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)
𝜃𝑚 = is the log mean temperature difference and it is expressed as Δ𝑇1− Δ𝑇2 𝜃𝑚 = 𝑙𝑛(Δ𝑇 Δ𝑇 ) 1 2
Where, Δ𝑇1 = 𝑇ℎ,𝑖𝑛 − 𝑇𝑐,𝑖𝑛 Δ𝑇2 = 𝑇ℎ,𝑜𝑢𝑡 − 𝑇𝑐,𝑜𝑢𝑡 for parallel flow Heat exchanger Δ𝑇1 = 𝑇ℎ,𝑖𝑛 − 𝑇𝑐,𝑜𝑢𝑡 Δ𝑇2 = 𝑇ℎ,𝑜𝑢𝑡 − 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) 𝑞 = 𝑚̇ 𝑔𝑎𝑠 × 𝑐𝑝,𝑔𝑎𝑠 × (𝑇𝑔𝑎𝑠,𝑖𝑛 𝑇𝑔𝑎𝑠,𝑜𝑢𝑡) 200𝑘𝑔 ℎ = ( ) ( ) ℎ 3600 𝑠 × (1.13 𝑘𝐽𝑘𝑔 − 𝐾) × (350 − 100)𝐾 = 15.694 𝑘𝑊

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




1400𝑘𝑔 ℎ 15.694 𝑘𝑊 = ( )( ) × (4.19 𝑘𝐽𝑘𝑔 − 𝐾) × (𝑡𝑒𝑥𝑖𝑡 − 10)𝐾 ℎ 3600 𝑠

Rearranging we have
15.694 𝑘𝑊 (0.389 × 4. ) = (𝑡𝑒𝑥𝑖𝑡 − 10)𝐾 19
𝑡𝑒𝑥𝑖𝑡 = 9.63 + 10 = 19.63 𝑜𝐶
The overall heat transfer coefficient is determined from the following equation
1 1 1 1 = + = 4.0 𝑚2 𝐾⁄𝑘𝑊 𝑎𝑛𝑑 𝑈 = = 0.25 𝑘𝑊𝑚2 𝐾 𝑈 0.3 1.5 4
The logarithmic mean effective temperature can be found (for parallel flow) by
Δ𝑇1− Δ𝑇2 𝜃𝑚 = 𝑙𝑛(Δ𝑇 Δ𝑇 ) 1 2
Δ𝑇1 = 𝑇ℎ,𝑖𝑛 − 𝑇𝑐,𝑖𝑛 Δ𝑇2 = 𝑇ℎ,𝑜𝑢𝑡 − 𝑇𝑐,𝑜𝑢𝑡 for parallel flow Heat exchanger
Δ𝑇1 = 𝑇ℎ,𝑖𝑛 − 𝑇𝑐,𝑖𝑛 = 350 − 10 = 340𝑜𝐶 Δ𝑇2 = 𝑇ℎ,𝑜𝑢𝑡 − 𝑇𝑐,𝑜𝑢𝑡 = 100 − 19.63 = 80.37𝑜𝐶
340 − 80.37 𝜃𝑚 = = 180.1𝑜𝐶 𝑙𝑛(34080.37)
The heat transfer can also be determined from
𝑞 = 𝑈𝐴𝜃𝑚 𝐴 = 𝜋𝑑𝐿 Rearranging the above equation in terms of L we have
𝑞 15.694 𝐿 = = = 1.48 𝑚 𝜋𝑑𝑈𝜃𝑚 𝜋 × 0.075 × 0.25 × 180.1

