## Heat Transfer Miscellaneous

#### Heat-Transfer

1. A long glass cylinder of inner diameter = 0.03 m and outer diameter = 0.05 m carries hot fluid inside. If the thermal conductivity of glass = 1.05 W/mK, the thermal resistance (K/W) per unit length of the cylinder is

1. Given data:
d1 = 0.03 m

 ∴ r1 d1 = 0.3 = 0.015 m 2 2

d2 = .05 m
∴ r2 = .025 m
 K = 1.05 W mK

Thermal resistance
 Rt = 1 loge r2 2πKL r1

 = 1 loge .025 2 × 3.14 × 1.05 × 1 0.015

= 0.077 W/m

##### Correct Option: B

Given data:
d1 = 0.03 m

 ∴ r1 d1 = 0.3 = 0.015 m 2 2

d2 = .05 m
∴ r2 = .025 m
 K = 1.05 W mK

Thermal resistance
 Rt = 1 loge r2 2πKL r1

 = 1 loge .025 2 × 3.14 × 1.05 × 1 0.015

= 0.077 W/m

1. Consider a long cylindrical tube of inner and outer radii, ri and r0, respectively, length L and thermal conductivity, k. Its inner and outer surfaces are maintained at Ti, and T0, respectively (Ti > T0). Assuming one dimensional steady state heat conduction in the radial direction, the thermal resistance in the wall of the tube is

1. Ar = 2πrL
From Fourier ’s Law

 qr = kAr dT dr

 qr = 2π krL dT dr

Boundary conditions:
T = Ti at r = ri
T = T0 at r = r0
 q = 2πkL(Ti - T0) In(r0/ri)

 Rth = In(r0/ri) 2πkL

##### Correct Option: C

Ar = 2πrL
From Fourier ’s Law

 qr = kAr dT dr

 qr = 2π krL dT dr

Boundary conditions:
T = Ti at r = ri
T = T0 at r = r0
 q = 2πkL(Ti - T0) In(r0/ri)

 Rth = In(r0/ri) 2πkL

1. A hollow cylinder has length L, inner radius r1 outer radius r2, and thermal conductivity k. The thermal resistance of the cylinder for radial conduction is

1.  In(r2/r1) 2πkL

##### Correct Option: A

 In(r2/r1) 2πkL

1. For flow of viscous fluid over a flat plate, if the fluid temperature is the same as the plate temperature, the thermal boundary layer is

1. NA

##### Correct Option: D

NA

1. An industrial gas (c = 1 kJ/kgK) enters a parallel flow heat exchange at 250°C with a flow rate of 2 kg/s to heat a water stream. The water stream (c = 4 kJ/kgK) enters the heat exchange at 50°C with a flow rate of 1 kg/s. The heat exchange has an effectiveness of 0.75. The gas stream exit temperature will be

1.  ε = Ch(Th1 - Th2) Cmin(Thi - Tci)

 0.75 = 2 ×(250 - Th2) 2 ×(250 - 50)

Th2 = 100° C

##### Correct Option: B

 ε = Ch(Th1 - Th2) Cmin(Thi - Tci)

 0.75 = 2 ×(250 - Th2) 2 ×(250 - 50)

Th2 = 100° C