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Latest Conduction with Uniform Heat Generation MCQ Objective Questions

Top Conduction with Uniform Heat Generation MCQ Objective Questions

Conduction with Uniform Heat Generation Question 1:

Heat is generated uniformly in a 4 cm diameter, 16 cm long solid bar [k = 24 W/m°c]. The temperatures at the centre and the surface of the bar are measured to be 210°C and 45°C respectively. The rate of heat generation within the bar is,

  1. 240 W
  2. 1013 W
  3. 7962 W
  4. 3.96 × 106 W

Answer (Detailed Solution Below)

Option 3 : 7962 W

Conduction with Uniform Heat Generation Question 1 Detailed Solution

Concept:

The temperature difference of the cylindrical bar, when the heat is generated internally, is given as,

where

0 = Temperature at the centre of the cylinder, Tw = Temperature at the surface of the cylinder, , k = thermal conductivity of the cylinder

Calculation:

Given

Diameter of bar, d = 4 cm, Length of the bar, L = 16 cm, The temperature at the centre, T0 = 210°C, The temperature at the surface, Tw = 45°C

Conduction with Uniform Heat Generation Question 2:

Consider a rod of uniform thermal conductivity whose one end (x = 0) is insulated and the other end (x = L) is exposed to flow of air at temperature T with convective heat transfer coefficient h. The cylindrical surface of the rod is insulated so that the heat transfer is strictly along the axis of the rod. The rate of internal heat generation per unit volume inside the rod is given as 

The steady-state temperature at the mid-location of the rod is given as TA. What will be the temperature at the same location, if the convective heat transfer coefficient increases to 2h?

  1. 2TA
  2. TA

Answer (Detailed Solution Below)

Option 3 : TA

Conduction with Uniform Heat Generation Question 2 Detailed Solution

Explanation:

Steady-state conduction with internal heat generation:

 ......(i)

It is given that the surface of the cylinder is insulated the heat will transfer only in x-direction i.e. 1-D heat transfer.

The temperature will remain the same even if we change the convective heat transfer coefficient.

Proof:

The rate of heat generation is 

To find the temperature profile we will differentiate equation (i) and will use boundary conditions.

Integrating equation we get,

 = 

 ....... (a)

substitute boundary condition:

at x = 0 , dT/dx = 0, sin (0) = 0

we get c1 = 0

Integrating again we get

 ........(b)

at x = L 

Heat conducted = Heat convected 

sin 2π = 0 , LHS = 0 hence

Tx = L = T

substituting in equation b

c2 = 

Hence the equation of temperature is:

It does not depend on convective heat transfer coefficient h.

Hence on there is no effect on temperature TA if convective heat transfer is increased to 2h.

Confusion Points

  • Generally, the equations of temperature with heat generation that we derive for wall, sphere, cylinder all depend on convective heat transfer.
  • But here due to the given value of heat generation, the general equation after integration gave different results.

Conduction with Uniform Heat Generation Question 3:

A long cylindrical rod of radius ‘R’ has a Surface heat flux  of q0 The uniform internal heat generation rate in the rod is

  1. 2q0

Answer (Detailed Solution Below)

Option 1 :

Conduction with Uniform Heat Generation Question 3 Detailed Solution

Explanation:

The temperature distribution in a solid cylinder of radius R with uniform internal heat generation is given by

where, 

Tw = temperature at the outer radius (R), q̇s = uniform heat generation, k = thermal conductivity

T = Temperature at radius r

At steady state, heat going out of the cylinder by conduction is absorbed by the atmosphere through convection whose temperature is T (Tw > T).

