 2 p m,o m,i o im c T T q q /4 D D Lπ − = = −& &

( )

( ) ( )

( ) ( )

( ) ( ) ( )

p m,o m,i

2 2 2 26 3o i

m c T T 0.1 kg/s 4178 J/kg K 20 CL

q /4 D D 10 W/m / 4 0.04m 0.02mπ π

− ⋅= =

⎡ ⎤− −⎢ ⎥⎣ ⎦

o&

&

L 8.87m.=


9/15


( )2 2

2 2o oi i s i 2 2 i si i

q r q r q q r r : T r T r n r C C r n r T .

4k 2k 4k 2k = = = − + + = − +

& && &l l

The temperature distribution and the maximum wall temperature (r = r o) are

( ) ( )2

2 2 osi

i

q r q r

T r r r n T4k 2k r = − − + +

&&

l

( ) ( )2

2 2 o ow,max o o si

i

q r r q T T r r r n T

4k 2k r = = − − + +

&&l

where Ts, the inner surface temperature of the wall at the exit, follows from

( ) ( ) ( )( )

2 2 2 2o oi i

s s m,oi i

q /4 D D L q D Dq h T T

D L 4 D

π

π

− −′′ = = = −

& &

where h is the local convection coefficient at the exit. With

( )D 6 2

i4 m 4 0.1 kg/sRe 7928 D 0.02m 803 10 N s/mπ μ π −

×= = =× ⋅

&

the flow is turbulent and, with (L/Di) = (8.87 m/0.02m) = 444 >> (xfd /D) ≈ 10, it is also fullydeveloped. Hence, from the Gnielinski correlation, Eq. 8.62,

i

k Dh1/ 2 2 / 3D

21/2 2/3

(f / 8)(Re 1000) Pr

1 12.7(f /8) (Pr 1)

0.617 W/m K (0.033618)(7928 - 1000)5.45 = = 1796 W/m K

0.02 m 1 + 12.7(0.033618) (5.45 - 1)

=⎡ ⎤−⎢ ⎥⎢ ⎥+ −⎣ ⎦

⎡ ⎤⋅⋅⎢ ⎥

⎢ ⎥⎣ ⎦

where from Eq. 8.21, f = (0.790 ln ReD-1.64)-2

= 0.0336. Hence, the inner surfacetemperature of the wall at the exit is

( ) ( ) ( )( )

2 26 32 2o i

s m,o 2i

10 W/m 0.04m 0.02mq D DT T 40 C 48.4 C

4 h D 4 1796 W/m K 0.02m

⎡ ⎤−− ⎢ ⎥⎣ ⎦= + = + =× ⋅

o o&

and ( ) ( )6 3

2 2w,max

10 W/mT 0.02m 0.01m

4 15 W/m K

⎡ ⎤= − −⎢ ⎥⎣ ⎦× ⋅

( )26 310 W/m 0.02m 0.02n 48.4 C 52.6 C.

2 15 W/m K 0.01+ + =

× ⋅o o

l


10/15

PROBLEM 8.46

KNOWN: Surface temperature and diameter of a tube. Velocity and temperature of air in

cross flow. Velocity and temperature of air in fully developed internal flow.

FIND: Convection heat flux associated with the external and internal flows.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform cylinder surface temperature, (3)

Fully developed internal flow, (4) For internal flow, air is an ideal gas with negligible viscous

dissipation and pressure variations.

PROPERTIES: Table A-4, Air (336 K): ν = 19.51 × 10-6 m2/s, k = 0.0290 W/m⋅K, Pr =0.702.

ANALYSIS: For the external and internal flows,

4mD -6 2

u DVD 30 m/s 0.05 mRe 7.69 10 .

