ENGR1166 8 Hwk7 Solutions PDF

Preview

Summary

Download ENGR1166 8 Hwk7 Solutions PDF

Description

ENGR 1166 #8 – Spring 2016 - Heat Transfer - Solutions

1. What is one dimensional conductive heat transfer and what are the necessary conditions required for this approximation to be a realistic one? One dimensional heat transfer means that a temperature gradient exists in only one direction. An example that was brought up multiple times is a large wall with constant temperature or heat flux boundary conditions on the sides. It is important to have a large wall because the effects of the top and bottom edges decrease when the wall is large. 2. What are the three modes of heat transfer? How do they differ from one another? (Explain both conceptually as well as with equations.) The three modes of heat transfer are: conduction, convection, and radiation. Conduction operates within a 𝑑𝑇 ′′ material or between materials where the one dimensional conductive heat flux is 𝑞𝑐𝑜𝑛𝑑,𝑥 = −𝑘 𝑑𝑥. Convection

′′ = ℎ(𝑇𝑠 − 𝑇∞ ). Finally operates between a fluid moving relative to a surface, the convective heat flux is 𝑞𝑐𝑜𝑛𝑣 radiation operates w/out a media. The radiative heat flux is determined as the net heat flux of the emitted and ′′ absorbed radiation i.e. 𝑞𝑟𝑎𝑑 = 𝜖𝜎(𝑇𝑠4 − 𝑇∞4 ).

3. In class we looked mainly at steady state heat transfer conditions. We briefly described the transient case by looking at the temperature profile of conduction through a wall. Imagine a wall that starts at a cold temperature 𝑇𝐶 . The right side of the wall is held at temperature 𝑇𝐶 and held there for all time. At time 𝑡 = 0 seconds the left wall is raised to temperature 𝑇𝐻 and held there for all time. Draw the temperature profile of the wall below for 3 different times (𝑡𝑜 , 𝑡1 , 𝑡∞ ). 𝑡𝑜 is the time 𝑡 = 0 seconds where the entire wall is at temperature 𝑇𝐶 except for the left wall. 𝑡∞ is the steady state time and 𝑡1 is a time between 𝑡𝑜 and 𝑡∞ . 𝑡0

𝑡1

𝑡∞

𝑇𝐻

𝑇𝐻

𝑇𝐻

𝑇𝐶

𝑇𝐶

𝑇𝐶

4. The radiation case was specified to be unique because the heat flux equations are not 1st order. For certain problems we did in class this did not pose as mathematically more difficult. For other problems this required an iterative process. What is the difference between these two types of problems? I.e. how can one identify if a heat transfer problem involving a radiative process requires an iterative process to solve? In the radiation heat flux equation the surface temperature and air temperature are raised to the 4th power. If one of these is the unknown and there is another surface/air temperature as part of the equation (i.e. convective or

conduction term) this would require an iterative process. However if one of these temperatures is the unknown but it does not show up anywhere else in the equation this is a mathematically simple problem. Also if the surface/air temperatures are not the unknown, this problem is again mathematically simple. 5. A concrete slab of a basement floor is 11 meters long, 8 meters wide, and 20 centimeters thick. During the winter, the top of the slab is held at a constant temperature of 17oC by a furnace that burns natural gas. The bottom of the slab is underground and remains at 10oC.

a. If the concrete has a thermal conductivity of 1.4 (Assume steady state conditions.) 𝑞𝑐𝑜𝑛𝑑,𝑥 = −𝑘𝐴

𝑊

𝑚𝐾

what is the rate of heat loss through the slab?

𝑑𝑇 𝑊 10𝑜 𝐶 − 17𝑜 𝐶 = − (1.4 ) (11𝑚 × 8𝑚) ( ) = 4312 𝑊 𝑑𝑥 𝑚𝐾 0.2 𝑚

b. If natural gas is $0.01/MJ what is the daily cost of maintaining the top surface at 17oC? Cost/day = 𝑞𝑐𝑜𝑛𝑑,𝑥 (

𝑋 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 𝑑𝑎𝑦

= (4312 𝑊) (

)(

1𝐽/𝑠 1𝑀𝐽 $0.01 ) ( 106𝐽) ( 1𝑊 ) 𝑀𝐽

24 ℎ𝑟𝑠 60𝑚𝑖𝑛 60𝑠𝑒𝑐 $0.01 1𝑀𝐽 1𝐽/𝑠 )( )( )( )( 6 )( ) = $3.73/𝑑𝑎𝑦 1 𝑑𝑎𝑦 1ℎ𝑟 1min 𝑀𝐽 10 𝐽 1𝑊

c. How much more would it cost per day to maintain the top surface at 1oC higher? First solve for the rate of heat loss if the top surface was at 18𝑜 𝐶 instead: 𝑑𝑇 𝑊 10𝑜 𝐶 − 18𝑜 𝐶 𝑞𝑐𝑜𝑛𝑑(1𝑜 𝐶 𝐻𝑖𝑔ℎ𝑒𝑟),𝑥 = −𝑘𝐴 ) = 4928 𝑊 = − (1.4 ) (11𝑚 × 8𝑚) ( 0.2 𝑚 𝑑𝑥 𝑚𝐾 Increase Cost/day = Δ𝑞𝑐𝑜𝑛𝑑,𝑥 ( = (4928𝑊 − 4312 𝑊) (

