8/2/2019 Heat Transfer Course Work 1

1/7

UNIVERSITY OF NOTTINGHAM

DEPARTMENT OF ARCHITECTURE AND BULT ENVIRONMENT

AUTUMN SEMESTER 2011-2012

Heat Transfer for Engineering-K13THT

Name: Jie JIN

ID: 4143804

Date: Nov.14th, 2011

8/2/2019 Heat Transfer Course Work 1

2/7

Nomenclature

Coefficient of volume expansion

TTemperature of PV panel fluid

T Temperature of the heat sink surfaceT Temperature of room air Thermal conductivity

Pr Prandtl number

v Kinematic viscosity

Ra Rayleigh number iGr Grashof number

g Acceleration of gravity, 9.8 m/s2 Characteristic length

Nu Nusselt number Heat transfer coefficient

Q Rate of heat transfer

Dynamic viscosity

Specific volume of water

Cp Specific heat capacity at constant p (pressure) Density of fluid

A Area of Flow A

A Area of SurfaceR Thermal Resistance

8/2/2019 Heat Transfer Course Work 1

3/7

1

Q1

Assumption: 1. the pressure is at 1 atm; 2. Air is ideal gas; 3. Steady operating conditions

exist; 4. the coefficient of volume expansion = 1/TSolution:

T = T+T 2 = 80+ 20 2 = 50 = 323K and at 1 atmBy the property table:

If T = 300 K = 2.62410kW/mKIf T = 325 K = 2.81610kW/mK

According to the linear relationship, if T = 323 K,

= 2.810kW/mKSimilarly,

Pr =0.7015

v = 1.788 10m/sThen

= 1/T = 1323 = 0.0031 K

Inclined Plate

The characteristic length = 1.2 m, the Rayleigh number is

Ra = Gr Pr = g T T

v Pr= 9.80.0031 80 20 1.2 0.7015 1.78810 = 6.911510

Based on the recommended empirical equations Ra Range: 109-10

13

8/2/2019 Heat Transfer Course Work 1

4/7

2

There is no need replacing g by g cos

and Nu = 0.1 Ra/ = 0.1 6.911510/ = 190.48

=

Nu/

= 2.8 10

190.48 1.2kW/m

K = 4.44 W/m

KHence,Q = AT T = 4.44 1.2 1 8 0 2 0 = 320 W

Q2

Solution:

T = T + T 2 = 15 + 85 2 = 50By the property table: = 0.101210 m/kg

= 64310kW/mK = 544 10kg/ms

Cp = 4.182 kJ/kg K(1) Reynolds number

Re = u D

Where,

- = 1 = 1 0.101210 m/kg = 988.14 kg/m- mass flow rate = 0.15 kg/s u = mass flow rate A = 0.15

988.14 3.14 0.0452m/s = 0.0955 m/s- D = Internal Diameter of Tube for circle cylinder tube, which is 0.045 m here

Hence,

Re = u D

= 988.140.09550.045544 10 = (2) Prandtl number

Pr = Cp

= 54410 4.182

64310 = .

8/2/2019 Heat Transfer Course Work 1

5/7

3

(3) Stanton numberSt =

u Cp

Q

= m CpT T =

AT T

And m = Au

St = u Cp =

AT TAT T =

D4 DL

T TT T =

DT T4LT T

= 0.04585154 1 5 9 0 5 0 = .

(4) Nusselt numberNu = Re Pr St = 7806 3.538 1.3125 10 =36.248

Q3

Solution:

(1) Overall heat transfer coefficient between the hot and cool fluidsAssume: 1. the overall heat transfer coefficient is based on the outside area of tube; 2.the

thermal conductivity of copper tube is p; 3. the length of tube is L.

R = 1aA =1

aDL

Rp =InDD

2p

L

8/2/2019 Heat Transfer Course Work 1

6/7

4

R =InDD2L

R = 1

aDL

Then,

R = R + Rp + R + R = 1aDL +InDD

2pL +InDD2L +

1aDL

Hence, the overall heat transfer coefficient based on the outside area of tube is

U =

1

A R =

1

DL 1aDL + In DD2pL + In D

D2L + 1aDL

= 1

D 1aD +In DD2p +

In DD2 +1

Da

=

+ +

+

+

+

( + + + )

+

+ +

(2) Critical thickness of insulation (neglecting the copper wall conduction resistance)Q = T TR =

T T1

aDL +InDD2L +

1aDL

The value D at which Q reaches a maximum is determined from the requirement thatdQ/dD = 0 (zero slope). Perform the differentiation and solving for D yields the criticaldiameter of insulation for the tube as following.

dQdD =

T TL[ 12D 1

aD]

1aDL +In DD2L +

1aDL

= 0

so 12D 1aD = 0 i.e.When D = 2a , Q reach its maximum.

8/2/2019 Heat Transfer Course Work 1

7/7

5

Hence,

critical thickness of insulation = D D 22 =

The variation of Q with the outer radius of the insulation D can be plotted.