GEEN 1121 Model Solution

Program: MSc Mechanical Engineering, Meng Mechanical Engineering Module
Title & Code: Advanced Thermo-fluid Application-GEEN1121
Level: 7
Date: April/May 2021

University of Greenwich

Questio nSolutionMark
Q1Solution (a)
πœƒπ‘š = is the log mean temperature difference and it is expressed as Δ𝑇1βˆ’ Δ𝑇2 πœƒπ‘š = 𝑙𝑛(Δ𝑇 ⁄Δ𝑇 ) 1 2
Where, Δ𝑇1 = π‘‡β„Ž,𝑖𝑛 βˆ’ 𝑇𝑐,𝑖𝑛 Δ𝑇2 = π‘‡β„Ž,π‘œπ‘’π‘‘ βˆ’ 𝑇𝑐,π‘œπ‘’π‘‘ for parallel flow Heat exchanger Δ𝑇1 = π‘‡β„Ž,𝑖𝑛 βˆ’ 𝑇𝑐,π‘œπ‘’π‘‘ Δ𝑇2 = π‘‡β„Ž,π‘œπ‘’π‘‘ βˆ’ 𝑇𝑐,𝑖𝑛 for counter flow Heat exchanger
Solution (b)
The heat exchangers usually operate for long periods of time with no change in their operating conditions, and then they can be modeled as steady-flow devices. As such, the mass flow rate of each fluid remains constant and the fluid properties such as temperature and velocity at any inlet and outlet remain constant. The kinetic and potential energy changes are negligible. The specific heat of a fluid can be treated as constant in a specified temperature range. Axial heat conduction along the tube is negligible. Finally, the outer surface of the heat exchanger is assumed to be perfectly insulated so that there is no heat loss to the surrounding medium and any heat transfer thus occurs is between the two fluids only.
Solution (c) Heat transfer from exhaust gas can be determined by (i) π‘ž = π‘šΜ‡ π‘”π‘Žπ‘  Γ— 𝑐𝑝,π‘”π‘Žπ‘  Γ— (π‘‡π‘”π‘Žπ‘ ,𝑖𝑛 βˆ’ π‘‡π‘”π‘Žπ‘ ,π‘œπ‘’π‘‘) 200π‘˜π‘” β„Ž = ( ) ( ) β„Ž 3600 𝑠 Γ— (1.13 π‘˜π½β„π‘˜π‘” βˆ’ 𝐾) Γ— (350 βˆ’ 100)𝐾 = 15.694 π‘˜π‘Š

Let the exit temperature of the water to be t exit, therefore,










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1400π‘˜π‘” β„Ž 15.694 π‘˜π‘Š = ( )( ) Γ— (4.19 π‘˜π½β„π‘˜π‘” βˆ’ 𝐾) Γ— (𝑑𝑒π‘₯𝑖𝑑 βˆ’ 10)𝐾 β„Ž 3600 𝑠

Rearranging we have
15.694 π‘˜π‘Š (0.389 Γ— 4. ) = (𝑑𝑒π‘₯𝑖𝑑 βˆ’ 10)𝐾 19
𝑑𝑒π‘₯𝑖𝑑 = 9.63 + 10 = 19.63 π‘œπΆ
The overall heat transfer coefficient is determined from the following equation
1 1 1 1 = + = 4.0 π‘š2 πΎβ„π‘˜π‘Š π‘Žπ‘›π‘‘ π‘ˆ = = 0.25 π‘˜π‘Šβ„π‘š2 𝐾 π‘ˆ 0.3 1.5 4
The logarithmic mean effective temperature can be found (for parallel flow) by
Δ𝑇1βˆ’ Δ𝑇2 πœƒπ‘š = 𝑙𝑛(Δ𝑇 ⁄Δ𝑇 ) 1 2
Δ𝑇1 = π‘‡β„Ž,𝑖𝑛 βˆ’ 𝑇𝑐,𝑖𝑛 Δ𝑇2 = π‘‡β„Ž,π‘œπ‘’π‘‘ βˆ’ 𝑇𝑐,π‘œπ‘’π‘‘ for parallel flow Heat exchanger
Δ𝑇1 = π‘‡β„Ž,𝑖𝑛 βˆ’ 𝑇𝑐,𝑖𝑛 = 350 βˆ’ 10 = 340π‘œπΆ Δ𝑇2 = π‘‡β„Ž,π‘œπ‘’π‘‘ βˆ’ 𝑇𝑐,π‘œπ‘’π‘‘ = 100 βˆ’ 19.63 = 80.37π‘œπΆ
340 βˆ’ 80.37 πœƒπ‘š = = 180.1π‘œπΆ 𝑙𝑛(340⁄80.37)
The heat transfer can also be determined from
π‘ž = π‘ˆπ΄πœƒπ‘š 𝐴 = πœ‹π‘‘πΏ Rearranging the above equation in terms of L we have
π‘ž 15.694 𝐿 = = = 1.48 π‘š πœ‹π‘‘π‘ˆπœƒπ‘š πœ‹ Γ— 0.075 Γ— 0.25 Γ— 180.1

