# 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

- Single phase liquid at entrance—at a temperature lower than the boiling point
- Saturated liquid—– at the boiling temperature
- Bubbly flow—vapors starts forming and going up with rest of the liquid.
- Slug flow regime: small vapors already formed change into bigger bubbles called slugs and hence the flow is called slug flow.
- 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.
- Mist flow: Dryness fraction becomes slightly greater than 0.25, the annular liquid disappears and vapors are carrying drops of liquid.
- 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)

∂^{2}t/∂r^{2} + (2/r) ∂t/∂r + (1/r^{2}sinϴ)∂/∂ϴ[sinϴ∂t/∂ϴ] + (1/ r^{2}sin^{2}ϴ) ∂^{2}t/∂Ф^{2} +q_{g}^{.}/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

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 ∂(r^{2}∂t/∂r) /∂r) dr

=[ –kdr rdϴ rsinϴ dФ/r^{2}] ∂(r^{2}∂t/∂r) /∂r)

= –k [δV / r^{2} ][ r^{2}∂^{2}t/∂^{2}r + 2r ∂t/∂r]

= —k{[ δV][ ∂^{2}t/∂^{2}r] + (1/r) ∂t/∂r}

(∂Q_{r}/∂r) dr = –k δV[∂^{2}t/∂^{2}r] + (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)rdϴ

(∂Q_{ϴ }/ r∂ϴ)rdϴ = (∂/rsinϴ[-(k/r) δV ∂t/∂ϴ] / r∂ϴ) = –(k/r^{2}sin ϴ) δV∂^{2}t/∂^{2}ϴ (2)

© Q_{Ф} = –k (rdϴ dr) ∂t/(rsinϴ∂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 (rdϴ dr)/(rsinϴ)∂^{2}t/∂^{2}ФX 1/r sinϴ (3)

= –k [δV/ r^{2}sin^{2}ϴ] ∂^{2}t/∂^{2}Ф

Rate of change of heat inside= ρ δVc_{p}∂t/∂ ȥ (4)

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

–k δV[∂^{2}t/∂^{2}r] + (1/r) ∂t/∂r]- –(k/r^{2}sin ϴ) δV∂^{2}t/∂^{2}ϴ–(k/r^{2}sin ϴ) δV∂^{2}t/∂^{2}ϴ+ q^{.}_{g} δV = ρ δVc_{p}∂t/∂ ȥ

Rearranging and dividing by k δV, we get

**∂ ^{2}t/∂r^{2} + (2/r) ∂t/∂r + (1/r^{2}sinϴ)∂/∂ϴ[sinϴ∂t/∂ϴ] + (1/ r^{2}sin^{2}ϴ) ∂^{2}t/∂Ф^{2} +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. f
_{12}=1 and f_{21}=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 E_{1} with α_{1}, Є_{1 }at temperature T_{1} and surface 2 emits E_{2} with α_{2}, Є_{2 }at temperature T_{2}

After number of reflections

Q_{1net} =E_{1}–[α1 (1– α_{2}) E_{1}+ α_{1} (1—α_{1}) (1—α_{2})^{2}E_{1} + α_{1} (1—α_{1})^{2} (1– α_{2})^{3} E_{1}+…………………….]

After number of reflections

Q_{1net} =E_{1}— After number of reflections

Q_{1net} =E_{1}— α_{1}(1– α_{2})E_{1}[1 + (1—α_{1}) (1—α_{2})+ (1—α_{1})^{2} (1– α_{2})^{2} +…………………………]

Q_{1net} =E_{1}— α_{1} (1– α_{2}) E1 [1 +Z+Z^{2} + Z^{3} + ………………………………]

Where Z= (1—α_{1}) (1—α_{2})

As Z<1 the sum of the series 1+Z+Z^{2}+Z^{3}+……………. =1/ (1—Z)

Q_{1net} =E_{1}— α_{1} (1– α_{2}) E1/ (1—Z)

Q_{1net} =E_{1} [1— α_{1} (1– α_{2})/ (1– (1—α_{1}) (1—α_{2})]

