Convection
Exam Duration: 45 Mins Total Questions : 25
Which non-dimensional number relates the thermal boundary layer and hydrodynamic boundary layer?
- (a)
Rayleigh number
- (b)
Peelet number
- (c)
Grashof number
- (d)
Prandtl number
\(\cfrac { { \delta }_{ t } }{ \delta } =\cfrac { 1 }{ \left( Pr \right) ^{ { 1 }/{ 3 } } } \)
Match List I (Non-dimensional number) with List II (Application) and select the correct answer using the codes given below the lists
List I | List II |
P. Grashof number | 1. Mass transfer |
Q. Stanton number | 2. Unsteady state heat conduction |
R. Sherwood number | 3. Free convection |
S. Fourier number | 4. Forced convection |
- (a)
P Q R S 4 3 1 2 - (b)
P Q R S 3 4 1 2 - (c)
P Q R S 4 3 2 1 - (d)
P Q R S 3 4 2 1
For a fluid having Prandtl number equal to unity, how are the hydrodynamic bondary layer thickness \({ \delta }_{ t }\) and the thermal boundary layer thickness \({ \delta }_{ t }\) related?
- (a)
\(\delta ={ \delta }_{ t }\)
- (b)
\(\delta >{ \delta }_{ t }\)
- (c)
\(\delta <{ \delta }_{ t }\)
- (d)
\({ \delta }_{ t }={ \delta }^{ 1/3 }\)
\(\cfrac { { \delta }_{ t } }{ \delta } =\cfrac { 1 }{ \left( PR \right) ^{ 1/3 } } \) as Pr = 1, \({ \delta }_{ t }=\delta \)
The Nusselt number is related to Reynolds number in laminar and turbulent flows respectivley a
- (a)
Re-1/2 and Re0.8
- (b)
Re1/2 and Re0.8
- (c)
Re-1/2 and Re-0.8
- (d)
Re1/2 and Re-0.8
For laminar flow, Nu = 0.332 (Re)1/2(Pr)1/3
For turbulent flow, Nu = 0.0288 (Re)0.8(Pr)1/3
Heat is lost from a 100 mm diameter steam pipe placed horizontally in ambient at 30"C. If the Nusselt number is 25 and thermal conductivity of air is 0.03 W/m-K, then the heat transfer coefficient will be
- (a)
7.5 W/m2-K
- (b)
16.2 W/m2-K
- (c)
25.2 W/m2-K
- (d)
30 W/m2-K
Nu = \(\cfrac { hL }{ K } \), if air flows over the plate
Nu = \(\cfrac { hd }{ K } \) , if air flows inside the plate
h = \(\cfrac { KNu }{ d } \) =7.5 W/m2 -k
Match List I (Process) with List II (Predominant parameter associated with the flow) and select the correct answer using the codes given below the lists
List I | List II |
P. Transient conduction | 1. Sherwood number |
Q. Mass transfer | 2. Biot number |
R. Forced convection | 3. Grashof number |
S. Free convection | 4. Reynolds number |
- (a)
P Q R S 2 4 3 1 - (b)
P Q R S 3 1 2 4 - (c)
P Q R S 2 1 4 3 - (d)
P Q R S 3 4 2 1
A 320 cm high vertical pipe at 150"C wall temperature is in a room with still air at 10"C. This pipe supplies heat at the rate of 8 kW into the room air by natural convection.
Assuming laminar flow, the height of the pipe needed to supply 1 kW only is
- (a)
10cm
- (b)
20 cm
- (c)
40 cm
- (d)
80 cm
Q1 = hA(TW - Ta)
= \(h\left( \pi d \right) { L }_{ 1 }\triangle T\)
Q2 = \(h\left( \pi d \right) { L }_{ 2 }\triangle T\)
\(\cfrac { { L }_{ 1 } }{ { L }_{ 2 } } =\cfrac { 8 }{ 1 } \Rightarrow \)L2=40 cm
The average Nusselt number in laminar natural convection from a vertical wall at 180oC with still air at 20 ° C is found to be 48. If the wall temperature becomes 30oC, all other
parameters remaining same, the average Nusselt number will be
- (a)
8
- (b)
16
- (c)
24
- (d)
32
For natural convection, Nu = (Gr·Pr )1/4
Gr = \(\cfrac { g\cdot \beta \cdot { L }^{ 3 }\cdot \triangle T }{ { v }^{ 2 } } ,\), Pr = \(\cfrac { \mu { C }_{ p } }{ K } \)
According to question, all parameters remains same besides temperature.
