Heat Exchanger Analysis MCQ Quiz in తెలుగు - Objective Question with Answer for Heat Exchanger Analysis - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

When one of the fluid in Heat Exchanger is undergoing a change of phase like,

1) The profile for Steam Generator is, 

2) The profile for Evaporator is,

In both the cases, the temperature of one fluid remains constant. 

So temperature differences interchange at inlet and outlet if changed direction but they do not change in magnitude.

Hence, both counter flow and parallel flow will have same LMTD. 

LMTDcounter = LMTDparallel

• For a concentric double pipe heat exchanger where one of the fluids undergo a phase change the direction of flow of the two fluids is of no consequences.
• Because the (LMTD) logarithmic mean temperature difference of the two fluids in both cases i.e. parallel and counter flow are the same so that the heat transfer remains the same in both the cases.

Concept:

For phase change effectiveness is 

ϵ = 1 - e-NTU

\(\rm NTU=\frac{UA}{(\dot mc_p)}_{small}\)

\(\rm \epsilon=\frac{Q_{actual}}{Q_{max}}=\frac{Q_{actual}}{(mC_p)_{small}(T_{hi}-T_{ci})}\)

Calculation:

Given:

T_hi = 120 °C, Tci = 20 °C, U =  310 W/m2K, D = 1.4 cm, L = 35 m, h_fg = 831.235 kJ/kg

Oil mass flow rate = 0.2 kg/s, Specific heat of oil = 2.13 kJ/kg K  

For phase change effectiveness is 

ϵ = 1 - e^-NTU

\(\rm NTU=\frac{UA}{(\dot mc_p)}_{small}\)

\(\rm NTU=\frac{\pi ×0.014×35×310}{0.2×2.13×10^3}=1.12\)

ϵ = 1 - e^-1.12 = 0.673

\(\rm \epsilon=\frac{Q_{actual}}{Q_{max}}=\frac{Q_{actual}}{(mC_p)_{small}(T_{hi}-T_{ci})}\)

Qactual = 28.703 mJ

Q = m × h_fg

m = 34.53 gram/sec

Concept:

For a heat exchanger with inlet temperature difference ΔT1 and outlet temperature difference ΔT2 ,

\(AMTD = \frac{{{\rm{θ}}{_1} \;+ \;{\rm{θ}}{_2}}}{2}\)

Counter-flow heat exchanger:

where θ₁ = T_h1 - T_c2, and θ₂ = T_h2 - T_c1

Calculation:

Given:

Th1 = 90°C, Th2 = 40°C, Tc1 = 20°, Tc2 = 40°C.

θ1 = Th1 - Tc2 = 90 - 40 ⇒ 50°C

θ2 = Th2 - Tc1 = 40 - 20 ⇒ 20°C

\(AMTD = \frac{{{\rm{θ}}{_1} \;+ \;{\rm{θ}}{_2}}}{2}\)

\(AMTD = \frac{50\;+\;20}{2}=35^{\circ}C\)

Additional Information Parallel flow heat exchanger:

where θ1 = Th1 - Tc1, and θ2 = Th2 - Tc2

Concept:

NTU is a measure of the effectiveness of the heat exchanger.

The NTU is a measure of the heat transfer size of the exchanger; larger the value of NTU, the closer the heat exchanger approaches its thermodynamic limit.

\(NTU = \frac{{UA}}{{{C_{min}}}}\)

C_min is the minimum heat capacity between the hot and cold fluid

Calculation:

Given:        

A = 50 m², U = 100 W/m² K,

C_max = C_min = 1000 W/K

Concept:

Heat exchanger effectiveness:

\(\epsilon= \frac{{{{\rm{Q}}_{{\rm{actual}}}}}}{{{{\rm{Q}}_{{\rm{max}}}}}}\)

\({\rm{\eta }} = \frac{{{{\rm{C}}_{\rm{h}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}} \right)}}{{{{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{C}}1}}} \right)}} = \frac{{{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}}}{{{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{c}}1}}}}\)

Calculation:

\({{\rm{Q}}_{{\rm{actul}}}} = {{\rm{\dot m}}_{\rm{h}}}{{\rm{c}}_{{\rm{ph}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}} \right) = {{\rm{\dot m}}_{\rm{c}}}{{\rm{c}}_{{\rm{pc}}}}\left( {{{\rm{T}}_{{\rm{c}}2}} - {{\rm{T}}_{{\rm{c}}1}}} \right)\)

\({{\rm{Q}}_{{\rm{max}}}} = {{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}} \right)\)

\(\epsilon= \frac{{{{\rm{Q}}_{{\rm{actual}}}}}}{{{{\rm{Q}}_{{\rm{max}}}}}} = \frac{{{{\rm{C}}_{\rm{h}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}} \right)}}{{{{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}} \right)}} = \frac{{{{\rm{C}}_{\rm{c}}}\left( {{{\rm{T}}_{{\rm{c}}2}} - {{\rm{T}}_{{\rm{c}}1}}} \right)}}{{{{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{c}}1}}} \right)}}\)