The logarithmic mean effective temperature can be found (for counter flow) by









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Δ𝑇1βˆ’ Δ𝑇2 πœƒπ‘š = 𝑙𝑛(Δ𝑇 ⁄Δ𝑇 ) 1 2
Δ𝑇1 = π‘‡β„Ž,𝑖𝑛 βˆ’ 𝑇𝑐,π‘œπ‘’π‘‘ Δ𝑇2 = π‘‡β„Ž,π‘œπ‘’π‘‘ βˆ’ 𝑇𝑐,𝑖𝑛 for counter flow Heat exchanger
Δ𝑇1 = π‘‡β„Ž,𝑖𝑛 βˆ’ 𝑇𝑐,π‘œπ‘’π‘‘ = 350 βˆ’ 19.63 = 330.37π‘œπΆ Δ𝑇2 = π‘‡β„Ž,π‘œπ‘’π‘‘ βˆ’ 𝑇𝑐,𝑖𝑛 = 100 βˆ’ 10 = 90π‘œπΆ
330.37 βˆ’ 90 πœƒπ‘š = = 184.84π‘œπΆ 𝑙𝑛(330.37⁄90)
The heat transfer can also be determined from
π‘ž = π‘ˆπ΄πœƒπ‘š 𝐴 = πœ‹π‘‘πΏ Rearranging the above equation in terms of L we have
π‘ž 15.694 𝐿 = = = 1.44π‘š πœ‹π‘‘π‘ˆπœƒπ‘š πœ‹ Γ— 0.075 Γ— 0.25 Γ— 184.84











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Q2
Solution (a) The vorticity is twice the average angular rotation of a fluid element. It can be calculated by taking the Curl of the velocity vector.
Solution (b-i) πœ™ = π‘šπΏπ‘›(π‘Ÿ)
πœ•πœ™ π‘š π‘£π‘Ÿ = πœ•π‘Ÿ = π‘Ÿ 1 πœ•πœ™ π‘£πœƒ = π‘Ÿ πœ•πœƒ = 0
Solution (b-ii) π‘š 1 πœ•πœ“ πœ•πœ“ π‘£π‘Ÿ = π‘Ÿ = π‘Ÿ πœ•πœƒ ⟹ πœ•πœƒ = π‘š


4 marks






4 marks

Integrate with respect to πœƒ:
πœ“ = π‘šπœƒ + 𝑓1(π‘Ÿ)

