Orange solution - Peter Nikrityuk Heat Transfer Fall 2018 PDF

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Peter Nikrityuk Heat Transfer Fall 2018...

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CHE 314 - Heat Transfer Midterm Exam (Fall 2018), Oct. 22, 11am-11:50am Student’s Name: Abdurachman Ibn Said Durichman Student’s ID: 1234567891111110

Instructions Write your name and student ID number on all sheets, number the sheets.

Variant-Orange • This exam consists of 3 problems; total of 100 points. • Duration of this exam is 50 min. • The midterm exam is closed book. allowed.

No any books and no any lecture notes are

• If something in the question is not clear, state your interpretation or assumptions concerning it and proceed to solve the problem. Do not seek clarification from anyone including the Instructors. • Only faculty approved non-programmable calculators are allowed. • ONE double side page (Letter Format, Width 27.94 cm, Height 21.59 cm) with ONLY own handwritten notes (not scanned and not photo-copied) is allowed.

Problem 1 - Theoretical Questions a. What heat transfer (mode) does play the most significant role between a surface and a fluid moving over the surface? (5 points) b. What is the difference between N u and Bi numbers? (5 points)

Solution a. convection b. N u uses kf and Bi uses ks , plus, sometimes they have different characteristic scales.

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Problem 2 • In a pharmaceutical plant, a copper pipe (kc = 400 W/mK) with inner diameter of D1 = 20 mm and wall thickness of 2.5 mm is used for carrying liquid oxygen to a storage tank, see Fig. 1. The liquid oxygen flowing in the pipe has an average temperature of TO2 = −200 o C and a convection heat transfer coefficient of hO2 = 150 W/m2 K. The condition surrounding the pipe has an ambient air temperature of T∞ = 20 o C and a convective heat transfer coefficient of h∞ = 20 W/m2 K. The emissivity of the outer surface of insulation is 10−3 . Thermal conductivity of insulation is kins = 0.05 W/m K. The copper surface temperature has the following value Tc = −192.3 o C, see Fig. 1. • Air transport properties are: P r = 0.7, ρair = 1.1 kg/m3 , µair = 1.8 · 10−5 kg/m s, kair = 0.026 W/m K, cp = 1007 J/kg K. a. Sketch a thermal circuit of the system. (10 points) b. If the dew point (dew point corresponds to a temperature at which condensation starts) is Ts,ins = 10 o C, determine the thickness of insulation around the copper pipe to avoid condensation on the outer surface of insulation. (40 points) • NOTE: Assume thermal contact resistance is negligible. The Stefan-Boltzmann constant is σ = 5.67 · 10−8 W/m2 K4 .

Insulation

Pipe

Tc

T s,ins

TO2 Liq. O2

8

T h

8

Surrounding Air

D1 D ins

Figure 1: Scheme for the problem#2.

Solution a. The thermal circuit scheme of this setup is shown in Fig. 2. b. Given geometry: – inner diameter of the copper pipe is d 1 = 20 · 10−3 m or radius is r1 = 10 · 10−3 m 2

T

conduction

conduction

convection

radiation

Ts,ins

Tc

Ts,1

8

TO2

8

T

convection

Figure 2: The thermal circuit scheme for the problem#2. – outer diameter of the copper pipe is d 2 = 25 · 10−3 m or radius is r2 = 12.5 · 10−3 m – outer diameter of the insulation is d ins m or radius is rins – where the thickness of insulation around the copper pipe can be calculated as rins − r2 after rins is defined. • The condensation on the outer surface of insulation can be avoided if the insulation surface temperature is Ts,ins ≥ 10 o C. Assuming Ts,ins = 10 o C we can calculate the heat flux between insulation and air as follows:  4  q 4 = − 200.054 W/m2 − T∞ q ′′ = = h∞ (Ts,ins − T∞ ) + σ ε Ts,ins {z } | As | {z } −200

−0.054

• outer surface area is As = 2π rins · L, thus, the heat rate per unit length is q = q ′′ · (2π(r2 + δ) · L) ; =⇒ q ′ = q ′′ · (2π rins )