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



Δ𝑇1− Δ𝑇2 𝜃𝑚 = 𝑙𝑛(Δ𝑇 Δ𝑇 ) 1 2
Δ𝑇1 = 𝑇ℎ,𝑖𝑛 − 𝑇𝑐,𝑜𝑢𝑡 Δ𝑇2 = 𝑇ℎ,𝑜𝑢𝑡 − 𝑇𝑐,𝑖𝑛 for counter flow Heat exchanger
Δ𝑇1 = 𝑇ℎ,𝑖𝑛 − 𝑇𝑐,𝑜𝑢𝑡 = 350 − 19.63 = 330.37𝑜𝐶 Δ𝑇2 = 𝑇ℎ,𝑜𝑢𝑡 − 𝑇𝑐,𝑖𝑛 = 100 − 10 = 90𝑜𝐶
330.37 − 90 𝜃𝑚 = = 184.84𝑜𝐶 𝑙𝑛(330.3790)
The heat transfer can also be determined from
𝑞 = 𝑈𝐴𝜃𝑚 𝐴 = 𝜋𝑑𝐿 Rearranging the above equation in terms of L we have
𝑞 15.694 𝐿 = = = 1.44𝑚 𝜋𝑑𝑈𝜃𝑚 𝜋 × 0.075 × 0.25 × 184.84

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) 𝜙 = 𝑚𝐿𝑛(𝑟)
𝜕𝜙 𝑚 𝑣𝑟 = 𝜕𝑟 = 𝑟 1 𝜕𝜙 𝑣𝜃 = 𝑟 𝜕𝜃 = 0
Solution (b-ii) 𝑚 1 𝜕𝜓 𝜕𝜓 𝑣𝑟 = 𝑟 = 𝑟 𝜕𝜃 𝜕𝜃 = 𝑚

4 marks

4 marks

Integrate with respect to 𝜃:
𝜓 = 𝑚𝜃 + 𝑓1(𝑟)

On the other hand, 𝑣 = 0 = − 𝜕𝜓 𝜃 𝜕𝑟 Therefore, 𝜓 is not function of 𝑟, so 𝑓1(𝑟) is a constant and the equation for 𝜓 becomes: 𝜓 = 𝑚𝜃 + 𝐶1
𝐶1 is an arbitrary constant. Also, 𝜓 = 0 for 𝜃 = 𝜋 3
Thus, 0 = 𝑚 × 𝜋 + 𝐶 ⟹ 𝐶 = −𝑚 × 𝜋 3 1 1 3
𝜋 𝜋 𝜓 = 𝑚𝜃 − 𝑚 = 𝑚(𝜃 − ) 3 3
Solution (b-iii) At point 𝐵, 𝑟 = √𝑥2 + 𝑦2 = √12 + 42 = 4.12
𝜃 = tan−1 𝑦) = tan−1(4) = 1.33 ( 𝑥
The value of the stream function at B:
𝜋 𝜓𝐵 = −0.71 = 𝑚(1.33 − 3) 𝑚 = −2.51

12 marks

5 marks
Q3Solution (a) (i)

T (oC)s (kJ/kg-K)h (kJ/kg)
100.9305 𝑘𝐽𝑘𝑔 − 𝐾256.58
12.230.93773 𝑘𝐽𝑘𝑔 − 𝐾258.69
200.9628 𝑘𝐽𝑘𝑔 − 𝐾265.86

Solution Q3 (a-ii)
Determine the properties at different states.
State1: Refer the following value for saturated vapour of Refrigerant 134a at 𝑇1 = −10.09𝑜𝐶 from property table, we can read. ℎ1 = ℎ𝑔 = 244.46 𝑘𝐽𝑘𝑔 𝑠1 = 𝑠𝑔 = 0.93773 𝑘𝐽𝑘𝑔 − 𝐾
State2: the compression process through compressor 1 is isentropic that is 𝑠2 = 𝑠1 = 0.93773 𝑘𝐽𝑘𝑔 − 𝐾 From superheated table at 5 bar (0.4 MPa) we read
By interpolation assuming linearity find the temperature and enthalpy 20 − 𝑇 0.9628 − 0.93773 = 20 − 10 0.9628 − 0.9305 20 − 𝑇 = 0.777 10 𝑇2 = 20 − (10 × 0.77) = 12.23𝑜𝐶 For enthalpy 265.86 − ℎ 0.9628 − 0.93773 = 265.86 − 256.59 0.9628 − 0.9305