On the other hand, 𝑣 = 0 = βˆ’ πœ•πœ“ πœƒ πœ•π‘Ÿ Therefore, πœ“ is not function of π‘Ÿ, so 𝑓1(π‘Ÿ) is a constant and the equation for πœ“ becomes: πœ“ = π‘šπœƒ + 𝐢1
𝐢1 is an arbitrary constant. Also, πœ“ = 0 for πœƒ = πœ‹ 3
Thus, 0 = π‘š Γ— πœ‹ + 𝐢 ⟹ 𝐢 = βˆ’π‘š Γ— πœ‹ 3 1 1 3
πœ‹ πœ‹ πœ“ = π‘šπœƒ βˆ’ π‘š = π‘š(πœƒ βˆ’ ) 3 3
Solution (b-iii) At point 𝐡, π‘Ÿ = √π‘₯2 + 𝑦2 = √12 + 42 = 4.12
πœƒ = tanβˆ’1 𝑦) = tanβˆ’1(4) = 1.33 ( π‘₯
The value of the stream function at B:
πœ‹ πœ“π΅ = βˆ’0.71 = π‘š(1.33 βˆ’ 3) π‘š = βˆ’2.51












12 marks











5 marks
Q3Solution (a) (i)








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T (oC)s (kJ/kg-K)h (kJ/kg)
100.9305 π‘˜π½β„π‘˜π‘” βˆ’ 𝐾256.58
12.230.93773 π‘˜π½β„π‘˜π‘” βˆ’ 𝐾258.69
200.9628 π‘˜π½β„π‘˜π‘” βˆ’ 𝐾265.86



(ii)
Solution Q3 (a-ii)
Determine the properties at different states.
State1: Refer the following value for saturated vapour of Refrigerant 134a at 𝑇1 = βˆ’10.09π‘œπΆ from property table, we can read. β„Ž1 = β„Žπ‘” = 244.46 π‘˜π½β„π‘˜π‘” 𝑠1 = 𝑠𝑔 = 0.93773 π‘˜π½β„π‘˜π‘” βˆ’ 𝐾
State2: the compression process through compressor 1 is isentropic that is 𝑠2 = 𝑠1 = 0.93773 π‘˜π½β„π‘˜π‘” βˆ’ 𝐾 From superheated table at 5 bar (0.4 MPa) we read
By interpolation assuming linearity find the temperature and enthalpy 20 βˆ’ 𝑇 0.9628 βˆ’ 0.93773 = 20 βˆ’ 10 0.9628 βˆ’ 0.9305 20 βˆ’ 𝑇 = 0.777 10 𝑇2 = 20 βˆ’ (10 Γ— 0.77) = 12.23π‘œπΆ For enthalpy 265.86 βˆ’ β„Ž 0.9628 βˆ’ 0.93773 = 265.86 βˆ’ 256.59 0.9628 βˆ’ 0.9305

265.86 βˆ’ β„Ž 9.27

= 0.777

β„Ž2 = 265.86 βˆ’ (9.27 Γ— 0.77) = 258.69 π‘˜π½β„π‘˜π‘”

state5: refrigerant is leaving the condensed as saturated liquid at 𝑃5 = 8 π‘π‘Žπ‘Ÿ

β„Ž5 = β„Žπ‘“ = 95.47 π‘˜π½β„π‘˜π‘”

State6: throttling of the refrigerant and enthalpy remains constant and therefore,

β„Ž6 = 95.47 π‘˜π½β„π‘˜π‘”

We can determine the quality of the steam by reading the enthalpy value from the property table at 𝑃6 = 4 π‘π‘Žπ‘Ÿ pressure.

β„Žπ‘“ = 63.94 π‘˜π½β„π‘˜π‘” β„Žπ‘“π‘” = 191.62 π‘˜π½β„π‘˜π‘”

(95.47 βˆ’ 63.94) π‘˜π½β„π‘˜π‘”

π‘₯6 =

191.62 π‘˜π½β„π‘˜π‘”

= 0.165

State7: Refer the following value for saturated liquid at 𝑃7 = 4 π‘π‘Žπ‘Ÿ

from property table:

β„Ž7 = β„Žπ‘“ = 63.94 π‘˜π½β„π‘˜π‘”

State8: Throttling occurs as the refrigerant flows through the expansion valve.

