LMTD Method MCQ Quiz in मल्याळम - Objective Question with Answer for LMTD Method - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

Mass flow rate, \({\dot m_H} = 1\;kg/s\), and,  \({\dot m_c} = 2\;kg/s\)

Specific heat, C_H = 10 kJ/kgK, and, C_c = 5 kJ/kgK

Heat capacity of hot fluid \(= {\dot m_{\dot H}}{c_H} = 1 \times 10 = 10\;\frac{{kJ}}{{k - s}}\)

Heat capacity of cold fluid \(= {\dot m_{\dot c}}{c_c} = 5 \times 2 = 10\;\frac{{kJ}}{{k - s}}\)

Since heat capacity is same so LMTD is difference of temperature of fluid at either ends.

LMTD = 60° - 30° = 30°C

Explanation:

LMTD calculation for Parallel flow heat exchanger:

The formula for calculation of LMTD (Log Mean Temperature Difference) for a parallel flow heat exchanger is given by-

\(LMTD_{parallel}=\frac{{{θ _1} - {θ _2}}}{{\ln \left( {\frac{{{θ _1}}}{{{θ _2}}}} \right)}}=\frac{{{θ _2} - {θ _1}}}{{\ln \left( {\frac{{{θ _2}}}{{{θ _1}}}} \right)}}\)

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

Additional InformationLMTD calculation for Counterflow heat exchanger:

\(LMTD_{counter}=\frac{{{θ _1} - {θ _2}}}{{\ln \left( {\frac{{{θ _1}}}{{{θ _2}}}} \right)}}=\frac{{{θ _2} - {θ _1}}}{{\ln \left( {\frac{{{θ _2}}}{{{θ _1}}}} \right)}}\)

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

Concept:

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

Energy conservation between hot fluid and cold fluid,

C_h (ΔT_h) = C_c (ΔT_c)

Calculation:

Given: Th_i = Inlet temperature of hot fluid = 100°C, Tc_i = Inlet temperature of cold fluid = 15°C, Tc_e = Exit temperature of cold fluid = 45°C

C_h = 2000 W/K, C_c = 1200 W/K

C_h (Δ T_h) = C_c (Δ T_c)

⇒ 2000 (100 – T­_he) = 1200 (45° - 15°)

⇒ T_he = 82°C

θ₁ = Th_i – Tc_i = 85°C; θ₂ = T_he – T_ce = 37°C

\(LMTD\left( {{\theta _m}} \right) = \frac{{{\theta _1} - {\theta _2}}}{{\ln \left( {\frac{{{\theta _1}}}{{{\theta _2}}}} \right)}} = \frac{{85 - 37}}{{\ln \left( {\frac{{85}}{{37}}} \right)}} = 57.71^\circ C\)

Concept:

The total heat transfer rate in the heat exchanger is given by

Q = U × A × θm

Where U = overall heat transfer coefficient in W/m2-K,

A = effective surface area of heat exchanger in m2,

θm = log mean temperature difference \(\left( {LMTD} \right) = \frac{{{\bf{\Delta }}{T_1} - {\bf{\Delta }}{T_2}}}{{{\bf{ln}}\left( {\frac{{{\bf{\Delta }}{T_1}}}{{{\bf{\Delta }}{T_2}}}} \right)}}\)

Also, Q = Cph (Thi-Th0) = Cpc (Tco-Tci)

i.e. heat lost by hot fluid = heat gain by the cold fluid

where Cph and Cpc are heat capacity of hot and cold fluid respectively in kJ/K

Since the heat capacity for hot and cold fluids are the same and it is a case of counter flow so the LMTD will be the difference in temperature of either side.

Calculation:

The temperature profiles are shown in the figure below

θ1 = θ2 = 45°C

Hence LMTD = 45°C

Concept:

Heat transfer in the heat exchanger is given by

Q = UA ΔTm

Where U is the overall heat transfer coefficient, A is the area of the heat exchanger, ΔTm is the logarithmic temperature difference.

