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Test: Conduction Level - 2 - Chemical Engineering MCQ


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20 Questions MCQ Test - Test: Conduction Level - 2

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Test: Conduction Level - 2 - Question 1

The temperature distribution, at a certain instant of time in a concrete slab during curing is given by T = 3x2 + 3x + 16, where x is in cm and T is in K. The rate of change of temperature with time is given by (assume thermal diffusivity of concrete to be 0.0003 cm2/s).

Detailed Solution for Test: Conduction Level - 2 - Question 1
Given

By general heat conduction equation for 1-D heat conduction and without heat generation in Cartesian coordinates

Test: Conduction Level - 2 - Question 2

A large concrete slab 1 m thick has one dimensional instantaneous temperature distribution

T = 4 - 10x + 20x2 + 10x3

Where T is temperature in °C and x is distance in m from one face towards the other face of the wall. If the slab material has thermal diffusivity of 2 X 10-3m2/hr, what is the rate of change of temperature at the other face of the wall for the same instant?

Detailed Solution for Test: Conduction Level - 2 - Question 2

Substitute the values in equation

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Test: Conduction Level - 2 - Question 3

Consider two hot lumps with the same initial temperature To. Both the lumps are dropped in a cold fluid at a constant temperature Tf. If ?1 & ?2 represent time constants of the two lumps, then based on the cooling curve which of the following options is correct.

Detailed Solution for Test: Conduction Level - 2 - Question 3
For a lump

Thus at the beginning of cooling process (i.e t = 0)

For both lumps, T0 & Tf represent initial temperature and fluid temperature respectively. Also -Ve sign represents fall in temperature with time because it is a cooling process.

∴ Greater is value of slower is rate of cooling.

As per graph, cools down slowly as compared to .

∴ ?1 > ?2

Test: Conduction Level - 2 - Question 4

Consider a large plane wall of thickness L, thermal conductivity k and surface area A. The left surface of the wall is exposed to the ambient air at temperature T with a heat transfer coefficient of h while the right surface is insulated. The variation of temperature in the wall for steady state one-dimensional heat conduction with no heat generation is

Detailed Solution for Test: Conduction Level - 2 - Question 4

As the system is at steady state, therefore

Also

But surface x = ? is insulated

Also there is no internal heat generation,

therefore

Thus

I.e., hA(T = T0)

By heat conduction equation at any point in the slab

As internal heat generation is zero,

and

∴ c1 = 0

As T = T0 at x = 0

∴ T = 0 X x + c2 = T0 = T

i.e., T = T

i.e., The entire slab is at a uniform temperature equal to fluid temperature.

Test: Conduction Level - 2 - Question 5

If thermal conductivity of a material of wall varies as k0 (1 - ∝T), the temperature at the center of the wall will be (∝ is + ve)

Detailed Solution for Test: Conduction Level - 2 - Question 5

if this material is at steady state with no internal heat generation

If k1 = k2 i.e. k does not depend upon temperature

Thus temperature profile is a straight line

But given k = k0(1 - ∝T), where ∝ is +Ve.

∴ k1 < />2

∵ T1 > T2

Thus temperature profile is as given

*Answer can only contain numeric values
Test: Conduction Level - 2 - Question 6

Thermal conductivity of a material varies linearly with temperature such that k = 20 W/m-°C at 100°C&k = 10 W/m-°C at 40°C. A plane slab 10 cm thick of same material has it’s left and right surfaces maintained a temperatures 20°C & 80°C. What is the rate of heat conduction through the slab in W/m2 ?


Detailed Solution for Test: Conduction Level - 2 - Question 6
Given

k = aT + b

k = 20W/m-°C at T = 40°C

k = 10W/m-°C at T = 40°C

Substituting the given conditions

20 = 100a = b

10 = 40a = b

For the slab

But as is a linear function of T

Where kavg is k for Tavg =

Thus in this case

Kavg = k50

Or 7 kW/m2

*Answer can only contain numeric values
Test: Conduction Level - 2 - Question 7

A hot fluid is being conveyed through a long pipe of 4 cm outer diameter and covered with 2 cm thick insulation. It is proposed to reduce the conduction heat loss to the surroundings to one-third of the present rate by further covering with same insulation. Calculate the additional thickness of insulation.