\(\cfrac { { Nu }_{ 1 } }{ { Nu }_{ 2 } } =\cfrac { { \left( \triangle T \right) }^{ 1/4 } }{ \left( \triangle T_{ 2 } \right) ^{ 1/4 } } \)
\(\Rightarrow\) \(\cfrac { 48 }{ N{ u }_{ 2 } } =\cfrac { \left( 180-20 \right) ^{ 1/4 } }{ \left( 30-20 \right) ^{ 1/4 } } \)
Nu2 = 24
In a thermal boundary region, thickness of thermal boundary layer is \({ \delta }_{ t }\) If air flow over the plate has temperature \({ T }_{ \infty }\) and T(y) = ay - by 2, where a and bare constant.
Which of the following is correct?
- (a)
\({ \delta }_{ t }\) > 0.5, if a=b
- (b)
\({ \delta }_{ t }\) = 0.5,if a=2,b =1
- (c)
\({ \delta }_{ t }\) = 0.5, if a = 1, b = 2
- (d)
\({ \delta }_{ t }\) < 0.5, if a < b
At the boundary layer, where y = \({ \delta }_{ t }\)
Q = -KA\(\cfrac { dT }{ dy } \) = 0
(Because after this point temperature does not change.)
\(\cfrac { d }{ dy } \left( ay-by^{ 2 } \right) { | }_{ y={ \delta }_{ t } }=0\)
a = 2b \(\cdot { \delta }_{ t }\)
\(\Rightarrow\) \({ \delta }_{ t }=0.5\left( \cfrac { a }{ b } \right) \)
For a = 2, b = l, \({ \delta }_{ t }=0.5\times \cfrac { 2 }{ 1 } =1\)
For a = 1, b = 2, \({ \delta }_{ t }\) = 0.05
If b > a, \(\cfrac { a }{ b } \) < 1, \({ \delta }_{ t }\)< 0.5
In a thermal boundary region, thickness of thermal boundary layer is \({ \delta }_{ t }\) If air flow over the plate has temperature \({ T }_{ \infty }\) and T(y) = ay - by 2, where a and bare constant.
If \({ T }_{ \infty }\)= 100oC, \({ \delta }_{ t }\) = 0.5 m, then value of a is (If b = 0)
- (a)
200°C/m
- (b)
100°C/m
- (c)
400°C/m
- (d)
300°C/m
T (y) = ay - by2
If \(y={ \delta }_{ t }\), \(T=\left( { \delta }_{ t } \right) ={ T }_{ \infty }\) (air temperature)
100 = a X \({ \delta }_{ t }\)
a =\(\cfrac { 100 }{ 0.5 } \) = 200oC/m
The temperature distribution within the thermal boundary layer over a heated isothermal flat plate is given by
\(\cfrac { T-{ T }_{ w } }{ { T }_{ \infty }-{ T }_{ w } } =\cfrac { 3 }{ 2 } \left( \cfrac { y }{ { \delta }_{ t } } \right) -\cfrac { 1 }{ 2 } { \left( \cfrac { y }{ { \delta }_{ t } } \right) }^{ 3 }\)
where, Tw and \({ T }_{ \infty }\) are the temperature of plate and free stream respectively and y is the normal distance measured from the plate. The local Nusselt number based on the thermal boundary layer thickness \({ \delta }_{ t }\) is given by
- (a)
1.33
- (b)
1.50
- (c)
2.0
- (d)
4.64
We know at surface of Nusselt number based on thermal boundary layer is as plate,
\(-KA\cdot \cfrac { \partial T }{ \partial y } |_{ y=0 }=h\cdot A\left( { T }_{ w }-{ T }_{ \infty } \right) \)
Tw and \({ T }_{ \infty }\) are plate surface temperature and air temperature respectively.