𝑋 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 𝑑𝑎𝑦

)(

$0.01 ) 𝑀𝐽

24 ℎ𝑟𝑠 60𝑚𝑖𝑛 60𝑠𝑒𝑐 $0.01 1𝑀𝐽 1𝐽/𝑠 ) = $0.53 𝑚𝑜𝑟𝑒/𝑑𝑎𝑦 )( )( )( )( 6 )( 1 𝑑𝑎𝑦 1ℎ𝑟 1min 𝑀𝐽 10 𝐽 1𝑊

6. A 25 meter long uninsulated industrial steam pipe of 100 mm diameter is routed overhead through a building whose walls and air are at 25oC. Pressurized steam maintains a pipe surface temperature of 150oC, and the coefficient associated with natural convection is ℎ =

10𝑊 . 𝑚2 𝐾

The surface emissivity is 𝜖 = 0.8.

a. What is the rate of heat loss from the steam line? 4 ) 𝑞 = 𝑞𝑐𝑜𝑛𝑣 + 𝑞𝑟𝑎𝑑 = ℎ𝐴(𝑇𝑠 − 𝑇∞ ) + 𝜖𝜎𝐴(𝑇𝑠4 − 𝑇𝑠𝑢𝑟

= (10

𝑊

𝑚2 𝐾

) (𝜋 (0.1𝑚)(25𝑚))(150𝑜 𝐶 − 25𝑜 𝐶) + 0.8 (5.67 × 10−8

= 18414𝑊 = 𝑞

𝑊

𝑚2 𝐾 4

) (𝜋 (0.1𝑚)(25𝑚))((150 + 273 𝐾 )4 − (25 + 273 𝐾)4 )

b. If the steam is generated in a gas-fired boiler that burns natural gas which is priced at $0.01/𝑀𝐽, what is the annual cost of heat loss from the line?

Cost per year = 𝑞 (𝑁

𝑠𝑒𝑐

)( 𝑦𝑒𝑎𝑟

= 18414𝑊 (365

$0.01 𝑀𝐽

1𝑀𝐽

) ( 106𝐽) (

1𝐽 𝑠

1𝑊

)

𝑑𝑎𝑦𝑠 ℎ𝑟𝑠 𝑚𝑖𝑛 𝑠𝑒𝑐 $0.01 1𝑀𝐽 1𝐽/𝑠 ) (24 ) (60 ) (60 )( )( 6 )( ) = $5807/𝑦𝑟 𝑦𝑒𝑎𝑟 𝑑𝑎𝑦 ℎ𝑟 min 𝑀𝐽 10 𝐽 1𝑊

7. A common procedure for measuring the velocity of an air stream involves insertion of an electrically heated wire into the air flow, with the axis of the wire pointed perpendicular to the flow direction. The electrical energy dissipated in the wire depends on the convection coefficient, which in turn, depends on the velocity of the air. Consider a wire of length 20 mm and diameter 0.5 mm, for which a calibration velocity 𝑈 = 6.25 × 10−5 ℎ2 has been determined. The velocity 𝑈 and the convection coefficient ℎ have units of

𝑚

𝑠

𝑊

and 𝑚2𝐾, respectively. In an

application involving air at a temperature of 𝑇∞ = 25𝑜 𝐶, the surface temperature of the wire is maintained at 𝑇𝑠 = 75𝑜 𝐶, with a voltage drop of 5V and an electric current of 0.1 A. (Electrical energy can be calculated as the product of voltage and current. 𝑃𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑖𝑡𝑦 = 𝑉𝐼. Use the conversion 1 𝐴𝑚𝑝 × 1 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 = 1 𝑊𝑎𝑡𝑡)

a. What is the convection coefficient?

∴ℎ=

𝑉𝐼

𝑃 = 𝑉𝐼 = ℎ𝐴(𝑇𝑠 − 𝑇∞ )

𝐴(𝑇𝑠 − 𝑇∞ )

=

𝑊 (5𝑉)(0.1𝐴) 2 = 318.31 𝑚 𝐾 𝜋0.0005𝑚(0.02𝑚)(75𝑜 𝐶 − 25𝑜 𝐶)

b. What is the velocity of the air? 𝑈 = 6.25 × 10−5(318.31)2 = 6.33

𝑚 𝑠...