19.71 10 m / s

×= = = = ×

×ν ν

From the Churchill-Bernstein relation for the external flow,

4 / 51/ 2 1/ 3 5 /8D D

D1/ 4

2 / 3

4 / 55 /8

4 1/ 2 1/ 3 4

1/ 42 /3

0.62Re Pr Re Nu 0.3 1

282,0001 (0.4 / Pr)

0.62(7.69 10 ) 0.702 7.69 10 0.3 1 180

282,0001 (0.4/ 0.702)

⎡ ⎤⎛ ⎞⎢ ⎥= + + ⎜ ⎟

⎢ ⎥⎝ ⎠⎡ ⎤ ⎣ ⎦+⎢ ⎥⎣ ⎦

⎡ ⎤⎛ ⎞× ×⎢ ⎥⎜ ⎟= + + =⎢ ⎥⎜ ⎟

⎡ ⎤ ⎝ ⎠+ ⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦

Hence, the convection coefficient and heat flux are

2D

k 0.0290 W/m K h Nu 180 104 W/m K

D 0.05 m

⋅= = × = ⋅

( ) ( )2 2sq h T T 104W/m K 100 25 C 7840 W/m .∞′′ = − = ⋅ − =o


11/15


and the heat flux is

( ) ( )2 2s mq h T T 94 W/m K 100 25 C 7040 W/m .′′ = − = ⋅ − =o


12/15

PROBLEM 8.48

KNOWN: Diameter and length of circular tube, liquid water flow rate, liquid water entrance

temperatures and tube surface temperatures.

FIND: Water outlet temperatures for (a) T m,i = 500 K, T s = 510 K and (b) T m,i = 300 K, T s = 310 K. (c)

Discuss whether the flow is laminar or turbulent for T m,i = 300 K, T s = 647 K.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties in parts (a) and (b), (3)

Negligible viscous dissipation.

PROPERTIES: Table A.6 , liquid water ( mT = 505 K, assumed): μ = 115.5×10-6 N⋅s/m2, Pr = 0.855,

k = 0.635 W/m⋅K, c p = 4700 J/kg⋅K. Liquid water ( mT = 305 K, assumed): μ = 769×10-6 N⋅s/m2, Pr =

5.20, k = 0.620 W/m⋅K, c p = 4178 J/kg⋅K.

ANALYSIS: (a) We begin by calculating the Reynolds number

6 2

4 4 0.1 kg/s11,014

0.1m 115.5 10 N s/m D

m Re

Dπ μ π −×

= = =× × × ⋅

&

Therefore, the flow is in a fully turbulent condition. Since L/D = 6m/0.1m = 60, we conclude thatentrance effects are not important. We may use Dittus-Boelter (Eq. 8.60) to determine the average heat

transfer coefficient and the mean outlet temperature may be found from Eq. (8.41b).

4 /5 0.4 4/5 0.4 2D

0.635W/m K 0.023 0.023 11,014 0.855 235 W/m K

0.1m

k h Re Pr

D

⋅⎡ ⎤ ⎡ ⎤= = × = ⋅⎣ ⎦ ⎣ ⎦

, ,

2

( )exp

0.1 m 6 m 510 K 10 K exp 235 W/m K 506.1 K

0.1 kg/s 4700 J/kg K

m o s s m i p

PLT T T T h

mc

π

⎛ ⎞= − − −⎜ ⎟⎜ ⎟

⎝ ⎠

⎛ ⎞× ×= − × − ⋅ =⎜ ⎟

× ⋅⎝ ⎠

&


13/15


Therefore, the flow is laminar. The thermal entrance length is x fd,t = 0.05 × D × Re D × Pr = 0.05 × 0.1m× 1655 × 5.20 = 43.0 m > L. Therefore, we expect entrance effects to be significant. With Pr > 5, we

may use Eq. (8.57) with Eq. (8.56) for the Graetz number, to estimate the value of h .

[ ]

[ ]

2 /3

2

2/ 3

0.0668( / )3.66

1 0.04 ( / )

0.620W/m K 0.0668(0.1 m / 6 m) 1655 5.20 = 3.66 51.0 W/m K

0.1 m 1 0.04 (0.1 m/ 6.0 m) 1655 5.20

D

D

k D L Re Pr h

D D L Re Pr

⎧ ⎫⎪ ⎪= +⎨ ⎬+⎪ ⎪⎩ ⎭

⎧ ⎫⋅ × ×⎪ ⎪+ = ⋅⎨ ⎬

+ × ×⎪ ⎪⎩ ⎭

Using Eq. (8.41b)

, ,

2

( )exp

0.1 m 6 m 310 K 10 K exp 51 W/m K 302.1 K

0.1 kg/s 4178 J/kg K

m o s s m i p

PLT T T T h

mc

π

⎛ ⎞= − − −⎜ ⎟⎜ ⎟

⎝ ⎠

⎛ ⎞× ×= − × − ⋅ =⎜ ⎟

× ⋅⎝ ⎠

&


14/15

PROBLEM 8.51

KNOWN: Gas-cooled nuclear reactor tube of 20 mm diameter and 780 mm length with helium

heated from 600 K to 1000 K at 8 × 10-3

kg/s.

