## Heat Transfer Miscellaneous

#### Heat-Transfer

1. A metal ball of diameter 60 mm is initially at 220°C. The ball is suddenly cooled by an air jet of 20°C. The heat transfer coefficient is 200 W/m2K. The specific heat, thermal conductivity and density of the metal ball are 400 J/kgK, 400 W/mK and 9000 kg/m3, respectively. The ball temperature (in °C) after 90 seconds will be approximately

1. CP = 400 J /kg-K
K = 400 W/mK
ρ = 9000 kg/m3
Time (τ) = 90 sec  hA τ ρVCP

 Ti - T∞ = e T - T∞

 Put A = 3 = 100 v R hA τ = 200 × 3 × 1 × 1 × 90 = 0.5 ρVCP 0.03 9000 400

 220 - 20 = e0.5 T - 20

By solving, we get T = 141.3°C

##### Correct Option: A

CP = 400 J /kg-K
K = 400 W/mK
ρ = 9000 kg/m3
Time (τ) = 90 sec  hA τ ρVCP

 Ti - T∞ = e T - T∞

 Put A = 3 = 100 v R hA τ = 200 × 3 × 1 × 1 × 90 = 0.5 ρVCP 0.03 9000 400

 220 - 20 = e0.5 T - 20

By solving, we get T = 141.3°C

1. A cylindrical steel rod, 0.01 m in diameter and 0.2 m in length is first heated to 750°C and then immersed in a water bath at 100°C. The heat transfer coefficient is 250 W/m2K. The density, specific heat and thermal conductivity of steel are ρ = 7801 kg/m3, c = 473 J/kgK, and k = 43 W/mK, respectively. The time required for the rod to reach 300°C is ______ seconds.

1. d = 0.01 m
L = 0.2 m, R = 7801 kg/m3
ti = 750 °C, ta = 100 °C
C = 473 J/Kg K, K = 43 W/mK
h = 250 W/m2K

 Lc = V = (π / 4)d2l = d A πdl 4 T - Ta = exp -hAτ = exp -hτ Ti - Ta ρVC ρCLc

 300 - 100 = exp -250 × 4 × τ 750 - 100 7801 × 473 × 0.01

⇒ τ = 43.49

##### Correct Option: C

d = 0.01 m
L = 0.2 m, R = 7801 kg/m3
ti = 750 °C, ta = 100 °C
C = 473 J/Kg K, K = 43 W/mK
h = 250 W/m2K

 Lc = V = (π / 4)d2l = d A πdl 4 T - Ta = exp -hAτ = exp -hτ Ti - Ta ρVC ρCLc

 300 - 100 = exp -250 × 4 × τ 750 - 100 7801 × 473 × 0.01

⇒ τ = 43.49

1. Two cylindrical shafts A and B at the same initial temperature are simultaneously placed in a furnace. The surfaces of the shafts remain at the furnace gas temperature at all times after they are introduced into the furnace. The temperature variation in the axial direction of the shafts can be assumed to be negligible. The data related to shafts A and B is given in the following Table.
 Quantity Shaft A Shaft B Diameter (m) 0.4 0.1 Thermal Conductivity (W/mK) 40 20 Volumetric heat capacity (J/m3K) 2 × 106 2 × 107

The temperature at the centerline of the shaft A reaches 400°C after two hours. The time required (in hours) for the centerline of the shaft B to attain the temperature of 400°C is ____.

1.  T - T∞ = e(-hAst / ρ V Cp)

(ρ Cp)A = 2 × 106 J/m3K,
(ρ Cp)B = 2 × 107 J/m3K, dA = 0.4 m, dB = 0.1 m
tA = 2hrs = 2 × 3600 = 7200 sec
hA = hB hAs × t = hAs × t ρVCp A ρVCp B

 ⇒ π × 0.4 × L × 7200 = π × 0.1 × L × t × t 2 × 106 × (π / 4) × (0.4)2 × L A 2 × 107 × (π / 4) × (0.1)2 × L B

⇒ tB = 2.5 hrs

##### Correct Option: B

 T - T∞ = e(-hAst / ρ V Cp)

(ρ Cp)A = 2 × 106 J/m3K,
(ρ Cp)B = 2 × 107 J/m3K, dA = 0.4 m, dB = 0.1 m
tA = 2hrs = 2 × 3600 = 7200 sec
hA = hB hAs × t = hAs × t ρVCp A ρVCp B

 ⇒ π × 0.4 × L × 7200 = π × 0.1 × L × t × t 2 × 106 × (π / 4) × (0.4)2 × L A 2 × 107 × (π / 4) × (0.1)2 × L B

⇒ tB = 2.5 hrs

1. A steel ball of 10 mm diameter at 1000 K is required to be cooled to 350 K by immersing it in a water environment at 300 K. The convective heat transfer coefficient is 1000 W/m2K. Thermal conductivity of steel is 40 W/m K. The time constant for the cooling process is 16 s. The time required (in s) to reach the final temperature is

1.  Biot Number = hLc = 0.458 K

 For sphere Lc = Volume = d surface area 6

 ∴ Bi = hd = 1000 × 0.01 6K 6 × 40

= 0.0416 < 0.1
Hence lumped heat analysis is used
 T - T∞ = exp -hAst = e -t Ti - T∞ ρ V Cp ṫ

Thermal time constant,
 t* = ρ V Cp = 16 sec hAs

 ∴ 350 - 300 = e -t 1000 - 300 16

⇒ t = 42.2249 sec

##### Correct Option: A

 Biot Number = hLc = 0.458 K

 For sphere Lc = Volume = d surface area 6

 ∴ Bi = hd = 1000 × 0.01 6K 6 × 40

= 0.0416 < 0.1
Hence lumped heat analysis is used
 T - T∞ = exp -hAst = e -t Ti - T∞ ρ V Cp ṫ

Thermal time constant,
 t* = ρ V Cp = 16 sec hAs

 ∴ 350 - 300 = e -t 1000 - 300 16

⇒ t = 42.2249 sec

1. Saturated vapor is condensed to saturated liquid in a condenser. The heat capacity ratio
 is Cr = Cmin . The effectiveness ( ε ) of the condenser is Cmax

1. ε of condenser is given by 1-exp(-NTU) Because

 Cr = Cmin = 0 ( as Cmax → ∞ ) Cmax

##### Correct Option: D

ε of condenser is given by 1-exp(-NTU) Because

 Cr = Cmin = 0 ( as Cmax → ∞ ) Cmax