Therefore,

β„Ž8 = β„Ž7 = 63.94 π‘˜π½β„π‘˜π‘”

State3: Refer the following value for saturated vapour at 𝑃3 = 4 π‘π‘Žπ‘Ÿ

from property table,

β„Ž3 = β„Žπ‘” = 259.55 π‘˜π½β„π‘˜π‘”

State 9: The fraction of the flow into the flash chamber at state 6 that exits as saturated vapour at state 3 is equal to the quality at state 6 (π‘₯6). The liquid leaving the flash chamber at state 7 is the fraction (1 βˆ’ π‘₯6). With these flow rates ratio,

0 = (1 βˆ’ π‘₯6)β„Ž2 + π‘₯6β„Ž3 βˆ’ 1β„Ž9 β„Ž9 = (1 βˆ’ π‘₯6)β„Ž2 + π‘₯6β„Ž3

β„Ž9 = (1 βˆ’ 0.165)(258.69 π‘˜π½β„π‘˜π‘”) + (0.165)(259.55 π‘˜π½β„π‘˜π‘”) = 258.83 π‘˜π½β„π‘˜π‘”

Use the table to obtain the value of specific entropy corresponding to 𝑃9 = 4 π‘π‘Žπ‘Ÿ and β„Ž9 = 258.83 π‘˜π½β„π‘˜π‘” ∢










From superheated table at 8 bar (0.8 MPa) we read

T (oC)s (kJ/kg-K)h (kJ/kg)
100.9305 π‘˜π½β„π‘˜π‘” βˆ’ 𝐾256.58
12.42
258.83














200.9628 π‘˜π½β„π‘˜π‘” βˆ’ 𝐾265.86



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By interpolation assuming linearity find the temperature and enthalpy 20 βˆ’ 𝑇 265.86 βˆ’ 258.83 = 20 βˆ’ 10 265.86 βˆ’ 256.58 20 βˆ’ 𝑇 = 0.758 10 𝑇2 = 20 βˆ’ (10 Γ— 0.758) = 12.42π‘œπΆ For enthalpy 0.9628 βˆ’ 𝑠 0.9628 βˆ’ 0.93773 = 0.9628 βˆ’ 0.9305 0.9628 βˆ’ 0.9305 0.9628 βˆ’ 𝑠 = 0.758 0.0323 𝑠9 = 0.9628 βˆ’ (0.0323 Γ— 0.758) = 0.9383 π‘˜π½β„π‘˜π‘” βˆ’ 𝐾
State4: The compression process through compressor 2 is isentropic that is 𝑠3 = 𝑠4 = 0.9383 π‘˜π½β„π‘˜π‘” βˆ’ 𝐾.
β„Ž4 = 273.5 π‘˜π½β„π‘˜π‘” ∢
From superheated table at 8 bar (0.8 MPa) we read T (oC) s (kJ/kg-K) h (kJ/kg) 31.31 0.9183 267.29 37.2 0.9383 273.5 40 0.9480 276.45
By interpolation assuming linearity find the temperature and enthalpy 40 βˆ’ 𝑇 0.9480 βˆ’ 0.9383 = 40 βˆ’ 31.31 0.9480 βˆ’ 0.9183 40 βˆ’ 𝑇 = 0.327 8.69 𝑇4 = 40 βˆ’ (8.69 Γ— 0.327) = 37.2π‘œπΆ For enthalpy 276.45 βˆ’ β„Ž 0.9480 βˆ’ 0.9383 = 276.45 βˆ’ 267.29 0.9480 βˆ’ 0.9183 276.45 βˆ’ β„Ž = 0.327 9.16 β„Ž4 = 276.45 βˆ’ (9.16 Γ— 0.327) = 273.5 π‘˜π½β„π‘˜π‘”
(iii)