Detailed Solution for Test: Conduction Level - 2 - Question 7
r1 = 2 cm and r2 = 4 cm

Heat loss with existing insulation,

Heat loss with additional insulation,

Where x is the additional thickness of insulation According to the given condition:

∴ x = 8 r1 -r2

= 8 X 2 - 4 = 12 cm

Test: Conduction Level - 2 - Question 8

For the body shown in the figure, find out the thermal resistance against heat conduction along the x-axis. Consider the body to be at steady state with no internal heat generation in the material. The material is homogenous

Detailed Solution for Test: Conduction Level - 2 - Question 8

Consider an element at a distance x with thickness dx and cross sectional area A.

The thermal resistance of element is given as

Where A =ab

Test: Conduction Level - 2 - Question 9

Heat is generated in a 3-cm-diameter spherical radio-active material uniformly at a rate of 15 W/cm3. Heat is dissipated to the surrounding medium at 25°C with a heat transfer coefficient of 120 W/m2-K. The surface temperature of the material in steady operation is

Detailed Solution for Test: Conduction Level - 2 - Question 9

At steady state,

Rate of heat generation inside the sphere

= Rate of heat dissipation from the surface

T8 = 650°C

Note:

In L.H.S, radius is put in cm, because is given in W/cm3 but in R.H.S, radius is put in m because h is given in W\m2-k.

*Answer can only contain numeric values
Test: Conduction Level - 2 - Question 10

A thick walled tube of stainless steel [18% Cr, 8% Ni,k = 19 W/m ͘· °C] with 2-cm inner diameter (ID) and 4-cm outer diameter (OD) is covered with a 3-cm layer of asbestos insulation [k = 0.2 W/m ͘· °C]. If the inside wall temperature of the pipe is maintained at 600°C calculate the tube-insulation interface temperature. Assume the outermost surface temperature to be 300°C


Detailed Solution for Test: Conduction Level - 2 - Question 10

given that

r1 = 1 cmR1 = 1 cm

r2 = 2 cm

r3 = 2 + 3 = 5 cm

At steady state,

heat transfer rate by conduction in stainless steel

= heat transfer rate by conduction in asbestos

Ti = 597.63°C

Test: Conduction Level - 2 - Question 11

A copper wire of radius 0.5 mm is insulated with a sheathing of thickness 1 mm having a thermal conductivity of 0.5 W/m-K. The outside surface convective heat transfer coefficient is 10 W/m2-K. If the thickness of insulation sheathing is raised by 10 mm, then the electrical current-carrying capacity of the wire will

Detailed Solution for Test: Conduction Level - 2 - Question 11

But according to the given data the initial value of outer radius of insulation is rin = 0.5 = 1 = 1.5 mm

If the radius of insulation is less than the critical radius, then by adding insulation, heat transfer rate increases. So by increasing the insulation by 10 mm, the radius of insulation will become 11.5 mm that means the heat transfer rate will also increase. That’s why the current carrying capacity of wire will also increase.

Test: Conduction Level - 2 - Question 12

A cylindrical pin fin of diameter 0.6 cm and length of 3 cm with negligible heat loss from the tip has an efficiency of 0.7. The effectiveness of this fin is

Detailed Solution for Test: Conduction Level - 2 - Question 12
d = 0.6 cm = 0.006 m and ?fin = 0.7

L = 3 cm = 0.003m

∵ Given that the fin is insulated from the tip,

hence the effectiveness

Test: Conduction Level - 2 - Question 13

The maximum temperature inside a solid sphere maintained at steady state with one dimensional temperature profile having internal heat generation rate and surface temperature is Tw is

Detailed Solution for Test: Conduction Level - 2 - Question 13
For spherical coordinates, the general heat conduction equation with internal heat generation is given by

By solving the above equation we get

Let

Tc = center temperature

Tw = surface temperature

= heat generated per unit volume

∵ at r = 0, tT = Tc

If we put this B.C in equation ① then we get finite value in L.H.S. and undefined in R.H.S.

Hence mathematically equation is true only if

C1, = 0, then the equation becomes

Hence,

Test: Conduction Level - 2 - Question 14

Two very long fins of same length L, perimeter P and area of cross-section A are attached on a surface with temperature To . The surface is exposed to a fluid with temperature Tf & coefficient of convection coefficient h. One fin is made of copper and the other of aluminium. Considering the fins to be at steady state with one dimensional heat transfer which one of the following options represents the correct temperature profile.

Detailed Solution for Test: Conduction Level - 2 - Question 14
For infinitely long fin, the temperature profile is given as

T - Tf = (To - Tf)e-me

Consider the slope of temperature profile at

x = 0

Thus greater is the value of m, more steep is the slope i.e., lower is the value of k. As kal < />cu, the slope is more steep for aluminium fins.