Also, Q = ṁCdT

Where ṁ is the mass flow rate, C is the specific heat and dT is the temperature difference between inlet and outlet of the fluid.

If there is a fluid whose phase changes in the heat exchanger, then

The temperature profile is as follows

Then, \(LMTD = \frac{{{\rm{\Delta }}{T_1} - \;{\rm{\Delta }}{T_2}}}{{\ln \left( {\frac{{{\rm{\Delta }}{T_1}}}{{{\rm{\Delta }}{T_2}}}} \right)}}\)

Calculation:

Given, C_c = 6 kJ/kgK

T_ci = 50°C, T_co = 100°C

T_steam = 125°C        

The temperature profile is shown below,

ΔT₁ = 125 – 50 = 75°C and ΔT₂ = 125 – 100 = 25°C

LMTD of this heat exchanger \( = \frac{{75\; - \;25}}{{\ln \;\left( {\frac{{75}}{{25}}} \right)}} = 45.51^\circ C\)

Now, since steam losses 100 kJ of heat, the same heat will be gained by the oil, therefore

Q = UA ΔT_m = 100 kW

⇒ U × 20 × 45.51 = 100 kW

⇒ \(U = 0.1098\frac{{kW}}{{{m^2}K}} = 109.8\;W/{m^2}K\)

Concept:

Apply energy balance:

Heat lost by hot = Heat gained by cold and use

\(LMTD = \frac{{\Delta {T_1} - \;\Delta {T_2}}}{{\ln \left( {\frac{{\Delta {T_1}}}{{\Delta {T_2}}}} \right)}}\)

Heat transfer rate, Q = UALMTD

Calculation:

ṁ_c = 3 g/s, T_ci = 30°C, t_co = 37°C

ṁ_h = 5 g/s, T_hi = 39°C

ṁ_hC_h(T_hi – T_ho) = ṁ_c(T_co - T_ci)

ΔT₁ = 39 – 37 = 2

ΔT₂ = 34.8 – 30 = 4.8

\({T_{ho}} = \frac{{ - {{\dot m}_c}\left( {{T_{c0}} - {T_{ci}}} \right) + {{\dot m}_h}\;{T_{hi}}}}{{{{\dot m}_h}}}\)

\({T_{ho}} = \frac{{ - 3\left( 7 \right) + 5 \times 39}}{5}\)        

∴ T_h0 = 34.8°C

\(LMTD = \frac{{{\rm{\Delta }}{T_1} - {\rm{\Delta }}{T_2}}}{{ln\left( {\frac{{{\rm{\Delta }}{T_1}}}{{{\rm{\Delta }}{T_2}}}} \right)}} = 3.198^\circ C = {\rm{\Delta }}{T_{lm}}\)

Q = m_cC_c = (T_co – T_c1)

UAΔT_lm = 0.003 × 3450 × 7

130 π × 0.05 L × 3.198 = 0.003 × 7 × 3450

∴ L = 1.11 m

Additional InformationBy energy balance, we can see that the outlet temperature of hot fluid is 34.8°C, and in the question, it has been given that the outlet of cold fluid is 37°C, which is higher than the hot fluid outlet and only valid for counter-flow.

In parallel flow, the outlet of hot fluid cannot be lesser than that of the outlet of cold fluid.

Concept:

Condenser can be treated as counter-flow or counter-current heat exchanger.

Calculate the logarithmic mean temperature difference

\(\Delta {T_{lm}} = \;\frac{{\Delta {T_1} - \;\Delta {T_2}}}{{\ln \left( {\frac{{\Delta {T_1}}}{{\Delta {T_2}}}} \right)}}\)

Using heat transfer, Q = hA∆T_lm

And equating this with heat lost from steam we can calculate the mass condensed per min.

