PTU-HEAT TRANSFER QUESTION

PAPER SOLUTION-SECTION-B

Solution of University question paper increases the

under standing to a larger extent. One recognizes

the importance of various topics of a subject.

 Q2. Explain different phases of flow boiling.

ANS:

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

Sub-cooled liquid is entering from below. It is getting heat from the tube heated externally at constant heat flux. Different phases of flow boiling with changing dryness fraction are 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. Hence the flow is slug flow.

5. Annular flow: Rate of vapor formation increases. The vapors start moving as a central core. 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 +q_g^./k =(1/ ) ∂t/∂ȥ

SPHERICAL COORDINATES ARE r, ϴ and Ф

Size of the spherical element will be dr, r dϴ 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

Q_r =-k A_r ∂t/∂r =  -k  (rdϴ)( r sinϴ dФ) ∂t/∂r

Q_r =  -k  (rdϴ)( r sinϴ dФ) ∂t/∂r

Energy balance for the element

• In the radial direction

Q_r+dr = Q_r + (∂ Q_r/∂r) dr

NET IN r direction =(∂ Q_r/∂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 ][ r²∂²t/∂²r + 2r ∂t/∂r]

= —k{[ δV][ ∂²t/∂²r] + (1/r) ∂t/∂r}

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

• In the ϴ direction ( r-Ф plane)

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

= –k [(rdϴ r sinϴ dФ dr) ∂t/(r∂ϴ)]

Q_ϴ  = —(k/r) δV ∂t/∂ϴ

Q_ϴ+dϴ = Q_ϴ +  (∂Q_ϴ / r∂ϴ)rdϴ

Substitute the value of Q_ϴ

We get  the net outflow in ϴ direction

Q_ϴ+dϴ — Q_ϴ = (∂Q_ϴ / r∂ϴ) rd ϴ

(∂Q_ϴ / r∂ϴ) rd ϴ = (∂/r sinϴ[-(k/r) δV ∂t/∂ϴ] / r∂ϴ)

                            = –(k/r²sin ϴ) δV∂²t/∂²ϴ  (2)

© Q_Ф = –k (r dϴ dr) ∂t/(r sinϴ ∂Ф)]

Similarly net flow in Ф direction will be

Q_Ф+dФ = Q_Ф + ∂Q_Ф/r sinϴ ∂Ф) r sinϴ ∂Ф

— Q_Ф =dQ_Ф = (dQ_Ф/rsinϴdФ) rsinϴdФ

(dQ_Ф/rsinϴdФ) rsinϴdФ  =–k (r dϴ dr)/(r sinϴ)∂²t/∂²ФX 1/r sinϴ (3)

                                             = –k [δV/ r²sin²ϴ] ∂²t/∂²Ф

Rate of change of heat inside= ρ δVc_p∂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^._gδV = ρδVc_p∂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 +q_g^./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 absorptivity 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

Q_1net =E₁–[α1 (1– α₂) E₁+ α₁ (1—α₁) (1—α₂)²E₁ + α₁ (1—α₁)² (1– α₂)³ E₁+…………………….]

After number of reflections

We get      Q_1net =E₁— After number of reflections

Firstly        Q_1net =E₁— α₁(1– α₂)E₁[1 + (1—α₁) (1—α₂)+ (1—α₁)² (1– α₂)² +…………]

Secondly     Q_1net =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)

Q_1net =E₁— α₁ (1– α₂) E1/ (1—Z)

Q_1net =E₁ [1— α₁ (1– α₂)/ (1– (1—α₁) (1—α₂)]

As per Kirchhoff’s law α=Є i.e. α₁=Є₁ and α₂=Є₂

Q_1net =E₁ [1— Є₁ (1– Є₂)/ (1– (1—Є₁) (1—Є₂)]

Q_1net =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 Q_2net

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

Q₁₂ =Є₁ 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/k_f

h = f(ρ,L,μ, k,β,g, ∆t)

In this, β,g, ∆t represent 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

Firstly          Π₁ = ρ^a1L^b1 μ^c1 k^d1, h

 Secondly     Π₂ = ρ^a2L^b2 μ^c2 k^d2C_p

 Thirdly        Π₃ = ρ^a3L^b3 μ^c3 k^d3β g ∆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

  L         0 = -3a₁+b₁–c₁+d₁

  T          0 = –c₁–3d₁—3

  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

  L         0 = -3a₂+b₂—c₂+d₂ + 2

  T          0 = –c₂–3d₂—2

  ϴ          0 = — d₂—1

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

we get

 Firstly       Π₁ = μC_p k^—1

 Secondly  Π₂ = μC_p/k

 Thirdly     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

  L         0 = -3a₃+b₃—c₃+d₃ + 1

 T          0 = –c₃–3d₃—2

  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:

∆t _cold = 20⁰C, ∆t_hot = 30°C,

ε =0.6,

A = 1m²,

U = 60 W/m2 °C

Heat gained by cold fluid = heat lost by hot fluid

We get

C_c 20 = C_h 30, Therefore C_c > C_h, C_h is minimum.

C =C _min/C _max = 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/C_min = 60 x 1/ C_min

C_min = 60/0.002 = 30000

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

 

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