PTU-HEAT TRANSFER PAPER SOLUTION-B

 

https://www.mesubjects.net/wp-admin/post.php?post=7462&action=edit        Dimensionless num HT

https://www.mesubjects.net/wp-admin/post.php?post=7420&action=edit        HT Symbols-2

https://www.mesubjects.net/wp-admin/post.php?post=7414&action=edit         HT Symbols-1

https://www.mesubjects.net/wp-admin/post.php?post=7368&action=edit          MCQ HT-1

https://www.mesubjects.net/wp-admin/post.php?post=7340&action=edit           MCQ HT-2

https://www.mesubjects.net/wp-admin/post.php?post=6434&action=edit             PTU HT Paper Sol A

https://www.mesubjects.net/wp-admin/post.php?post=6261&action=edit            Fin HT

PTU-HEAT TRANSFER PAPER SOLUTION-B

Solution of University question paper increases the under standing to a larger extent. One is to recognize the importance of various topics of a subject.

SECTION-B

 Q2. Explain different phases of flow boiling.

ANS:

Flow boiling is much more complex than nucleate boiling. Here it will be a two phase heat transfer. Consider a vertical tube as shown in figure.

 

Sub-cooled liquid is entering from below. As it moves upwards, getting heat from the tube which is being heated externally at constant heat flux, different phases of flow boiling with changing dryness fraction will be observed. These phases are

  1. Single phase liquid at entrance—at a temperature lower than the boiling point
  2. Saturated liquid—– at the boiling temperature
  3. Bubbly flow—vapors starts forming and going up with rest of the liquid.
  4. Slug flow regime: small vapors already formed change into bigger bubbles called slugs and hence the flow is called slug flow.
  5. Annular flow: Rate of vapor formation increases and the vapors start moving as a central core while liquid is moving in an annulus around the vapors flow.
  6. Mist flow: Dryness fraction becomes slightly greater than 0.25, the annular liquid disappears and vapors are carrying drops of liquid.
  7. Only dry saturated vapors flow.

Q3. Derive three dimensional heat conduction equations in spherical coordinates. Reduce the equation to one dimension, steady state without internal heat generation.

SOLUTION

The final equation will be in Spherical coordinates as given below:

Second order differential equation (3 Dimensional)

2t/∂r2 + (2/r) ∂t/∂r + (1/r2sinϴ)∂/∂ϴ[sinϴ∂t/∂ϴ] + (1/ r2sin2ϴ) ∂2t/∂Ф2 +qg./k =(1/ ) ∂t/∂ȥ

SPHERICAL COORDINATES ARE

r,ϴ andФ

Size of the spherical element will be dr, rdϴ and r sinϴ dФ

This is the size of a spherical element.

[It is exactly similar to a piece which a water melon seller gives to a customer who is interested to by a water melon. It is done to insure that seller is selling only sweet melon).

Inside the sphere there is a heat generation per unit volume = q.g

Total heat generated= q.g  δV

Qr =-k Ar ∂t/∂r =  -k  (rdϴ)( r sinϴ dФ) ∂t/∂r

Qr =  -k  (rdϴ)( r sinϴ dФ) ∂t/∂r

Energy balance for the element

  • In the radial direction

Qr+dr = Qr + (∂ Qr/∂r) dr

NET IN r direction =(∂ Qr/∂r) dr =(∂(-k  (rdϴ)( r sinϴ dФ) ∂t/∂r) /∂r) dr

= –k dϴ sinϴ dФ ∂(r2∂t/∂r) /∂r) dr

=[ –kdr rdϴ rsinϴ dФ/r2] ∂(r2∂t/∂r) /∂r)

= –k [δV / r2 ][ r22t/∂2r + 2r ∂t/∂r]

= —k{[ δV][ ∂2t/∂2r] + (1/r) ∂t/∂r}

(∂Qr/∂r) dr       = –k δV[∂2t/∂2r] + (1/r) ∂t/∂r]     (1)

  • In the ϴ direction ( r-Ф plane)

Qϴ =  -k  [( r sinϴ dФ dr) ∂t/(r∂ϴ)][ rdϴ]