FIND: (a) Uniform tube wall temperature required to heat the helium, (b) Outlet temperature and

required flow rate to achieve same removal rate and wall temperature if the coolant gas is air.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Ideal gas with negligible viscous dissipation and

pressure variation, (3) Fully developed conditions.

PROPERTIES: Table A-4, Helium ( )mT 800K, 1 atm := ρ = 0.06272 kg/m3, c p = 5193 J/kg⋅K, k =

0.304 W/m⋅K, μ = 382 × 10-7

N⋅s/m2, ν = 6.39 × 10

-4 m

2/s, Pr = 0.654; Air ( )mT 800K, 1 atm := ρ

= 0.4354 kg/m3, c p = 1099 J/kg⋅K, k = 57.3 × 10

-3 W/m⋅K, ν = 84.93 × 10

-6 m

2/s, Pr = 0.709.

ANALYSIS: (a) For helium and a constant wall temperature, from Eq. 8.41b,

s m,o

s m,i p

T T PLhexp

T T m c

⎛ ⎞−= −⎜ ⎟

⎜ ⎟− ⎝ ⎠&

where P = πD. For the circular tube,

34

D -7 2

4m 4 8 10 kg/sRe 1.333 10

D 0.020 m 382 10 N s/mπ μ π

−× ×= = = ×

× × × ⋅

&

and using the Dittus-Boelter correlation for turbulent, fully developed flow,

( ) ( )4 / 5 0.44/5 0.4 4

D Nu 0.023 Re Pr 0.023 1.333 10 0.654 38.7= = × =

2h Nu k/D 38.7 0.304 W/m K/0.02 m 588 W/m K.= ⋅ = × ⋅ = ⋅

Hence, the surface temperature is

( ) 2s-3

s

0.020 m 0.780 m 588 W/m K T 1000 K exp 0.500

T 600 K 8 10 kg/s 5193 J/kg K

π ⎡ ⎤× × ⋅−⎢ ⎥= − =

− ⎢ ⎥× × ⋅⎣ ⎦

sT 1400 K.=


15/15


(b) For the same heat removal rate (q) and wall temperature (Ts) with air supplied at Tm,i, the relevant

relations are

( )a p m,o m,iq 16,620 W m c T T= = −& (1)

s m,o as m,i a p

T T PLhexpT T m c

⎡ ⎤− = −⎢ ⎥− ⎢ ⎥⎣ ⎦&

(2)

4/5 0.4aD

4m hDRe 0.023 Re Pr

D k π μ = =

& (3,4)

where Tm,o and m& are unknown. An iterative solution is required: assume a value of Tm,o and find

&m from Eq. (1); use m& in Eqs. (3) and (4) to find h and then Eq. (2) to evaluate Tm,o; compare

results and iterate. Using thermophysical properties of air evaluated at mT = 800K, the above

relations, written in the order they would be used in the iteration, become

am,o

15.1mT 600

=−

& (5)

4/5a ah 5600m= & (6)

( )-5m,o a aT 1400 K 800 K exp 4.459 10 h / m⎡ ⎤= − × − ×⎢ ⎥⎣ ⎦& (7)

Results of the iterative solution are

Trial Tm,o (K) m& (kg/s) ( )2ah W/m K ⋅ Tm,o (K)(Assumed) Eq. (5) Eq. (6) Eq. (7)

1 1000 3.781 × 10-2

407 905

2 950 4.321 × 10-2

453 899

3 900 5.041 × 10-2

513 891

4 890 5.215 × 10-2

527 890

Hence, we find

2a m,om 5.22 10 kg/s T 890 K.

−= × =&

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Solution Ch8