\(\Rightarrow\) \(-K\cdot \cfrac { d }{ dy } \left[ { T }_{ w }+\left( { T }_{ \infty }-{ T }_{ w } \right) \left\{ \cfrac { 3 }{ 2 } \cfrac { y }{ { \delta }_{ t } } -\cfrac { 1 }{ 2 } \cfrac { { y }^{ 3 } }{ { { \delta }_{ t }^{ 3 } } } \right\} \right] _{ y=0 }\)
=\(h\cdot \left( { T }_{ w }-{ T }_{ \infty } \right) \)
\(\Rightarrow\) \(K\cdot \left( \cfrac { 3 }{ 2\delta _{ t } } -\cfrac { 3 }{ 2 } \cfrac { { y }^{ 2 } }{ { \delta }_{ t }^{ 3 } } \right) _{ y=0 }=h\)
h = \(\cfrac { K\left( 1.5 \right) }{ { \delta }_{ t } } \)
Nu = \(\cfrac { hu }{ K } \)
Nu = \(\cfrac { h{ \delta }_{ t } }{ K } \)
Nu = \(\cfrac { 1.5K{ \delta }_{ t } }{ { \delta }_{ t }K } =1.5\)
A straight tube having a diameter of 40 mm carries water with a velocity of 10 m/s.The temperature of the tube surfaceis 45°C and the flowing water is heated from the inlet temperature Ti= 10 oC to an outlet temperature To = 20oC. Physical properties of water at its mean bulk temperature are
v = 1.006 x 10-6 m2/s
K = 59.86 x 10-2 W/m-K
Cp = 4183 J/kg-K; Pr = 0.702
The coefficient of heat transfer from the tube surface to water will be
- (a)
52.4 kW/m2-K
- (b)
9.2 kW/m2-K
- (c)
524.13 kW/m2-K
- (d)
91.5 kW/m2-K
\({ R }_{ e }=\cfrac { \rho vD }{ \mu } =\cfrac { vD }{ V } \)
=\(\cfrac { 10\times 0.04 }{ 1.006\times { 10 }^{ -6 } } \)
= 3.976 X 106
\(\therefore\) Flow is turbulent, because Reynolds number is greater than 2500, hence
Nu = 0.023 Re0.8 (Pr)0.33
= 0.023(3.976 X 105)0.8(0.702)0.33
= 617.41
Now,Nu = \(\cfrac { hd }{ K } \)
\(\therefore\) h = \(\cfrac { Nu\times h }{ d } \)
=\(\cfrac { 617.41\times 59.86\times { 10 }^{ -2 } }{ 0.04 } \)
= 9239.6 w/m2-K
= 9.2 kW/m2-K
A straight tube having a diameter of 40 mm carries water with a velocity of 10 m/s .The temperature of the tube surfaceis 45°C and the flowing water is heated from the inlet temperature Ti= 10 oC to an outlet temperature To = 20oC. Physical properties of water at its mean bulk temperature are
v = 1.006 x 10-6 m2/s
K = 59.86 x 10-2 W/m-K
Cp = 4183 J/kg-K; Pr = 0.702
Length of the tube will be
- (a)
9.05 m
- (b)
15.05 m
- (c)
21.05 m
- (d)
27.05 m
Heat transfer from tube = h \(\pi \)dL (Ts - Tb)
= Change in internal energy of water
= m Cp (To - Ti)
\(h\pi dL\left( { T }_{ s }-\cfrac { { T }_{ o }+{ T }_{ i } }{ 2 } \right) =\rho \times \cfrac { \pi }{ 4 } { d }^{ 2 }\left( \cfrac { L }{ T } \right) { C }_{ p }\left( { T }_{ 0 }-{ T }_{ i } \right) \)
\(h\cdot L\left( { T }_{ s }-\cfrac { { T }_{ o }+{ T }_{ i } }{ 2 } \right) =\cfrac { \rho }{ 4 } \times { d }\times u\times { C }_{ p }\left( { T }_{ 0 }-{ T }_{ i } \right) \)
Putting h = 9239.6 W/m2-K
Ts = 45°C, To = 20°C
1i = 10°C, P = 1000 kg/rn3
d = 0.04 rn, u = 10 rnIs
Cp = 4.83 J/kg-K
We get L = 15.05 m
Air flows through a 10 cm internal diameter tube at the rate of 75 kg/h. Measurement indicate that at a particular point in the tube, the pressure and temperature of air are 15 bar and 350 K respectively, while the tube wall temperature is 400 K. General non dimensional correlation for turbulent the tube is
Nu = 0.023 Reo.8 PrO.4
where fluid properties are evaluated at the bulk temperature.