= –k [(rdϴ r sinϴ dФ dr) ∂t/(r∂ϴ)]

Qϴ  = —(k/r) δV ∂t/∂ϴ

Qϴ+dϴ = Qϴ +  (∂Qϴ / r∂ϴ)rdϴ

Substitute the value of Qϴ

We get  the net outflow in ϴ direction

Qϴ+dϴ — Qϴ = (∂Qϴ / r∂ϴ)rdϴ

(∂Qϴ / r∂ϴ)rdϴ = (∂/rsinϴ[-(k/r) δV ∂t/∂ϴ] / r∂ϴ) = –(k/r2sin ϴ) δV∂2t/∂2ϴ  (2)

© QФ = –k (rdϴ dr) ∂t/(rsinϴ∂Ф)]

Similarly net flow in Ф direction will be

QФ+dФ = QФ + ∂QФ/r sinϴ ∂Ф) r sinϴ ∂Ф

— QФ =dQФ = (dQФ/rsinϴdФ) rsinϴdФ

(dQФ/rsinϴdФ) rsinϴdФ  =–k (rdϴ dr)/(rsinϴ)∂2t/∂2ФX 1/r sinϴ (3)

= –k [δV/ r2sin2ϴ] ∂2t/∂2Ф

Rate of change of heat inside= ρ δVcp∂t/∂ ȥ  (4)

Now combining equations (1), (2), (3) and (4), we get

–k δV[∂2t/∂2r] + (1/r) ∂t/∂r]- –(k/r2sin ϴ) δV∂2t/∂2ϴ–(k/r2sin ϴ) δV∂2t/∂2ϴ+ q.g  δV = ρ δVcp∂t/∂ ȥ

Rearranging and dividing by k δV, we get

2t/∂r2 + (2/r) ∂t/∂r + (1/r2sinϴ)∂/∂ϴ[sinϴ∂t/∂ϴ] + (1/ r2sin2ϴ) ∂2t/∂Ф2 +qg./k =(1/ ) ∂t/∂ȥ

Q4. Derive relation of emissive power for non-black long parallel plates.

PROOF: FOR INFINITE PARALLEL SURFACES

Assumptions used

  • Configuration factor for either surface is unity i.e. f12=1 and f21=1
  • There is a non absorbing medium like air between these surfaces.
  • Properties like emissivity, reflectivity and absorbtivity are constant over the entire surfaces.

Let surface 1 emits E1 with α1, Є1 at temperature T1 and surface 2 emits E2 with α2, Є2 at temperature T2

After number of reflections

Q1net =E1–[α1 (1– α2) E1+ α1 (1—α1) (1—α2)2E1 + α1 (1—α1)2 (1– α2)3 E1+…………………….]

After number of reflections

Q1net =E1— After number of reflections

Q1net =E1— α1(1– α2)E1[1 + (1—α1) (1—α2)+ (1—α1)2 (1– α2)2 +…………………………]

Q1net =E1— α1 (1– α2) E1 [1 +Z+Z2 + Z3 + ………………………………]

Where Z= (1—α1) (1—α2)

As Z<1 the sum of the series 1+Z+Z2+Z3+……………. =1/ (1—Z)

Q1net =E1— α1 (1– α2) E1/ (1—Z)

Q1net =E1 [1— α1 (1– α2)/ (1– (1—α1) (1—α2)]

As per Kirchhoff’s law α=Є i.e. α11 and α22

Q1net =E1 [1— Є1 (1– Є2)/ (1– (1—Є1) (1—Є2)]

Q1net =E1 [1— (1– Є1) (1– Є2) — Є1 (1– Є2)]/ [(1– (1—Є1) (1—Є2)]

It is net energy going from body 1 towards body 2.

Similarly Q2net

It is net energy going from body 2 towards body 1.

Radiation Exchange between surfaces 1 and 2

Q12 = Q1 –Q2 =

Now E1 = σ T14       and E2= σ T24

Q121 Є2σ (T14 – T24)/( Є1 + Є2 Є1 Є2)

Q12 =f12 σ (T14 —  T24)

where f12 is interchange factor or configuration factor for grey bodies

 Q5. Prove by dimensional analysis for natural convection, Nu = f(Gr, Pr).