\({{\rm{\dot m}}_{\rm{n}}}{{\rm{C}}_{{\rm{pn}}}} = {{\rm{\dot m}}_{\rm{c}}}{{\rm{C}}_{{\rm{pc}}}} \Rightarrow {{\rm{C}}_{\rm{h}}} = {{\rm{C}}_{\rm{c}}} = {{\rm{C}}_{{\rm{min}}}} = {{\rm{C}}_{{\rm{max}}}}\)

\({\rm{\eta }} = \frac{{{{\rm{C}}_{\rm{h}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}} \right)}}{{{{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{C}}1}}} \right)}} = \frac{{{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}}}{{{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{c}}1}}}}\)

\({\rm{\eta }} = \frac{{{{\rm{C}}_{\rm{h}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}} \right)}}{{{{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{C}}1}}} \right)}} = \frac{{{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}}}{{{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{c}}1}}}}\)

\({\rm{\eta }} =\frac{76-47}{76-26}=0.58\)

Concept:

For counterflow heat exchanger:

Log mean temperature difference is given by:

\(LMTD = \frac{{{\rm{Δ }}{T_i}\; - \;{\rm{Δ }}{T_e}}}{{\ln \left( {\frac{{{\rm{Δ }}Ti}}{{{\rm{Δ }}Te}}} \right)}}\)

Where ΔTi = Thi – Tce

ΔTe = The - Tci

Thi & Tci = Temperature at which hot or and cold water enters respectively

The & Tce = Temperature at which hot air & cold water exits respectively

In the case of the balanced counter flow heat exchanger, 

ṁhch = ṁccc = C

ṁh and ṁc are mass flow rates of hot and cold fluid, and ch and cc specific heat capacity of hot and cold fluid.

LMTD for counter-flow heat exchanger, ΔTi = ΔTe

Explanation:

Given:

Tci = T₂, Tce = T4, Thi = T1, The = T4

ΔTi = T1 – T3

ΔTe = T4 – T2

\(LMTD = \frac{{{\rm{Δ }}{T_i}\; - \;{\rm{Δ }}{T_e}}}{{\ln \left( {\frac{{{\rm{Δ }}Ti}}{{{\rm{Δ }}Te}}} \right)}}\)

\(LMTD = \rm \frac{(T_1-T_3)-(T_4-T_2)}{In\frac{(T_1-T_3)}{(T_4-T_2)}}\)

Concept:

The heat exchanger effectiveness (ϵ) is defined as the ratio of actual heat transfer to the maximum possible heat transfer.

\(\epsilon = \frac{{{Q_{act}}}}{{{Q_{max}}}}\)

\(\epsilon = \frac{{{C_h}\left( {{T_{h1}} - {T_{h2}}} \right)}}{{{C_{min}}\left( {{T_{h1}} - {T_{c1}}} \right)}} = \frac{{{C_c}\left( {{T_{c2}} - {T_{c1}}} \right)}}{{{C_{min}}\left( {{T_{h1}} - {T_{c1}}} \right)}}\)

Qact = ṁhCph (Th1 – Th2) = ṁcĊpc (Tc2 – Tc1)

Fluid Capacity Rate, C : Ch = ṁhCph, Cc = ṁc Cpc

Qmax = C_min(Th1 – Tc1) 

Q = C_h(Th1 – Th2) = C_c(Tc2 – Tc1)

Q = Ch × 25 = Cc × 15

∴ Ch is less than Cc

The formula used will be:

\(\epsilon = \frac{{{C_h}\left( {{T_{h1}} - {T_{h2}}} \right)}}{{{C_{min}}\left( {{T_{h1}} - {T_{c1}}} \right)}} = \frac{{{}\left( {{T_{h1}} - {T_{h2}}} \right)}}{{{{}}\left( {{T_{h1}} - {T_{c1}}} \right)}}\)

\(\epsilon = \frac{{{}\left( {{65} - {{40}}} \right)}}{{{{}}\left( {{{65}} - {{15}}} \right)}}=0.5\)

Concept:

Heat exchanger effectiveness:

\(ϵ= \frac{{{{\rm{Q}}_{{\rm{actual}}}}}}{{{{\rm{Q}}_{{\rm{max}}}}}}\)

\({{\rm{Q}}_{{\rm{actul}}}} = {{\rm{\dot m}}_{\rm{h}}}{{\rm{c}}_{{\rm{ph}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}} \right) = {{\rm{\dot m}}_{\rm{c}}}{{\rm{c}}_{{\rm{pc}}}}\left( {{{\rm{T}}_{{\rm{c}}2}} - {{\rm{T}}_{{\rm{c}}1}}} \right)\)

\({{\rm{Q}}_{{\rm{max}}}} = {{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}} \right)\)