Given,\(\mu \) = 1.967 x 10-5 WIm2-K,
K = 0.02792 W/m-K, Pr = 0.713
The value of convective heat transfer coefficient will be
- (a)
7.296 W/m2-K
- (b)
9.76 W/m2-K
- (c)
11.29 W/m2-K
- (d)
None of these
When you are given mass flow rate then,
m = density x volume
m = \(\rho \)AXl
m =\(\rho A\times \cfrac { l }{ T } =\rho Au\)
\({ R }_{ e }=\cfrac { \rho ud }{ \mu } =\cfrac { m }{ A } \cdot \cfrac { d }{ \mu } =\cfrac { 4m }{ \pi d\mu } \)
\(\cfrac { 4\times \left( \cfrac { 75 }{ 3600 } \right) }{ \pi \times { \left( 0.1 \right) }^{ 2 }\times \left( 1.967\times { 10 }^{ -5 } \right) } \)
= 13492
Nu = 0.023 (13492)0.8 (0.713)0.4
= 40.46
h = \(\cfrac { Nu\times K }{ d } =\cfrac { 40.46\times 0.02792 }{ 0.1 } \)
= 11.296 W/m2-K
Air flows through a 10 cm internal diameter tube at the rate of 75 kg/h. Measurement indicate that at a particular point in the tube, the pressure and temperature of air are 15 bar and 350 K respectively, while the tube wall temperature is 400 K. General non dimensional correlation for turbulent the tube is
Nu = 0.023 Reo.8 PrO.4
where fluid properties are evaluated at the bulk temperature.
Given, \(\mu \) = 1.967 x 10-5 W/m2-K,
K = 0.02792 W/m-K, Pr = 0.713
The heat transfer rate from one metre length in the region of this point is
- (a)
95.35 W
- (b)
105.35 W
- (c)
150.35 W
- (d)
177.35 W
Heat flow rate
= hA\(\Delta \)T
= 11.296 x (\(\pi \)x 0.1 x 1)(375 -325)
= 177.35 W
Air passes through the every face of the plate (Ts = 60oC). For air,
Ta=20oC
\(\rho \) =1.09kg/m3
\(\mu \) = 20.1X120-6N-s/m2
K = 0.027W/m-K
Pr = 0.7
u = 30 m/s
Assume flow is turbulent with NuL = 0.036 \({ Re }_{ L }^{ 0.8 }\) Pr1/3
Value of NuL is=
- (a)
2200
- (b)
2237.8
- (c)
2507
- (d)
2001
\({ Re }_{ L }=\cfrac { \rho vL }{ \mu } =\cfrac { 1.09\times 30\times 0.7 }{ 20.1\times 10^{ -6 } } \)
NUL = 0.036 (ReL)0.8 (Pr)1/3= 2237.8
ir passes through the every face of the plate (Ts = 60oC). For air,
Ta=20oC
\(\rho \) =1.09kg/m3
\(\mu \) = 20.1X120-6N-s/m2
K = 0.027W/m-K
Pr = 0.7
u = 30 m/s
Assume flow is turbulent with NuL = 0.036 \({ Re }_{ L }^{ 0.8 }\) Pr1/3
Rate of heat transfer from all surface is
- (a)
1.754 kW
- (b)
1754 kW
- (c)
1712W
- (d)
1892W
\({ Re }_{ L }=\cfrac { h\cdot L }{ K } \Rightarrow \)h = 86.32 W/m2-K
Total surface area = 2(lb + bh + hl)
= 0.508m2
Heat transfer = \(hA\left( { T }_{ s }-{ T }_{ \infty } \right) \)
= 86.32 X 0.508X(60-20)
= 1754 W
\(Re=\cfrac { \rho vL }{ \mu } =8.59\times { 10 }^{ 4 }\)
A flat plate is 2 m long, 0.8 m wide and 3 mm thick. Density of plate is 3000 kg/m3. Specific heat of plate material is 700 J/kg-K. Its initial temperature is 90oC. A stream of air at 30oC is blow over both surfaces of the plate along its width, at a velocity 2 m/s.
Properties of air, \(\rho \) = 1.09 kg/m3, K = 0.028 W/m-K,
Pr = 0.698, \(\mu \) = 2.03 x 10-5 kg/m-s,
Nu = 0.664 (Re)1/2 (Pr)1/3
Initial rate of cooling is
- (a)
0.0058 oCIs
- (b)
0.0058oCIs
- (c)
0.0116 oC/min
- (d)
None of these
Heat transfer from the plate = Change in internal energy of plate
= \(m{ C }_{ p }\Delta T\)
= \(\rho lbt{ C }_{ p }\Delta T\)
= 3000 X 2 X 0.8 X 0.03 X 700\(\cdot \Delta \)T
= 100800\(\Delta \)T J/K
Q = mCp\(\Delta \)T
\(Q=m{ C }_{ p }\cfrac { \Delta T }{ t } =1173\)
(100800)\(\cfrac { \Delta T }{ t } \)=1173
\(\cfrac { \Delta T }{ t } =\cfrac { 1173 }{ 100800 } \) = 0.0116oC/s
Re = \(\cfrac { \rho vD }{ \mu } \)
Air flows between the space of two concentric pipes as shown in figure. Air is at 2 atm and 200"C and flows with a velocity of 12 m/s. Properties of air
Pr = 0.681, \(\mu \)= 2.57 x 10-5 kg/m-s, K = 0.0386 WIm-K and Cp = 1.025 kJ/kg-K, NUd = 0.023 (Re)0.8 (Pr)O.4. Assume wall temperature is 20°C above the air temperature all along the length of tube .