• At the same time, the heat rate per unit length can be evaluated as follows: TO2 − Tc

q′ =

1 2π r1 hO2

+

ln(r2 /r1 ) 2π kco

• Finally, we can write q ′′ · (2π rins ) = q ′ =

= −72.51 W/m

TO2 − Tc 1 2π r1 hO2

+

ln(r2 /r1 ) 2π kco

• Thus, rins =

T − Tc q′  O2 = ′′ q 2π q ′′ · 2π · 2π r11 hO + 2

• Insulation thickness is rins − r2 = 0.0452 m where 1 2π r1 hO2

= 0.0577;

ln(r2 /r1 ) 2π kco

 = 0.0577 m

ln(r2 /r1 ) = 8.879 · 10−5 ; q ′ = −72.51 W/m 2π kco

————————————————————

3

• Alternative Solution: TO2 − Tc

q′ =

1 2π r1 hO2

+

ln(r2 /r1 ) 2π kco

= −72.51 W/m

at the same time: q′ =

Tc − Ts,ins ln(r3 /r1 ) 2π kins

; =⇒ ln(r3 /r2 ) =

Tc − Ts,ins q′ 2π kins

Finally, r3 = r2 · exp

Tc − Ts,ins q′ 2π kins

= 0.8765

!

r3 = (0.0125) · 2.4025 = 0.03m • Insulation thickness is rins − r2 = 0.0175 m • Both solutions are correct! The discrepancy comes from unbalanced value of Tc .

4

Problem 3 • Velocity of air flowing over a sphere1 is to be determined by observing the temperature-time history of a sphere fabricated from pure copper. The sphere, which is 12.7 mm in diameter, is at 66 0 C before it is inserted into an airstream having a temperature of 27 0 C . After the sphere is inserted into the airstream, a thermocouple on the outer surface of the sphere indicates 55 0 C corresponding to the time of t = 69 s. • Properties of copper: ρ = 8900 kg/m3 , cp = 390 J/kg K, k = 400 W/m K • Properties of the air: ρ = 1.2 kg/m3 , cp = 1000 J/kg K, k = 0.03 W/m K, µ = 2 · 10−5 N s/m2 , P r = 0.7 a. Assume and then justify that the sphere behaves as a spacewise isothermal object, and calculate the velocity of air. (30 points) b. Calculate absolute value of the temperature gradient inside the sphere at the sphere surface at the time t = 69 s for the same conditions given in the problem if the velocity of air corresponds to Bi = 10. (10 points)

Solution a. Using LCM approach the temperature of the sphere varies in time following this equation:   ρ · V ol · cp V ol T (t) − T∞ t r0 , τt = = exp − = , τt 3 As Ti − T∞ h · As • Knowing τt we find h: τt =

ln



−t T (t)−T∞ Ti −T∞

=

ln

−69   = 208.2345 s 55−27 66−27

• Finally, ρ · r0 · cp 8900 · (0.0063) · 390 = 35 W/mK = 3 · 208.2345 3τt Here we used characterist length Ls = r30 , r0 = D/2 = 0.0063 m. h=

• To justify that the sphere behaves as a spacewise isothermal object we calculate Bi: Bi =

35 · (0.0063) h r0 = = 1.8375e − 04 < 0.1 3k 3 · 400

• Due to Bi < 0.1 the assuption that the sphere behaves as a spacewise isothermal object is valid. • To calculate air velocity we find Re from N u relation for a sphere: 1/2

N u = 2 + 0.6Re

Pr

1/3

; =⇒ Re =

Nu = 1

Volume of a sphere is

4 πR3 , 3



(N u − 2) 0.6P r 1/3

2

=



hD 35 (0.0127) = 14.8 = 0.03 kf

Surface of a sphere is 4πR 2

5

(14.8 − 2) 0.6(0.7)1/3

2

= 577.3

Re =

ρV∞ D Re µ 577.3 (2 · 10−5 ) = 0.7576 m/s , V∞ = = 1.2 (0.0127) µ ρD

b. LCM model is not valid for Bi > 1. Bi can be defined using Lc from previous case or using r0 . h=

−ks

3k Bi 3(400)10 = = 1.9 · 106 r0 0.0063

dT  1.9 · 106 dT h = h(Ts − T∞ ); =⇒ |  | = (Ts − T∞ ) = (55 − 27) = 1.33 · 105 K/m  dr solid dr solid ks 400

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