3 Engineering Example 2

3.1 Heat flow in an insulated metal plate

Introduction

Thermal insulation is important in many domestic (e.g. central heating) and industrial (e.g cooling and heating) situations. Although many real situations involve heat flow in more than one dimension, we consider only a one dimensional case here. The flow of heat is determined by temperature and thermal conductivity. It is possible to model the amount of heat Q (J) crossing point x in one dimension (the heat flow in the x direction) from temperature T 2 (K) to temperature T 1 (K) (in which T 2 > T 1 ) in time t s by

Q t = λ A T 2 T 1 x ,

where λ is the thermal conductivity in W m 1 K .

Problem in words

Suppose that the upper and lower sides of a metal plate connecting two containers are insulated and one end is maintained at a temperature T 2 (K) (see Figure 7).

The plate is assumed to be infinite in the direction perpendicular to the sheet of paper.

Figure 7:

{A laterally insulated metal plate}

  1. Find a formula for T .
  2. If λ = 205  ( W m 1 K 1 ),  T 1 = 300 (K),  A = 0.004 m 2 ),  x = 0.5 (m), calculate the value of T 2 required to achieve a heat flow of 100 J s 1 .

Mathematical statement of the problem

  1. Given Q t = λ A T 2 T 1 x express T 2 as the subject of the formula.
  2. In the formula found in part
    1. substitute λ = 205 , T 1 = 300 , A = 0.004 x = 0.5 and Q t = 100 to find T 2 .

Mathematical analysis

  1. Q t = λ A T 2 T 1 x

    Divide both sides by λ A

    Q t λ A = T 2 T 1 x

    Multiply both sides by x

    Q x t λ A = T 2 T 1

    Add T 1 to both sides

    Q x t λ A + T 1 = T 2

    which is equivalent to

    T 2 = Q x t λ A + T 1

  2. Substitute λ = 205 , T 1 = 300 , A = 0.004 , x = 0.5 and Q t = 100 to find T 2 :

    T 2 = 100 × 0.5 205 × 0.004 + 300 60.9 + 300 = 360.9

    So the temperature in container 2 is 361 K to 3 sig.fig.

Interpretation

The formula T 2 = Q x t λ A + T 1  can be used to find a value for T 2 that would achieve any desired heat flow. In the example given T 2 would need to be about 361 K ( 7 8 C ) to produce a heat flow of 100 J s 1 .

Exercises
  1. The formula for the volume of a cylinder is V = π r 2 h . Find V when r = 5 cm and h = 15 cm.
  2. If R = 5 p 2 , find R when
    1. p = 10 ,
    2. p = 16 .
  3. For the following formulae, find y at the given values of x .
    1. y = 3 x + 2 , x = 1 , x = 0 , x = 1 .
    2. y = 4 x + 7 , x = 2 , x = 0 , x = 1 .
    3. y = x 2 , x = 2 , x = 1 , x = 0 , x = 1 , x = 2 .
  4. If P = 3 Q R find P if Q = 15 and R = 0.300 .
  5. If y = x z find y if x = 13.200 and z = 15.600 .
  6. Evaluate M = π 2 r + s when r = 23.700 and s = 0.2 .
  7. To convert a length measured in metres to one measured in centimetres, the length in metres is multiplied by 100. Convert the following lengths to cm.  
    1. 5 m,   
    2. 0.5 m,   
    3. 56.2 m.
  8. To convert an area measured in m 2 to one measured in cm 2 , the area in m 2 is multiplied by 1 0 4 . Convert the following areas to cm 2 .
    1. 5 m 2 ,
    2. 0.33 m 2 ,
    3. 6.2 m 2 .
  9. To convert a volume measured in m 3 to one measured in cm 3 , the volume in m 3 is multiplied by 1 0 6 . Convert the following volumes to cm 3 .
    1. 15 m 3 ,
    2. 0.25 m 3 ,
    3. 8.2 m 3 .
  10. If η = 4 Q P π d 2 L n evaluate η when Q P = 0.0003 , d = 0.05 , L = 0.1 and n = 2 .
  11. The moment of inertia of an object is a measure of its resistance to rotation. It depends upon both the mass of the object and the distribution of mass about the axis of rotation. It can be shown that the moment of inertia, J , of a solid disc rotating about an axis through its centre and perpendicular to the plane of the disc, is given by the formula

    J = 1 2 M a 2

    where M is the mass of the disc and a is its radius. Find the moment of inertia of a disc of mass 12 kg and diameter 10 m. The SI unit of moment of inertia is kg m 2 .