Initially determine the mass flow rates to determine the compressor power. For the evaporator, 𝑄̇𝑖𝑛 = π‘šΜ‡ 1(β„Ž1 βˆ’ β„Ž8) 𝑄̇𝑖𝑛 10 π‘‘π‘œπ‘›π‘  211 π‘˜π½β„π‘šπ‘–π‘› 1 π‘šπ‘–π‘› π‘šΜ‡ 1 = (β„Ž βˆ’ β„Ž ) = (244.46 βˆ’ 63.94) π‘˜π½β„π‘˜π‘” [ 1 π‘‘π‘œπ‘› ] ( 60 𝑠 ) 1 8 = 0.195 π‘˜π‘”β„π‘ 
Also, since (1 βˆ’ π‘₯6) is the fraction of the total flow passing through the evaporator,
π‘šΜ‡1 = (1 βˆ’ π‘₯ ) π‘šΜ‡ 3 6 π‘šΜ‡ 1 0.195 π‘˜π‘”β„π‘  π‘šΜ‡ 3 = (1 βˆ’ π‘₯ ) = (1 βˆ’ 0.165) = 0.234 π‘˜π‘”β„π‘  6
(iv)
Power input to compressor 1 π‘ŠΜ‡ π‘π‘œπ‘šπ‘,1 = π‘šΜ‡ 1(β„Ž2 βˆ’ β„Ž1) = (0.195 π‘˜π‘”β„π‘ )(258.69 βˆ’ 244.46) π‘˜π½β„π‘˜π‘” = 2.77 π‘˜π‘Š Power input to compressor 2 π‘ŠΜ‡ π‘π‘œπ‘šπ‘,2 = π‘šΜ‡ 2(β„Ž4 βˆ’ β„Ž3) = (0.234 π‘˜π‘”β„π‘ )(273.5 βˆ’ 259.55) π‘˜π½β„π‘˜π‘” = 3.26 π‘˜π‘Š
(v)
The coefficient of performance is 𝑄̇𝑖𝑛 𝐢. 𝑂. 𝑃 = π‘ŠΜ‡ π‘π‘œπ‘šπ‘,1 + π‘ŠΜ‡ π‘π‘œπ‘šπ‘,2 10 π‘‘π‘œπ‘›π‘  211 π‘˜π½β„π‘šπ‘–π‘› 1 π‘šπ‘–π‘› 1 π‘˜π‘Š = (10.31 + 6.64) π‘˜π½β„π‘˜π‘” [ 1 π‘‘π‘œπ‘› ] ( 60 𝑠 )(1 π‘˜π½β„π‘ ) = 5.8

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Q4Solution (a)

A heat exchanger is classified as being compact if  > 700 m2/m3 or (200 ft2/ft3) where  is the ratio of the heat transfer surface area to its volume which is called the area density. The area density for double-pipe heat exchanger can not be in the order of 700.

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Solution (b) No, it cannot be classified as a compact heat exchanger. As the heat transfer area to volume ratio is below 700 m2/m3

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Solution (c)

In counter-flow heat exchangers, the hot and the cold fluids move
parallel to each other, but both enter the heat exchanger at2
opposite ends and flow in opposite direction. In cross-flow heat
exchangers, the two fluids usually move perpendicular to each
other.
Solution (d)

1
The cross-flow is said to be unmixed when the plate fins force the
fluid to flow through a particular inter fin spacing and prevent it
from moving in the transverse direction. When the fluid is free to
move in the transverse direction, the cross-flow is said to be mixed.

Solution (e)

In the shell and tube exchangers, baffles are commonly placed in5
the shell to force the shell side fluid to flow across the shell to
enhance heat transfer and to maintain uniform spacing between
the tubes. Baffles disrupt the flow of fluid, and an increased
pumping power will be needed to maintain flow. On the other hand,
baffles eliminate dead spots and increase heat transfer rate.
Solution (f)
Open question
Q5
Solution (a)
Applying Bernoulli’s equation from far upstream to just upstream the rotor:
1 1 π‘ƒπ‘œ + πœŒπ‘ˆ2 = 𝑃 + πœŒπ‘’2 2 π‘œ 2 It is also valid to apply Bernoulli’s equation from just behind the rotor to the far downstream wake.
1 1 𝑃 βˆ’ βˆ†π‘ƒ + πœŒπ‘’2 = π‘ƒπ‘œ + πœŒπ‘’2 2 2 1

.