If D = 5 cm, d = 2 cm, then heat transfer per unit length of tube is
- (a)
140 W/m
- (b)
136.79 W/m
- (c)
171 W/m
- (d)
None of these
\(Re=\cfrac { \rho vD }{ \mu } \)
But neither v nor \(\rho \) is given, we have to calculate p by
pV= mRT
p = \(\rho \)RT
\(\rho =\cfrac { \rho }{ RT } \)
Putting,
R = 287 J/kg-K
1 atm = 105 Pa = 105 N/m2
\(\rho =\cfrac { 2\times 10^{ 5 } }{ 287\times 473 } \)
=1.493 kg/m3
\(Re=\cfrac { \rho vD }{ \mu } \)=20914>4000
Hence, turbulent flow, Nu = 0.023 (Re)0.8 (Pr)0.4
Nu = \(\cfrac { hD }{ K } \)
But air flows in the space between both pipes. Hence, we have to calculate equivalent diameter.
\({ D }_{ equ }=\cfrac { 4{ A }_{ e } }{ P } =\cfrac { 4\cfrac { \pi }{ 4 } \left( { D }^{ 2 }-{ d }^{ 2 } \right) }{ \pi D+\pi d } \)
Dequ = D - d = 5 - 2 = 3 cm
h=\(\cfrac { 56.404\times 0.0386 }{ 0.03 } \)=75.27
\(\cfrac { \\ Q }{ L } =h\pi d\left( { T }_{ s }-{ T }_{ \infty } \right) \)=136.9 W/m
(Given, wall temperature is 20°C above the air. \(\therefore\)Ts- \({ T }_{ \infty }\) 20°C)
Air flows between the space of two concentric pipes as shown in figure . Air is at 2 atm and 200"C and flows with a velocity of 12 m/s. Properties of air
Pr = 0.681, \(\mu \)= 2.57 x 10-5 kg/m-s, K = 0.0386 WIm-K and Cp = 1.025 kJ/kg-K, NUd = 0.023 (Re)0.8 (Pr)O.4. Assume wall temperature is 20°C above the air temperature all along the length of tube .
If D = 5 cm, d = 2 cm, then heat transfer per unit length of tube is
Temperature of air after 4 m length of tube
- (a)
42.15oC
- (b)
242.15oC
- (c)
157.85oC
- (d)
Data is insufficient
\(Re=\cfrac { \rho vD }{ \mu } \)
But neither v nor \(\rho \) is given, we have to calculate p by
pV= mRT
p = \(\rho \)RT
\(\rho =\cfrac { \rho }{ RT } \)
Putting,
R = 287 J/kg-K
1 atm = 105 Pa = 105 N/m2
\(\rho =\cfrac { 2\times 10^{ 5 } }{ 287\times 473 } \)
=1.493 kg/m3
\(Re=\cfrac { \rho vD }{ \mu } \)=20914>4000
Hence, turbulent flow, Nu = 0.023 (Re)0.8 (Pr)0.4
Nu = \(\cfrac { hD }{ K } \)
But air flows in the space between both pipes. Hence, we have to calculate equivalent diameter.
\({ D }_{ equ }=\cfrac { 4{ A }_{ e } }{ P } =\cfrac { 4\cfrac { \pi }{ 4 } \left( { D }^{ 2 }-{ d }^{ 2 } \right) }{ \pi D+\pi d } \)
Dequ = D - d = 5 - 2 = 3 cm
h=\(\cfrac { 56.404\times 0.0386 }{ 0.03 } \)=75.27
Let temperaure of air after 4 m length is T. During this length, rate of change in internal energy of air
=mCp\(\Delta \)T
=\(\rho \cfrac { \pi }{ 4 } { d }_{ equ }^{ 2 }\mu \Delta T\times { C }_{ p }\)
=1.493 X \(\cfrac { \pi }{ 4 } \) X (0.03)2 X 12 X 1.02 X 103\(\Delta \)T
It will be equal to heat transfer rate during this length
= 136.79 x 4 = 547.16 W
By equating both, we get
\(\Delta \)T = 42.15°C
As wall temperature is always higher than air temperature.