  12. Transpose the given formulae to make the given variable the subject.
    1. y = 3 x 7 , for x ,
    2. 8 y + 3 x = 4 , for x ,
    3. 8 x + 3 y = 4 for y ,
    4. 13 2 x 7 y = 0 for x .
  13. Transpose the formula P V = R T for
    1. V ,
    2. P ,
    3. R ,
    4. T .
  14. Transpose v = x + 2 y ,
    1. for x ,
    2. for y .
  15. Transpose 8 u + 4 v 3 w = 17 for each of u , v and w .
  16. When a ball is dropped from rest onto a horizontal surface it will bounce before eventually coming to rest after a time T where

    T = 2 v g 1 1 e

    where v is the speed immediately after the first impact, and g is a constant called the acceleration due to gravity. Transpose this formula to make e , the coefficient of restitution, the subject.

  17. Transpose q = A 1 2 g h ( A 1 A 2 ) 2 1 for A 2 .
  18. Make x the subject of
    1. y = r + x 1 r x ,
    2. y = x 1 x + 1 .
  19. In the design of orifice plate flowmeters, the volumetric flowrate, Q ( m 3 s 1 ), is given by

    Q = C d A o 2 g Δ h 1 A o 2 A p 2

    where C d is a dimensionless discharge coefficient, Δ h (m) is the head difference across the orifice plate and A o ( m 2 ) is the area of the orifice and A p ( m 2 ) is the area of the pipe.

    1. Rearrange the equation to solve for the area of the orifice, A o , in terms of the other variables.
    2. A volumetric flowrate of 100 cm 3 s 1 passes through a 10 cm inside diameter pipe. Assuming a discharge coefficient of 0.6 , calculate the required orifice diameter, so that the head difference across the orifice plate is 200 mm.

      [Hint: be very careful with the units!]

  1. 1178.1 cm 3
    1. 500 ,
    2. 1280
    1. 1 , 2 , 5 ,
    2. 15 , 7 , 3 ,
    3. 5 , 3 , 1 , 0 ,
  2. P = 0.667
  3. y = 0.920
  4. M = 0.067
    1. 500 cm ,
    2. 50 cm ,
    3. 5620 cm .
    1. 50000 cm 2 ,
    2. 3300 cm 2 ,
    3. 62000 cm 2 .
    1. 15000000 cm 3 ,
    2. 250000 cm 3 ,
    3. 8200000 cm 3 .
  5. η = 0.764 .
  6. 150 kg m 2
    1. x = y + 7 3 ,
    2. x = 4 8 y 3 ,
    3. y = 4 8 x 3 ,
    4. x = 13 7 y 2
    1. V = R T P ,
    2. P = R T V ,
    3. R = P V T ,
    4. T = P V R
    1. x = v 2 2 y ,
    2. y = v 2 x 2
  7. u = 17 4 v + 3 w 8 , v = 17 8 u + 3 w 4 , w = 8 u + 4 v 17 3
  8. e = 1 2 v g T
  9. A 2 = ± A 1 2 q 2 2 A 1 2 g h + q 2
    1. x = y r 1 + y r ,
    2. x = 1 + y 2 1 y 2
    1. A 0 = Q A p Q 2 + 2 g Δ h A p 2 C d 2
    2. Q = 100 cm 3 s 1 = 1 0 4 m 2 s 1 A p = π 0 . 1 2 4 = 0.007854 m 2 C d = 0.6 Δ h = 0.2 m g = 9.81 m s 2 Substituting in answer
      1. gives A o = 8.4132 × 1 0 5 m 2 so diameter = 4 A o π = 0.01035 m = 1.035 cm