Combining the two equations yields: 𝑃 + 1 πœŒπ‘ˆ2 + 𝑃 βˆ’ βˆ†π‘ƒ + 1 πœŒπ‘ˆ2 + 1 πœŒπ‘’2 = 𝑃 + 𝑃 + 1 πœŒπ‘’2 + 1 πœŒπ‘’2 π‘œ 2 π‘œ 2 π‘œ 2 π‘œ 2 2 1
1 1 ⟹ πœŒπ‘ˆ2 βˆ’ βˆ†π‘ƒ = πœŒπ‘’2 2 π‘œ 2 1 ⟹ βˆ†π‘ƒ = 1 𝜌(π‘ˆ2 βˆ’ 𝑒2) 2 π‘œ 1
Therefore the thrust on the rotor is: 1 𝑇 = 𝜌𝐴(π‘ˆ2 βˆ’ 𝑒2) 2 π‘œ 1
Now applying the rate of change of momentum:
𝑇 = π‘šΜ‡ (π‘ˆπ‘œ βˆ’ 𝑒1)
Since π‘šΜ‡ = πœŒπ‘ˆπ‘œπ΄π‘œ = πœŒπ‘’π΄ = πœŒπ‘’1𝐴1
Implies 𝑇 = πœŒπ‘’π΄(π‘ˆπ‘œ βˆ’ 𝑒1)
Combining the two equations for the thrust yields:
1 𝜌𝐴(π‘ˆ2 βˆ’ 𝑒2) = πœŒπ‘’π΄(π‘ˆ βˆ’ 𝑒 ) 2 π‘œ 1 π‘œ 1
1 ⟹ (π‘ˆπ‘œ βˆ’ 𝑒1)(π‘ˆπ‘œ + 𝑒1) = 𝑒(π‘ˆπ‘œ βˆ’ 𝑒1) ⟹ 2
1 𝑒 = (π‘ˆπ‘œ + 𝑒1) 2
Solution (b) π‘ˆπ‘œ = 12π‘šπ‘ βˆ’1 π‘Ž = 0.21
π‘ˆπ‘œ βˆ’ 𝑒1 π‘Ž = ⟹ π‘Žπ‘ˆπ‘œ = π‘ˆπ‘œ βˆ’ 𝑒 π‘ˆπ‘œ
𝑒 = π‘ˆπ‘œ βˆ’ π‘Žπ‘ˆπ‘œ = π‘ˆπ‘œ(1 βˆ’ π‘Ž)
𝑒 = 12(1 βˆ’ 0.21)


















12 marks




















4 marks


𝑒 = 9.48π‘šπ‘ βˆ’1
Solution (c)
1 𝑒 = (π‘ˆπ‘œ + 𝑒1) ⟹ 2𝑒 = π‘ˆπ‘œ + 𝑒1 2
𝑒1 = 2𝑒 βˆ’ π‘ˆπ‘œ
𝑒1 = 2π‘ˆπ‘œ(1 βˆ’ π‘Ž) βˆ’ π‘ˆπ‘œ = 2π‘ˆπ‘œ βˆ’ 2π‘Žπ‘ˆπ‘œ βˆ’ π‘ˆπ‘œ
𝑒1 = π‘ˆπ‘œ βˆ’ 2π‘Žπ‘ˆπ‘œ ⟹ 𝑒1 = π‘ˆπ‘œ(1 βˆ’ 2π‘Ž)
𝑒1 = 12(1 βˆ’ 2 Γ— 0.21) ⟹ 𝑒1 = 6.96π‘šπ‘ βˆ’1
Solution (c) πœ‹π·2 𝑇 = πœŒπ‘’π΄(π‘ˆπ‘œ βˆ’ π‘ˆ1) ⟹ 𝑇 = πœŒπ‘’ 4 (π‘ˆπ‘œ βˆ’ 𝑒1) 1.2 Γ— 9.48 Γ— 3.14 Γ— 62 Γ— (12 βˆ’ 6.96) 𝑇 = ⟹ 4

𝑇 = 1620.28𝑁












6 marks





3 marks



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