Test: Conduction Level - 2 - Question 15

The temperature distribution in a fin (thermal conductivity 0.17 W/cm-°C) with uniform cross-sectional area of 2 cm2 and length of 1 cm exposed to ambient of 40°C (with a surface heat transfer coefficient of 0.0025 W/cm2-°C) is given by (T - T ) = 3x2 - 5x + 6, where T is in °C and x is in cm. If the base temperature is 100°C , then the heat dissipated by the fin surface will be

Detailed Solution for Test: Conduction Level - 2 - Question 15

∵ T - T= 3x2 - 5x + 6

A = 2 cm2

∵ The rate of heat dissipated by the fin will be equal to the rate of heat conducted into the fin from base surface.

Test: Conduction Level - 2 - Question 16

A small solid copper ball of mass 500 grams, when quenched in a water bath at 30°C cools from 530°C to 430°C in 10 seconds. What will be the temperature of the ball after the next 10 seconds?

Detailed Solution for Test: Conduction Level - 2 - Question 16
As the ball is small & is made of copper (high k), therefore the ball can be treated as a lump. Therefore variation of its temperature with time is given as

We need to find T after next 10 seconds i.e.

At t = 10 + 10

= 20 seconds.

T = 350°C

Test: Conduction Level - 2 - Question 17

A spherical thermocouple junction of diameter 0.706 mm is to be used for the measurement of temperature of a gas stream. The convective heat transfer co-efficient on the bead surface is 400 W/m2K. Thermo-physical properties of thermo couple material are k = 20 W/m-K,C = 400 J/kg- K and

? = 8500 kg/ m3. If the thermocouple initially at 20°C is placed in a hot stream of 500°C, the time taken by the bead to reach 400°C, is

Detailed Solution for Test: Conduction Level - 2 - Question 17
Consider the Biot’s number of thermocouple

= 0.0023

Which is much less than 0.1 Thus the thermocouple can be solved as lump & variation of it’s temperature with time is given as

⇒ t = 1.57 s

Test: Conduction Level - 2 - Question 18

A lump body cools from 90°C to 80°C in 5 minutes. Under the same external conditions, to cool from 80°C to 70°C the body will take

Detailed Solution for Test: Conduction Level - 2 - Question 18
For a lump, the temperature decreases exponentially with time and the rate of fall of temperature in the beginning will be faster and then it slows down.

Therefore, if time taken by the body in cooling from 90°C to 80°C is 5 min, then definitely time taken by the body in cooling from 80°C to 70°C will be more than 5 minutes.

*Answer can only contain numeric values
Test: Conduction Level - 2 - Question 19

Determine the heat transfer rate from the rectangular fin of length 20 cm, width 40 cm and thickness 2 cm. The tip of the fin is not insulated and the fin has a thermal conductivity of 150 W/m-K. The base temperature is 100°C and the fluid is at 20°C.

The heat transfer coefficient between the fin and the fluid is 30 W/m2-K.


Detailed Solution for Test: Conduction Level - 2 - Question 19
The extended length

A = 40 X 2 = 80 cm2 = 0.008 m2

As = LcP = 0.21 X 0.84 = 0.1764 m2

= 4.583 X 150 X 0.008 X 80

X tanh( 4.583 X 0.21)

= 328.0 W

= 0.775 X 30 X 0.1764(100 - 20) = 328 W.

*Answer can only contain numeric values
Test: Conduction Level - 2 - Question 20

An exterior wall of a house may be approximated by a 10 cm layer of common brick (k = 0.75 W/m-°C) followed by a 4 cm layer of gypsum plaster (k = 0.5w/m-°C). What thickness of loosely packed rock wool insulation (k = 0.065 W/m-°C) should be added to reduce the heat loss or gain through the wall by 75% ?


Detailed Solution for Test: Conduction Level - 2 - Question 20
Thermal resistance for a plane wall of thickness ?, area A and thermal conductivity k is prescribed by the relation Rt = ?/kA. Considering unit area of wall perpendicular to heat flow direction,

Resistance of brick work

Resistance of gypsum plaster

Resistance of rock wool insulation

Where x is thickness of rock wool insulation in cm

Case I: Rockwool insulation is not used:

Heat loss or gain

Case II: Rockwool insulation is used:

Heat loss or gain

or (0.133 + 0.08 + 0.1538 x)

∴ Thickness of rock wool, x

= 4.153 cm

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