8/9/2019 Meen461spring2015 Hw1 Solution

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PROBLEM 1.4

KNOWN: Dimensions, thermal conductivity and surface temperatures of a concrete slab. Efficiencyof gas furnace and cost of natural gas.

FIND: Daily cost of heat loss.

SCHEMATIC:

ASSUMPTIONS: (1) Steady state, (2) One-dimensional conduction, (3) Constant properties.

ANALYSIS: The rate of heat loss by conduction through the slab is

( ) ( )1 2T T 7 C

q k LW 1.4 W / m K 11m 8m 4312 Wt 0.20m

= = =

8/9/2019 Meen461spring2015 Hw1 Solution

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PROBLEM 1.21

KNOWN: Long, 30mm-diameter cylinder with embedded electrical heater; power required

to maintain a specified surface temperature for water and air flows.

FIND: Convection coefficients for the water and air flow convection processes, hwand ha,

respectively.

SCHEMATIC:

ASSUMPTIONS: (1) Flow is cross-wise over cylinder which is very long in the direction

normal to flow.

ANALYSIS: The convection heat rate from the cylinder per unit length of the cylinder has

the form

( ) ( )q = h D T Ts

and solving for the heat transfer convection coefficient, find

( )q

h = .D T Ts

Substituting numerical values for the water and air situations:

Water

( )

28 10 W/mh = = 4,570 W/m K

0.030m 90-25 C

32

w

o

8/9/2019 Meen461spring2015 Hw1 Solution

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PROBLEM 1.28

KNOWN: Length, diameter, surface temperature and emissivity of steam line. Temperatureand convection coefficient associated with ambient air. Efficiency and fuel cost for gas firedfurnace.

FIND: (a) Rate of heat loss, (b) Annual cost of heat loss.

SCHEMATIC:

ASSUMPTIONS: (1) Steam line operates continuously throughout year, (2) Net radiationtransfer is between small surface (steam line) and large enclosure (plant walls).

ANALYSIS: (a) From Eqs. (1.3a) and (1.7), the heat loss is

( ) ( )4 4conv rad s s sur q q q A h T T T T = + = +

where ( ) 2A DL 0.1m 25m 7.85m .= = =

Hence,

( ) ( )2 2 8 2 4 4 4 4q 7.85m 10 W/m K 150 25 K 0.8 5.67 10 W/m K 423 298 K = +

( ) ( )2 2q 7.85m 1,250 1,095 W/m 9813 8592 W 18,405 W= + = + =

8/9/2019 Meen461spring2015 Hw1 Solution

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PROBLEM 1.81

KNOWN: Conditions associated with surface cooling of plate glass which is initially at 600C.Maximum allowable temperature gradient in the glass.

FIND: Lowest allowable air temperature, T.

SCHEMATIC:

ASSUMPTIONS: (1) Surface of glass exchanges radiation with large surroundings at Tsur= T, (2)

One-dimensional conduction in the x-direction.ANALYSIS: The maximum temperature gradient will exist at the surface of the glass and at the

instant that cooling is initiated. From the surface energy balance, Eq. 1.13, and the rate equations,

Eqs. 1.1, 1.3a and 1.7, it follows that

( ) ( )4 4s s sur dT-k h T T T T 0dx

=

or, with (dT/dx)max= -15C/mm = -15,000C/m and Tsur= T,

( )C

2

W W1.4 15,000 5 873 T K

m K mm K

=

o

8 4 4 42 4

W0.8 5.67 10 873 T K .

m K

+

Tmay be obtained from a trial-and-error solution, from which it follows that, for T= 618K,

21 000 1275 19 730, , .W

m

W

m

W

m2 2 2 +

Hence the lowest allowable air temperature is

T K = 345 C. 618 o

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PROBLEM 2.5

KNOWN: Symmetric shape with prescribed variation in cross-sectional area, temperature

distribution and heat rate.

FIND: Expression for the thermal conductivity, k.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x-direction, (3)

No internal heat generation.

ANALYSIS: Applying the energy balance, Eq. 1.12c, to the system, it follows that, since

,E Ein out

xq Constant f x .

Using Fouriers law, Eq. 2.1, with appropriate expressions for A xand T, yields

x x

2 3

dTq k Adx

d K6000W=-k 1-x m 300 1 2x-x .

dx m

Solving for k and recognizing its units are W/mK,

22-6000 20

k= .1 x 2 3x1-x 300 2 3x