265.86 − ℎ 9.27

= 0.777

2 = 265.86 − (9.27 × 0.77) = 258.69 𝑘𝐽𝑘𝑔

state5: refrigerant is leaving the condensed as saturated liquid at 𝑃5 = 8 𝑏𝑎𝑟

5 = ℎ𝑓 = 95.47 𝑘𝐽𝑘𝑔

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

6 = 95.47 𝑘𝐽𝑘𝑔

We can determine the quality of the steam by reading the enthalpy value from the property table at 𝑃6 = 4 𝑏𝑎𝑟 pressure.

𝑓 = 63.94 𝑘𝐽𝑘𝑔 ℎ𝑓𝑔 = 191.62 𝑘𝐽𝑘𝑔

(95.47 − 63.94) 𝑘𝐽𝑘𝑔

𝑥6 =

191.62 𝑘𝐽𝑘𝑔

= 0.165

State7: Refer the following value for saturated liquid at 𝑃7 = 4 𝑏𝑎𝑟

from property table:

7 = ℎ𝑓 = 63.94 𝑘𝐽𝑘𝑔

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


8 = ℎ7 = 63.94 𝑘𝐽𝑘𝑔

State3: Refer the following value for saturated vapour at 𝑃3 = 4 𝑏𝑎𝑟

from property table,

3 = ℎ𝑔 = 259.55 𝑘𝐽𝑘𝑔

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 (𝑥6). The liquid leaving the flash chamber at state 7 is the fraction (1 − 𝑥6). With these flow rates ratio,

0 = (1 − 𝑥6)2 + 𝑥63 − 1ℎ9 9 = (1 − 𝑥6)2 + 𝑥63

9 = (1 − 0.165)(258.69 𝑘𝐽𝑘𝑔) + (0.165)(259.55 𝑘𝐽𝑘𝑔) = 258.83 𝑘𝐽𝑘𝑔

Use the table to obtain the value of specific entropy corresponding to 𝑃9 = 4 𝑏𝑎𝑟 and ℎ9 = 258.83 𝑘𝐽𝑘𝑔 ∶

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

T (oC)s (kJ/kg-K)h (kJ/kg)
100.9305 𝑘𝐽𝑘𝑔 − 𝐾256.58

200.9628 𝑘𝐽𝑘𝑔 − 𝐾265.86



By interpolation assuming linearity find the temperature and enthalpy 20 − 𝑇 265.86 − 258.83 = 20 − 10 265.86 − 256.58 20 − 𝑇 = 0.758 10 𝑇2 = 20 − (10 × 0.758) = 12.42𝑜𝐶 For enthalpy 0.9628 − 𝑠 0.9628 − 0.93773 = 0.9628 − 0.9305 0.9628 − 0.9305 0.9628 − 𝑠 = 0.758 0.0323 𝑠9 = 0.9628 − (0.0323 × 0.758) = 0.9383 𝑘𝐽𝑘𝑔 − 𝐾
State4: The compression process through compressor 2 is isentropic that is 𝑠3 = 𝑠4 = 0.9383 𝑘𝐽𝑘𝑔 − 𝐾.
4 = 273.5 𝑘𝐽𝑘𝑔 ∶
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 − 𝑇 0.9480 − 0.9383 = 40 − 31.31 0.9480 − 0.9183 40 − 𝑇 = 0.327 8.69 𝑇4 = 40 − (8.69 × 0.327) = 37.2𝑜𝐶 For enthalpy 276.45 − ℎ 0.9480 − 0.9383 = 276.45 − 267.29 0.9480 − 0.9183 276.45 − ℎ = 0.327 9.16 ℎ4 = 276.45 − (9.16 × 0.327) = 273.5 𝑘𝐽𝑘𝑔