Heat conducted:

At r = R

Heat convected:

At steady-state heat conducted is equal to heat convected:

Given heat flux is q0;

Conduction with Uniform Heat Generation Question 4:

Uniform internal heat generation at  is occurring in a cylindrical nuclear reactor fuel rod of 50 mm diameter. Under steady state conditions, the temperature varies as T(r) = a + br2, where T is in °C and r is in meters, while a = 800°C and b = -4.167 × 105 °C/m2. The fuel rod properties are K = 30 W/m-K,  and CP = 800 J/kg-K

  1. The rate of heat transfer per unit length of the rod at r = 0 (the centreline) is 0
  2. The rate of heat transfer per unit length of the rod at r = 25 mm (the surface) is 98 kW/m
  3. If the reactor power level is suddenly increases to  the initial time rate of temperature change at r = 0 and r = 25 mm is 56.82 K/s
  4. If the reactor power level is suddenly increases to  the initial time rate of temperature change at r = 0 and r = 25 mm is 45.56 K/s

Answer (Detailed Solution Below)

Option :

Conduction with Uniform Heat Generation Question 4 Detailed Solution

Concept:

For cylindrical coordinates, rate of heat conduction is given as

Calculation:

T(r) = a + br2

At r = 0,

At r = r0

 (-4.167 × 105) (0.025) = -0.208 × 105 K/m

= 0.980 × 105 W/m

Calculating the initial time rate of temperature change when the reactor power suddenly increases

General heat conduction equation in cylindrical coordinates is given as

Initially at t = 0, the temperature distribution is given by the prescribed form as

T(r) = 800 - 4.167 × 105 r2

Conduction with Uniform Heat Generation Question 5:

For the three-dimensional object shown in the figure below, five faces are insulated. The sixth face (PQRS), which is not insulated, interacts thermally with the ambient, with a convective heat transfer coefficient of 10 W/m2.K. The ambient temperature is 30°C. Heat is uniformly generated inside the object at the rate of 100 W/m3. Assuming the face PQRS to be at uniform temperature, its steady state temperature is

  1. 10°C
  2. 20°C
  3. 30°C
  4. 40°C

Answer (Detailed Solution Below)

Option 4 : 40°C

Conduction with Uniform Heat Generation Question 5 Detailed Solution

Concept:

Newtons Law of cooling:

The law states that the rate heat of transfer by convection between a solid body and the surrounding fluid is directly proportional to the temperature difference between them and is also directly proportional to the area of contact or area of exposure between them.

Q = hA(Ts - T)

If, qG = Heat generation per unit volume (W/m3)

Then, Q = qG × V

Where Ts = Temperature of the surface PQRS, T =  Ambient Temperature, h = Convective heat transfer coefficient, A = Area of exposure, V = Volume of the object.

Calculation:

Given:

h = 10 W/m2K, T = 30°C, qG = 100 W/m3, V = 2 × 1 × 2 = 4 m3

Q = qG × V = 100 × 4 = 400 W

Convection heat transfer from face PQRS

Q = hA(Ts - T)

400 = 10 × 2 × 2 × (Ts - 30)

Ts = 40°C

∴ Temperature of the surface PQRS is 40°C

Conduction with Uniform Heat Generation Question 6:

One dimensional unsteady state heat transfer equation for a sphere with heat generation at the rate ‘qg’, can be written as

Answer (Detailed Solution Below)

Option 2 :

Conduction with Uniform Heat Generation Question 6 Detailed Solution

Explanation:

General Heat Conduction Equations

In general, the temperature gradient may exist in all three directions of a solid. There may be internal heat generation. The temperature can also vary with the time (unstrady state).

The general Heat conduction equation with heat generation for a Slab is

The general Heat conduction equation with heat generation for a Cylinder is

The general Heat conduction equation with heat generation for a Sphere is

One Heat conduction equation with heat generation for a Slab is

One  Heat conduction equation with heat generation for a Cylinder is

One Heat conduction equation with heat generation for a Sphere is

Conduction with Uniform Heat Generation Question 7:

Which one of the following expresses the thermal diffusivity of a substance in terms of thermal conductivity k, mass density ρ and specific heat C?

  1. ρ2kC

Answer (Detailed Solution Below)

Option 3 :

Conduction with Uniform Heat Generation Question 7 Detailed Solution

Concept:

Thermal diffusivity (α ) of material:

Thermal diffusivity is the thermal conductivity divided by density and specific heat capacity at constant pressure.