 Nu = h L/kf

h = f(ρ,L,μ, k,β,g,∆t)

In this, β,g,∆t represet buoyant force and has the dimensions of LT—2

Therefore total number of variables = 7

Total number of fundamental dimensions are M,L,T andϴ = 4

Therefore number of π terms = 7—4 = 3

Π1 = ρa1Lb1 μc1 kd1, h

 Π2 = ρa2Lb2 μc2 kd2Cp

 Π3 = ρa3Lb3 μc3 kd3βg∆t

 For Π1 term

 M0L0T0ϴ0 = (ML-3) a1 (L)b1 (ML-1T-1)c1(MLT-3ϴ-1)d1 (ML-3 ϴ-1)

 Now equate the powers of M,L,T and ϴ on both sides

We get

For M          0 = a1+b1+c1+d1+1

 For L         0 = -3a1+b1–c1+d1

 For T          0 = –c1–3d1—3

 For ϴ          0 = — d1—1

 Therefore a1=0,  b1=1, c1 = 0 and  d1 = –1

 Π1 = Lk—1h

 Π1 = hL/k                                          (1)

 For Π2 term

 M0L0T0ϴ0 = (ML-3) a2 (L)b2 (ML-1T-1)c2(MLT-3ϴ-1)d2 (ML-3 ϴ-1)

 Now equate the powers of M,L,T and ϴ on both sides

We get

For M          0 = a2+c2+d2

 For L         0 = -3a2+b2—c2+d2 + 2

 For T          0 = –c2–3d2—2

 For ϴ          0 = — d2—1

 Therefore a2=0,  b2=1, c2 = 1 and  d2 = –1

 Π2 = μCp k—1

 Π2 = μCp/k

 For Π3 term

 M0L0T0ϴ0 = (ML-3) a3 (L)b3 (ML-1T-1)c3(MLT-3ϴ-1)d3 (ML-3 ϴ-1)

 Now equate the powers of M,L,T and ϴ on both sides

We get

For M          0 = a3+c3+d3

 For L         0 = -3a3+b3—c3+d3 + 1

 For T          0 = –c3–3d3—2

 For ϴ          0 = — d3

 Therefore a3=2,  b3=3, c3 = –2 and  d3 = 0

 Π3 = ρ2  L3 μ–2 (βg ∆t)

 Π3 = ρ2  L3 (βg ∆t)/  μ2 =  (βg ∆t)/ν2

 Nu = f(Pr, Gr)

 Nu = (Pr, Gr)m hence proved

 Q6. The temperature rise of cold fluid in a heat exchanger is 20°C and temperature drop of hot fluid is 30°C. The effectiveness of heat exchanger is 0.6. The heat exchanger area is 1m2and U = 60 W/m2 °C. Find the rate of heat transfer.

ANS:

∆tcold = 200C, ∆thot = 30°C, ε =0.6, A = 1m2, U = 60 W/m2 °C

Heat gained by cold fluid = heat lost by hot fluid

We get

Cc 20 = Ch 30, Therefore Cc > Ch, Ch is minimum.

C =Cmin/Cmax = 3/2

ASSUMING THE HEAT EXCHANGER TO BE PARALLEL FLOW

We know

ε =0.6 = {1—-exp[-NTU(1+C)]}/(1+C)  = {1—-exp[-NTU(1+1.5)]}/(1+1.5)

  0.6 = {1—-exp[–2.5NTU]}/(2.5)

1.5 = 1—-exp[–2.5NTU]

—exp[–2.5NTU] = –0.5

[–2.5NTU] = 0.5

Taking log on both sides

We get

–2.5 NTU = 0.4990 –1 = –0.5010

NTU = 0.002 = UA/Cmin = 60 x 1/ Cmin

Cmin = 60/0.002 = 30000

q. = Cmin 20 = 30000 x 20 = 600000 kJ/h =167 k/s = 167 kW