Initially determine the mass flow rates to determine the compressor power. For the evaporator, 𝑄̇𝑖𝑛 = 𝑚̇ 1(1 − ℎ8) 𝑄̇𝑖𝑛 10 𝑡𝑜𝑛𝑠 211 𝑘𝐽𝑚𝑖𝑛 1 𝑚𝑖𝑛 𝑚̇ 1 = (ℎ − ℎ ) = (244.46 − 63.94) 𝑘𝐽𝑘𝑔 [ 1 𝑡𝑜𝑛 ] ( 60 𝑠 ) 1 8 = 0.195 𝑘𝑔𝑠
Also, since (1 − 𝑥6) is the fraction of the total flow passing through the evaporator,
𝑚̇1 = (1 − 𝑥 ) 𝑚̇ 3 6 𝑚̇ 1 0.195 𝑘𝑔𝑠 𝑚̇ 3 = (1 𝑥 ) = (1 − 0.165) = 0.234 𝑘𝑔𝑠 6
Power input to compressor 1 𝑊̇ 𝑐𝑜𝑚𝑝,1 = 𝑚̇ 1(2 − ℎ1) = (0.195 𝑘𝑔𝑠)(258.69 − 244.46) 𝑘𝐽𝑘𝑔 = 2.77 𝑘𝑊 Power input to compressor 2 𝑊̇ 𝑐𝑜𝑚𝑝,2 = 𝑚̇ 2(4 − ℎ3) = (0.234 𝑘𝑔𝑠)(273.5 − 259.55) 𝑘𝐽𝑘𝑔 = 3.26 𝑘𝑊
The coefficient of performance is 𝑄̇𝑖𝑛 𝐶. 𝑂. 𝑃 = 𝑊̇ 𝑐𝑜𝑚𝑝,1 + 𝑊̇ 𝑐𝑜𝑚𝑝,2 10 𝑡𝑜𝑛𝑠 211 𝑘𝐽𝑚𝑖𝑛 1 𝑚𝑖𝑛 1 𝑘𝑊 = (10.31 + 6.64) 𝑘𝐽𝑘𝑔 [ 1 𝑡𝑜𝑛 ] ( 60 𝑠 )(1 𝑘𝐽𝑠) = 5.8


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.


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


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
Solution (d)

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
Solution (a)
Applying Bernoulli’s equation from far upstream to just upstream the rotor:
1 1 𝑃𝑜 + 𝜌𝑈2 = 𝑃 + 𝜌𝑢2 2 𝑜 2 It is also valid to apply Bernoulli’s equation from just behind the rotor to the far downstream wake.
1 1 𝑃 − ∆𝑃 + 𝜌𝑢2 = 𝑃𝑜 + 𝜌𝑢2 2 2 1


Combining the two equations yields: 𝑃 + 1 𝜌𝑈2 + 𝑃 − ∆𝑃 + 1 𝜌𝑈2 + 1 𝜌𝑢2 = 𝑃 + 𝑃 + 1 𝜌𝑢2 + 1 𝜌𝑢2 𝑜 2 𝑜 2 𝑜 2 𝑜 2 2 1
1 1 ⟹ 𝜌𝑈2 − ∆𝑃 = 𝜌𝑢2 2 𝑜 2 1 ⟹ ∆𝑃 = 1 𝜌(𝑈2 − 𝑢2) 2 𝑜 1
Therefore the thrust on the rotor is: 1 𝑇 = 𝜌𝐴(𝑈2 − 𝑢2) 2 𝑜 1
Now applying the rate of change of momentum:
𝑇 = 𝑚̇ (𝑈𝑜 − 𝑢1)
Since 𝑚̇ = 𝜌𝑈𝑜𝐴𝑜 = 𝜌𝑢𝐴 = 𝜌𝑢1𝐴1
Implies 𝑇 = 𝜌𝑢𝐴(𝑈𝑜 − 𝑢1)
Combining the two equations for the thrust yields:
1 𝜌𝐴(𝑈2 − 𝑢2) = 𝜌𝑢𝐴(𝑈 − 𝑢 ) 2 𝑜 1 𝑜 1
1 ⟹ (𝑈𝑜 − 𝑢1)(𝑈𝑜 + 𝑢1) = 𝑢(𝑈𝑜 − 𝑢1) ⟹ 2
1 𝑢 = (𝑈𝑜 + 𝑢1) 2
Solution (b) 𝑈𝑜 = 12𝑚𝑠−1 𝑎 = 0.21
𝑈𝑜 − 𝑢1 𝑎 = ⟹ 𝑎𝑈𝑜 = 𝑈𝑜 − 𝑢 𝑈𝑜
𝑢 = 𝑈𝑜 − 𝑎𝑈𝑜 = 𝑈𝑜(1 − 𝑎)
𝑢 = 12(1 − 0.21)