\(\therefore\) \(\Delta \)T = TL - Ti = TL - 200
TL = 242.15°C
Air having temperature 200°C flows over a plate whose surface temperature is 50° C. For plate, L = 10 cm, b = 5 cm, t = 2 cm, K = 10 W/m-K, value of heat transfer coefficient at distance x from leading edge is as h(x)=Coex/L
Here, Co is a constant.
Value of average heat transfer coefficient is
- (a)
\(\bar { h } \)= 1.71, Co:= 2
- (b)
\(\bar { h } \)= 3.43, Co = 1
- (c)
\(\bar { h } \)= 0.85, Co = 0.5
- (d)
None of these
Average heat transfer coefficient
=\(\cfrac { 1 }{ L } \int _{ 0 }^{ L }{ h(x)\cdot dx } \)
=\(\cfrac { 1 }{ L } \cdot \int _{ 0 }^{ L }{ { C }_{ 0 }\cdot { e }^{ x/L }dx } \)
=\(\cfrac { 1 }{ L } { C }_{ 0 }L{ \left[ { e }^{ x/L } \right] }_{ o }^{ L }\)
= C0(e-1)
=1.71 Co
Co = 0.5, \(\bar { h } \) = 0.85
Air having temperature 200°C flows over a plate whose surface temperature is 50° C. For plate, L = 10 cm, b = 5 cm, t = 2 cm, K = 10 W/m-K, value of heat transfer coefficient at distance x from leading edge is as h (x)=Coex/L
Here, Co is a constant.
At steady state, temperature of point P will be (if Co = 15 W/m2-K)
- (a)
5.1oC
- (b)
45oC
- (c)
42.30oC
- (d)
50oC
At steady state, heat transfer at plate surface = Heat passing through P
\(\bar { hA } \cdot \left( { T }_{ \infty }-{ T }_{ s } \right) =KA\cfrac { \left( { T }_{ s }-{ T }_{ P } \right) }{ t } \)
(\(\bar { h } \)= 1.71 Co = 1.71 x 15 = 25.65 W/m2-K)
25.65X(200-50) = \(\cfrac { 10\times \left( 50-{ T }_{ P } \right) }{ 0.02 } \)
Tp = 42.30oC
Properties of air are as follows:
\(\rho \) = 1.02 kg/m3, v = 18 x 10-6 m2/s, Cp = 1.005 kJ,/Kg-K,
\(\mu \) = 20 x 10-6 N-s/m, K = 0.029 W/m-K and
g = 9.8 m/s2, Ts = 100oC, \({ T }_{ \infty }\) = 30oC
Prandtl number is
- (a)
6.93
- (b)
0.00693
- (c)
0.0693
- (d)
0.693
\(Pr=\cfrac { \mu { C }_{ p } }{ K } =\cfrac { 20\times 10^{ -6 }\times 1.005\times { 10 }^{ 3 } }{ 0.029 } =0.693\)
Coefficient of volumetric expansion \(\beta \) is
- (a)
\(\beta =-\rho \left( \cfrac { \partial \rho }{ \partial T } \right) \)
- (b)
\(\beta =-\cfrac { 1 }{ \rho } \left( \cfrac { \partial \rho }{ \partial T } \right) _{ T=c }\)
- (c)
\(\beta =-\cfrac { 1 }{ \rho } \left( \cfrac { \partial \rho }{ \partial T } \right) _{ p=c }\)
- (d)
\(\beta =-\cfrac { 1 }{ T } \left( \cfrac { \partial \rho }{ \partial T } \right) _{ p=c }\)
A plate is maintained at 74oC and this surface is facing an air having temperature of 30oC, then what will be the value of coefficient of volumetric expansion \(\beta \) ?
- (a)
325/K
- (b)
0.0192/K
- (c)
3.07 x 10-3/K
- (d)
None of these
\({ T }_{ m }=\cfrac { 74+30 }{ 2 } =\cfrac { 104 }{ 2 } \)= 50oC
Tm = 52 + 273 = 325 K
\(\beta =\cfrac { 1 }{ { T }_{ m } } =\cfrac { 1 }{ 325 } \) = 3.07 X 10-3/K