PTUHEAT TRANSFER QUESTION PAPER SOLUTIONSECTION B
PTUHEAT TRANSFER QUESTION
PAPER SOLUTIONSECTIONB
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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.
Subcooled 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

Single phase liquid at entrance at a temperature lower than the boiling point

Saturated liquid at the boiling temperature

Bubbly flowvapors starts forming and going up with rest of the liquid.

Slug flow regime: small vapors already formed change into bigger bubbles called slugs. Hence the flow is slug flow.

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.

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 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 ∂(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 ( rФ 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/r^{2}sin ϴ) δV∂^{2}t/∂^{2}ϴ (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)∂^{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 = ρδVc_{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 nonblack 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 absorptivity 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
We get Q_{1net} =E_{1}— After number of reflections
Firstly 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} +…………]
Secondly 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 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 Π_{1} = ρ^{a1}L^{b1} μ^{c1} k^{d1}, h
Secondly Π_{2} = ρ^{a2}L^{b2} μ^{c2} k^{d2}C_{p}
Thirdly Π_{3} = ρ^{a3}L^{b3} μ^{c3} k^{d3}β g ∆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
L 0 = 3a_{1}+b_{1}–c_{1}+d_{1}
T 0 = –c_{1}–3d_{1}—3
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}
L 0 = 3a_{2}+b_{2}—c_{2}+d_{2} + 2
T 0 = –c_{2}–3d_{2}—2
ϴ 0 = — d_{2}—1
Therefore a_{2}=0, b_{2}=1, c_{2} = 1 and d_{2} = 1
we get
Firstly Π_{1} = μC_{p} k^{—1}
Secondly Π_{2} = μC_{p}/k
Thirdly 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}
L 0 = 3a_{3}+b_{3}—c_{3}+d_{3} + 1
T 0 = –c_{3}–3d_{3}—2
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
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