12 marks

4 marks

𝑢 = 9.48𝑚𝑠−1
Solution (c)
1 𝑢 = (𝑈𝑜 + 𝑢1) ⟹ 2𝑢 = 𝑈𝑜 + 𝑢1 2
𝑢1 = 2𝑢 − 𝑈𝑜
𝑢1 = 2𝑈𝑜(1 − 𝑎) − 𝑈𝑜 = 2𝑈𝑜 − 2𝑎𝑈𝑜 − 𝑈𝑜
𝑢1 = 𝑈𝑜 − 2𝑎𝑈𝑜 ⟹ 𝑢1 = 𝑈𝑜(1 − 2𝑎)
𝑢1 = 12(1 − 2 × 0.21) ⟹ 𝑢1 = 6.96𝑚𝑠−1
Solution (c) 𝜋𝐷2 𝑇 = 𝜌𝑢𝐴(𝑈𝑜 − 𝑈1) ⟹ 𝑇 = 𝜌𝑢 4 (𝑈𝑜 − 𝑢1) 1.2 × 9.48 × 3.14 × 62 × (12 − 6.96) 𝑇 = ⟹ 4

𝑇 = 1620.28𝑁

6 marks

3 marks

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Assessment 1: The Journal of Biological Chemistry

Guide abstract All cells within the body are bathed in a fluid. The composition of this fluid is tightly regulated to contain specific levels of various ions. Disruption in the balance of these ions can lead to cell death and damaging inflammation. However, standard surgical practice fails to consider these

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Due: Oct 1, before midnight. This document first provides the aims of this project. It then lists the requirements as explicitly as possible. It then hints at how these requirements can be met. Finally, it describes how it can be submitted. Aims The aims of this project are as follows:

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ASSESSMENT BRIEF 3: Calculations Summary Title Calculations Due Date Monday the 3rd of October at 11:59 (NSW) AEST – End of Week 5 Length 1000-word (does not include equations, charts, graphs or tables) Weighting  35% Submission  WORD document submitted to Turnitin Unit Learning Outcomes  ULO 3: apply basic statistical analysis

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ASSESSMENT TWO: LITERATURE REVIEW – DATA MINING TO RESEARCH PROPOSAL – DESIGN RESEARCH INSTRUMENT – PRESENTATION) Task Length: Mini Report (750 words) – Research Proposal) Presentation – Not more than 15 Slides –  (Templates Provided for both the mini report and presentation)  Duration of the Video Presentation – Not more

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VU22977 – Practice in a Legal Environment

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KIT714 ICT Research Principles: Assignment 1

Practical Qualitative Research Exercise                                                                                                                                         Type:                 In-Semester, Individual Assignment Task Length:   minimum 2,000 words Weighting:     20% of total assessment for this unit Due Date:       Friday 5 August 2022 – 11:55 pm (Week05) Submission:    electronic submission on MyLO (WORD or PDF) Description:    This practical exercise will engage students in a qualitative research

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