\(ϵ= \frac{{{{\rm{Q}}_{{\rm{actual}}}}}}{{{{\rm{Q}}_{{\rm{max}}}}}} = \frac{{{{\rm{C}}_{\rm{h}}}\left( {{{\rm{T}}_{{\rm{h}}1}} \;- \;{{\rm{T}}_{{\rm{h}}2}}} \right)}}{{{{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} \;- \;{{\rm{T}}_{{\rm{c}}1}}} \right)}} = \frac{{{{\rm{C}}_{\rm{c}}}\left( {{{\rm{T}}_{{\rm{c}}2}} \;-\; {{\rm{T}}_{{\rm{c}}1}}} \right)}}{{{{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} \;- \;{{\rm{T}}_{{\rm{c}}1}}} \right)}}\)

Calculation:

Given:

ϵ = 0.25, T_h1 = 200°C, T_h2 = 75°C, T_c1 = 40°C

Heat capacities of gases and air are not given, so for calculating cold fluid outlet temperature, we have to assume the heat capacity of cold air is minimum.

Because then only, the effectiveness formula includes cold air outlet temperature.

\(ϵ =\frac{\left( {{T}_{ce}}-{{T}_{ci}} \right)}{\left( {{T}_{hi}}-{{T}_{ci}} \right)}\)

\(0.25=\frac{{{T}_{ce}}-40}{200-40}\)

T_ce - 40 = 40

Explanation:

Overall heat transfer coefficient (OHTC):

• OHTC is defined as the quantity of rate of heat transfer takes place due to any mode per unit area for unit temperature difference.
• It doesn’t have any physical meaning, it’s an experimentally determined coefficient value.
• The concept of OHTC is valid for a steady-state. 
• In the case of a heat exchanger, there is a combined mode of heat transfer or multiple resistances present therefore we use,

​\(Q = \;UA{\bf{\Delta }}T \Rightarrow UA = \frac{1}{{{R_{Total}}}}\), 

where R Total is total resistance and U = OHTC

• Unit of ‘U’ is W/m2K
• U indicates the quantity of rate of heat transfer takes place due to any mode per unit area for the unit temperature difference.

Fouling factor indicates the resistance offered due to chemical deposits and scaling.

So, \(\frac{1}{{{U_{with\;fouling}}}} = \frac{1}{{{U_{without\;fouling}}}} + F\)

Calculation:

Given:

Uo = 400 W/m2K, \(F=\frac{1}{{2000}}\frac{m^2K}{{{}W}}\)

\(\begin{array}{l} \therefore \frac{1}{{{U_{with\;fouling}}}} = \frac{1}{{400}} + \frac{1}{{2000}}\\ \Rightarrow {U_{with\;fouling}} = 333\frac{W}{{{m^2}k}} \end{array}\)

\({\dot m_c} = 2{\dot m_h}\)

\({C_h} = 2{C_c}\)

By energy balance

\({m_h}{C_h}\left( {{T_{{h_1}}} - {T_{{h_2}}}} \right) = {m_c}{C_c}\left( {{T_{{c_2}}} - {T_{{c_1}}}} \right)\)

\({T_{{h_1}}} - {T_{{h_2}}} = {T_{{c_2}}} - {T_{{c_1}}}\)

⇒ \({T_{{h_1}}} - {T_{{c_2}}} = {T_{{h_2}}} - {T_{{c_1}}}\)

Or, θ₁ = θ₂

∴ \(LMTD = \frac{{\left( {{T_{{h_1}}} - {T_{{c_2}}}} \right) - \left( {{T_{{h_2}}} - {T_{{c_1}}}} \right)}}{{{\rm{ln}}\left( {\frac{{{T_{{h_1}}} - {T_{{c_2}}}}}{{{T_{{h_2}}} - {T_{{c_1}}}}}} \right)}}\)

\(\Rightarrow LMTD = \frac{{{\theta _1} - {\theta _2}}}{{\ln \left( {\frac{{{\theta _1}}}{{{\theta _2}}}} \right)}} = {\theta _m} \)

Let \(\frac{{{\theta _1}}}{{{\theta _2}}} = x\)

Then, \({\theta _m} = \frac{{{\theta _2}\left( {x - 1} \right)}}{{lnx}} \Rightarrow {\theta _m} = \mathop {\lim }\limits_{x \to 1} \frac{{{\theta _2}\left( {x - 1} \right)}}{{\ln x}}\; \Rightarrow since\frac{0}{0}\)

Applying L’ Hospital rule,we get

\(\Rightarrow \theta_m=\frac{\theta_2}{(\frac{1}{x})}\)

θ_m = θ₂ = θ₁

Given θ_m = 20° C

\({T_{{h_1}}} - {T_{{c_2}}} = 20^\circ C \Rightarrow 100 - {T_{{c_2}}} = 20^\circ C\)

\({T_{{c_2}}} = 80^\circ C\)