Calculation:

ΔT₁ = 30 – 14 = 16°C

ΔT₂ = 30 – 22 = 8°C

\(\Delta {T_{lm}} = \;\frac{{\Delta {T_1} - \;\Delta {T_2}}}{{\ln \left( {\frac{{\Delta {T_1}}}{{\Delta {T_2}}}} \right)}}\)

ΔT_lm = 11.54° C    

Q = hA ΔT_lm

Q = 2200 × 50 × 11.54

Q = 12.69.4 kW

Now,

Q = ṁh

\(\dot m = \frac{{1269.4}}{{2453}} = 0.52\;kg/s\)

Concept:

In counter-flow heat exchanger, the flowing streams flow in the opposite direction.

LMTD = \(\frac{{{\rm{Δ }}{{\rm{T}}_{\rm{i}}} - {\rm{Δ }}{{\rm{T}}_{\rm{o}}}}}{{\ln \frac{{{\rm{Δ }}{{\rm{T}}_{\rm{i}}}}}{{{\rm{Δ }}{{\rm{T}}_{\rm{o}}}}}}}\) …………….(1)

ΔT_i = T_hi - T_co  and ΔT_o = T_ho - T_ci

Heat loss by hot flow = heat gain by cold flow

Therefore, m_h C (T_hi - T_ho) = m_c C (T_co – T_ci)……….(2)

Calculation:

Given:

For Hot stream, m_h = 25 liter/s, C = 4186 J/kgK, T_hi = 46°C, T_ho = ?

For Cold stream, m_c = 19 liter/s, C = 4186 J/kgK, T_ci = 10°C, T_co = 38 °C

From equ. (2)

\(25 \times 4186 \times \left( {46 - {{\rm{T}}_{{\rm{ho}}}}} \right) = 19 \times 4186 \times \left( {38 - 10} \right)\)

\(\therefore {{\rm{T}}_{{\rm{ho}}}} = 24.72{\rm{}}^\circ {\rm{C}}\)

\({\rm{Δ }}{{\rm{T}}_{\rm{i}}} = 46 - 38 = 8{\rm{}}^\circ {\rm{C~and~Δ }}{{\rm{T}}_{\rm{o}}} = 24.72 - 10 = 14.72{\rm{}}^\circ {\rm{C}}\)

From equ. (1)

LMTD = \(\frac{{8 - 14.72}}{{\ln \frac{8}{{14.72}}}} = 11.02{\rm{}}^\circ {\rm{C}}\) 

Concept:

For counterflow heat exchanger:

Log mean temperature difference is given by:

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

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 the mass flow rate 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

Calculation:

Given:

Thi = 120°C, The= 80°C , Tci = 60°C

Since the heat capacities of the two fluids are equal.

ṁhch(Thi - The) = ṁccc(Tce - Tci)

120 - 80 = T_ce - 60

T_ce = 100°C

∴ LMTD is given by, LMTD = ΔTi = ΔTe 

⇒ LMTD = 120 - 100 = 20°C

∴ The mean temperature difference between the two fluids is 20°C

Concept:

\({\rm{Effectiveness\;}}\left(\epsilon \right) = \frac{{{Q_{actual}}}}{{{Q_{maximum}}}} = \frac{{{{\left( {mC} \right)}_{hot/cold}}{{\left( {{\rm{\Delta }}T} \right)}_{hot/cold}}}}{{{{\left( {mC} \right)}_{min}}\;{{\left( {{\rm{\Delta }}T} \right)}_{max}}}}\)

Calculation:

Using energy balance,

(mc)_c [120 - 100] = (mc)h [200 - 130]

(mc)h < (mc)c

(mc)_min = (mc)_hot

\(\epsilon = \frac{{{{\left( {mc} \right)}_{hot}}\left[ {{T_{h1}} - {T_{h2}}} \right]}}{{{{\left( {mc} \right)}_{hot}}\left[ {{T_{h1}} - {T_{c1}}} \right]}}\)

\(\epsilon = \frac{{200 - 130}}{{200 - 100}}\)

ϵ = 0.7