Heat Transfer Miscellaneous
Direction: Water (specifie heat, c = 4.18 kJ/kgK) enters a pipe at a rate 0.01 kg/s and a temperature of 20°C. The pipe, of diameter 50 mm and length 3 m, is subjected to a wall heat flux q"w in W/m2.
- If q"w = 2500 x, where x is in m and in the direction of flow (x = 0 at the inlet), the bulk means temperature of the water leaving the pipe in °C is
-
View Hint View Answer Discuss in Forum
At the inlet of pipe
x = O
q "wi = 2500 × 0 = 0
At the exit of pipe,
x = lq "wi = 2500 × 1 × 3 = 7500 W m2
Average heat transfer,q "avg q "wi + q "we = 0 + 7500 = 3750 w 2 2 m2
Net heat transfer
Q = q "avg × surface area = 3750 × πdl
= 3750 × 3.14 × 0.05 × 3 = 1766.25 W
Also Q = mCp (T0 – Ti)
1766.25 = 0.01× 4180 (T0 –20)
By solving above equation, we get
T0 = 62° CCorrect Option: B
At the inlet of pipe
x = O
q "wi = 2500 × 0 = 0
At the exit of pipe,
x = lq "wi = 2500 × 1 × 3 = 7500 W m2
Average heat transfer,q "avg q "wi + q "we = 0 + 7500 = 3750 w 2 2 m2
Net heat transfer
Q = q "avg × surface area = 3750 × πdl
= 3750 × 3.14 × 0.05 × 3 = 1766.25 W
Also Q = mCp (T0 – Ti)
1766.25 = 0.01× 4180 (T0 –20)
By solving above equation, we get
T0 = 62° C
- 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
-
View Hint View Answer Discuss in Forum
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 secCorrect 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
- Consider the radiation heat exchange inside an annulus between two very long concentric cylinders. The radius of the outer cylinder is R0 and that of the inner cylinder is Ri. The radiation view factor of the outer cylinder onto itself is
-
View Hint View Answer Discuss in Forum
F1– 1 = 0
F1– 2 = 1
Reciprocacy theorem
A2 F2– 1 = A1 F1– 2⇒ F2– 1 = A1 F1– 2 A2 = 2πRiL = Ri 2πRoL Ro
F2– 1 + F2– 2 = 1∴ F2– 2 = 1 - F2– 1 = 1 - Ri Ro
Correct Option: D
F1– 1 = 0
F1– 2 = 1
Reciprocacy theorem
A2 F2– 1 = A1 F1– 2⇒ F2– 1 = A1 F1– 2 A2 = 2πRiL = Ri 2πRoL Ro
F2– 1 + F2– 2 = 1∴ F2– 2 = 1 - F2– 1 = 1 - Ri Ro
- Sphere 1 with a diameter of 0.1 m is completely enclosed by another sphere 2 of diameter 0.4 m. The view factor F12 is
-
View Hint View Answer Discuss in Forum
F11 = 0
F12 = 1Correct Option: D
F11 = 0
F12 = 1
- Three sets of parallel plates LM, NR and PQ are given in Figures 1, 2 and 3. The view factor Fij is defined as the fraction of radiation leaving plate I that is intercepted by plate J. Assume that the val ues of FL M and FN R ar e 0.8 and 0.4, respectively. The value of FPQ (round off to one decimal place) is _____.
-
View Hint View Answer Discuss in Forum
FLM = F(I) + F(II) + F(III)
Where F is shape factor of surface area I, II and III of M with surface area of ‘L’
0.8 = F(I) + F(II) + F(III)
∵ (I) and (III) are symmetrical in shape so, shape factor of these surface will be same i.e. F(I) = F(III)
So, 0.8 = 2F(I) + F(II)... (1)
In this area (II) is same as fig. (I) area II.
i.e. FNR = F(II) = 0.4... (2)
From equation (1),F(III) = 0.8 - 0.4 = 0.2 2 
Here, FPQ = F(II) + F(III)
= 0.4 + 0.2 = 0.6
Correct Option: B
FLM = F(I) + F(II) + F(III)
Where F is shape factor of surface area I, II and III of M with surface area of ‘L’
0.8 = F(I) + F(II) + F(III)
∵ (I) and (III) are symmetrical in shape so, shape factor of these surface will be same i.e. F(I) = F(III)
So, 0.8 = 2F(I) + F(II)... (1)
In this area (II) is same as fig. (I) area II.
i.e. FNR = F(II) = 0.4... (2)
From equation (1),F(III) = 0.8 - 0.4 = 0.2 2
Here, FPQ = F(II) + F(III)
= 0.4 + 0.2 = 0.6