As per Kirchhoff’s law α=Є i.e. α_{1}=Є_{1} and α_{2}=Є_{2}

Q_{1net} =E_{1} [1— Є_{1} (1– Є_{2})/ (1– (1—Є_{1}) (1—Є_{2})]

Q_{1net} =E_{1} [1— (1– Є_{1}) (1– Є_{2}) — Є_{1} (1– Є_{2})]/ [(1– (1—Є_{1}) (1—Є_{2})]

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

Q_{12} = Q_{1} –Q_{2} =

Now E_{1} = σ T_{1}^{4} and E_{2}= σ T_{2}^{4}

**Q _{12} =Є_{1} Є_{2}**

**σ**

**(T**

_{1}^{4}– T_{2}^{4})/( Є_{1}**+**

**Є**

_{2}**—**

**Є**

_{1}Є_{2}**)**

Q_{12} =f_{12} σ (T_{1}^{4} — T_{2}^{4})

where f_{12} 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 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**

**Π _{1} = ρ^{a1}L^{b1} μ^{c1} k^{d1}, h**

** ****Π _{2} = ρ^{a2}L^{b2} μ^{c2} k^{d2}C_{p}**

** ****Π _{3} = ρ^{a3}L^{b3} μ^{c3} k^{d3}βg∆t**

** ****For Π _{1} term**

** ****M ^{0}L^{0}T^{0}ϴ^{0} = (ML^{-3})^{ a1} (L)^{b1} (ML^{-1}T^{-1})^{c1}(MLT^{-3}ϴ^{-1})^{d1} (ML^{-3 }ϴ^{-1})**

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

**We get**

**For M 0 = a _{1}+b_{1}+c_{1}+d_{1}+1**

** ****For L 0 = -3a _{1}+b_{1}–c_{1}+d_{1}**

** ****For T 0 = –c _{1}–3d_{1}—3**

** ****For ϴ 0 = — d _{1}—1**

** ****Therefore a _{1}=0, b_{1}=1, c_{1} = 0 and d_{1} = –1**

** ****Π _{1} = Lk^{—1}h**

** ****Π _{1} = hL/k (1)**

** ****For Π _{2} term**

** ****M ^{0}L^{0}T^{0}ϴ^{0} = (ML^{-3})^{ a2} (L)^{b2} (ML^{-1}T^{-1})^{c2}(MLT^{-3}ϴ^{-1})^{d2} (ML^{-3 }ϴ^{-1})**

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

**We get**

**For M 0 = a _{2}+c_{2}+d_{2}**

** ****For L 0 = -3a _{2}+b_{2}—c_{2}+d_{2} + 2**

** ****For T 0 = –c _{2}–3d_{2}—2**

** ****For ϴ 0 = — d _{2}—1**

** ****Therefore a _{2}=0, b_{2}=1, c_{2} = 1 and d_{2} = –1**

** ****Π _{2} = μC_{p} k^{—1}**

** ****Π _{2} = μC_{p}/k **

** ****For Π _{3} term**

** ****M ^{0}L^{0}T^{0}ϴ^{0} = (ML^{-3})^{ a3} (L)^{b3} (ML^{-1}T^{-1})^{c3}(MLT^{-3}ϴ^{-1})^{d3} (ML^{-3 }ϴ^{-1})**

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

**We get**

**For M 0 = a _{3}+c_{3}+d_{3}**

** ****For L 0 = -3a _{3}+b_{3}—c_{3}+d_{3} + 1**

** ****For T 0 = –c _{3}–3d_{3}—2**

** ****For ϴ 0 = — d _{3}**

** ****Therefore a _{3}=2, b_{3}=3, c_{3} = –2 and d_{3} = 0**

** ****Π _{3} = ρ^{2} L^{3} μ^{–2} (βg ∆t)**

** ****Π _{3} = ρ^{2} L^{3} (βg ∆t)/ μ^{2} = (βg ∆t)/ν^{2}**

** ****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 1m ^{2}and U = 60 W/m2 °C. Find the rate of heat transfer.**

ANS:

∆t_{cold} = 20^{0}C, ∆t_{hot} = 30°C, ε =0.6, A = 1m^{2}, 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**