Thermal diffusivity (α ) of material is given as

 

where k is thermal conductivity in W/m-kρ is density in kg/m3 and c specific heat capacity in J/kg-K

It is the property of the material. Larger the value of α, faster heat will diffuse through the material. A high value of α could result either from a high value of thermal conductivity or low value of thermal heat capacity ρc. Thermal diffusivity α has units of square meters per second.

unit of as Watt is J/s.

Hence from the option, the correct answer is

unit of 

Conduction with Uniform Heat Generation Question 8:

A 2-kW resistance heater wire whose thermal conductivity is 15 W/m.°C has a diameter of 4 mm and a length of 0⋅5 m, is used to boil the water. If the outer surface temperature of the resistance wire is 105°C, what is the temperature at the centre of the wire?

  1. 136°C
  2. 126°C
  3. 146°C
  4. 156°C

Answer (Detailed Solution Below)

Option 2 : 126°C

Conduction with Uniform Heat Generation Question 8 Detailed Solution

Concept:

Steady State One Dimensional Heat conduction equation with Heat Generation for a Cylinder is given by

where T = Temperature at radius r from the centerline, TS = Surface Temperature, qg = Heat Generation per Unit Volume, k = Thermal Conductivity,R = Radius of Cylinder

for the centerline temperature of the wire, we have to put r = 0

So, 

Calculation:

Given:

TS105°C, qg = 2000 W, R = 2 mm, k = 15 W/m.°C, L = 0.5 m

To -105 = 21.23

To = 126.23°C

The temperature at the centre of the wire is 126.23°C.

Conduction with Uniform Heat Generation Question 9:

In a 1 m thick wall, the temperature distribution at a given instant is  where T is in °C and x is in m. The constants are: Co = 800°C, C1 = –250°C/m, and C2 = –40°C/m2. The thermal conductivity of the wall is 50 W/mK and the wall area is 5 m2. If there is a heat source generating uniform volumetric heating at the rate of 500 W/m3 inside the wall, then, rate of change of energy storage in the wall, in kW is ________.

Answer (Detailed Solution Below) -17.5

Conduction with Uniform Heat Generation Question 9 Detailed Solution

Concept:

According to Fourier’s law, the rate of heat flow, Q through a homogeneous solid is directly proportional to the area A normal to the surface, and to the temperature gradient (dT/dx) along the path of heat flow.

(As we know that heat flows from high temperature to low temperature hence dt/dx is negative along x-direction)

The rate of change of energy storage in the wall may be determined by applying an overall energy balance to the wall.

Calculation:

Given:

Co = 800°C, C1 = –250°C/m and C2 = –40°C/m2

A = 5 m2, k = 50 W/mk, q̇gen = 500 W/m3.

Now,

qin = Qx=0 = -kAC1 = -50 × 5 × (-250) = 625000 W

Qin = 62.5kW

qout = Qx=1 = -kA (C1 + 2C2)

qout = -50 × 5 [-250 + 2(-40)] 

qout = 82500 W = 82.5 kW

The rate of change of energy storage in the wall may be determined by applying an overall energy balance to the wall.

62.5 + 2.5 - 82.5 = 

 = -17.5 kW

Mistake Points

Do Not Convert Temperature from °C into K, as we have given that where T is in °C.

Conduction with Uniform Heat Generation Question 10:

In case of one dimensional heat conduction in a medium with constant properties, T is a temperature at position x, at time t. then is proportional to

Answer (Detailed Solution Below)

Option 4 :

Conduction with Uniform Heat Generation Question 10 Detailed Solution

Explanation:

General heat conduction equation:

For a 3-dimensional body with variable properties the heat conduction equation is given by

For constant properties 'k', 'ρ' and 'c' can be taken out-

where 

If the heat conduction is one-dimension with constant properties and without heat generation.

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