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,









5















5



3

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








3





















5


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










4
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)








6
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



8
























6

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
3



2
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.

2

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

2


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



Universal Assignment (May 22, 2024) GEEN 1121 Model Solution. Retrieved from https://universalassignment.com/geen-1121-model-solution/.
"GEEN 1121 Model Solution." Universal Assignment - May 22, 2024, https://universalassignment.com/geen-1121-model-solution/
Universal Assignment May 1, 2022 GEEN 1121 Model Solution., viewed May 22, 2024,<https://universalassignment.com/geen-1121-model-solution/>
Universal Assignment - GEEN 1121 Model Solution. [Internet]. [Accessed May 22, 2024]. Available from: https://universalassignment.com/geen-1121-model-solution/
"GEEN 1121 Model Solution." Universal Assignment - Accessed May 22, 2024. https://universalassignment.com/geen-1121-model-solution/
"GEEN 1121 Model Solution." Universal Assignment [Online]. Available: https://universalassignment.com/geen-1121-model-solution/. [Accessed: May 22, 2024]

Please note along with our service, we will provide you with the following deliverables:

  • A premium expert will be assigned to complete your assignment.
  • Quality Control team will check the assignment on a regular basis before the delivery.
  • Plagiarism-free assignment will be provided to you with the Turnitin report.
  • Free revision policy will be provided in case you need any changes or amendments upto 15 days of final submission.
  • We strictly follow the assignment's guideline.
  • Your assignment will be delivered before the deadline provided.
  • We are here to help you 24X7 around the clock, 365 days a year.

Please do not hesitate to put forward any queries regarding the service provision.

We look forward to having you on board with us.
Recent Assignments
Categories

Get 90%* Discount on Assignment Help

Most Frequent Questions & Answers

Universal Assignment Services is the best place to get help in your all kind of assignment help. We have 172+ experts available, who can help you to get HD+ grades. We also provide Free Plag report, Free Revisions,Best Price in the industry guaranteed.

We provide all kinds of assignmednt help, Report writing, Essay Writing, Dissertations, Thesis writing, Research Proposal, Research Report, Home work help, Question Answers help, Case studies, mathematical and Statistical tasks, Website development, Android application, Resume/CV writing, SOP(Statement of Purpose) Writing, Blog/Article, Poster making and so on.

We are available round the clock, 24X7, 365 days. You can appach us to our Whatsapp number +1 (613)778 8542 or email to info@universalassignment.com . We provide Free revision policy, if you need and revisions to be done on the task, we will do the same for you as soon as possible.

We provide services mainly to all major institutes and Universities in Australia, Canada, China, Malaysia, India, South Africa, New Zealand, Singapore, the United Arab Emirates, the United Kingdom, and the United States.

We provide lucrative discounts from 28% to 70% as per the wordcount, Technicality, Deadline and the number of your previous assignments done with us.

After your assignment request our team will check and update you the best suitable service for you alongwith the charges for the task. After confirmation and payment team will start the work and provide the task as per the deadline.

Yes, we will provide Plagirism free task and a free turnitin report along with the task without any extra cost.

No, if the main requirement is same, you don’t have to pay any additional amount. But it there is a additional requirement, then you have to pay the balance amount in order to get the revised solution.

The Fees are as minimum as $10 per page(1 page=250 words) and in case of a big task, we provide huge discounts.

We accept all the major Credit and Debit Cards for the payment. We do accept Paypal also.

See More

Popular Assignments

EDM9780M CEEL Summative Assignment 2023-2024

Below you will find instructions on completing each of the four parts of your final summative assignment. Part 1 – Personal/professional area of interest in education (1000 – 1,500 words max) For this part of the assignment, you will need to: How to complete this part (Part 1): 1. Choose

Read More Β»

AT1 PREPARATION REFLECTIONΒ TEMPLATE

Weighting: 5 marks (10%) of the assignment. COMPLETE & SUBMIT INDIVIDUALLY. This is the second of THREE documents required for submission for the assignment. Complete the following, describing and reflecting upon your involvement with the preparation for the Group Presentation, including your interaction with other members of your team in

Read More Β»

SUMMATIVE ASSIGNMENT – Mathematics for Science

IMPORTANT INFORMATION 1 Electric power is widely used in industrial, commercial and consumer applications. The latter include laboratory equipment for example water baths, spectrophotometers, and chromatographs. If you have 17.3 kA and 5.5 MV, what is the power? Give the appropriate unit.                                                                                                               (3 marks) 2 Oil immersion objective lenses

Read More Β»

Assignment CW 2. Foundations of Biology

The instructions in RED are the ones which are mark-bearing and need to be answered as part of the assignment. The instructions in BLACK tell you how to carry out the simulation Diffusion simulation: Results table Use Excel to calculate the mean and standard deviation. The functions are AVERGAGE and

Read More Β»

MA Education Dissertation Proposal

Student Name Click here to enter text. Student ID                       Proposed title of research project Click here to enter text.       State the background references on which your research is based (ideally 4 or 5) Click here to

Read More Β»

Assignment: Implement five dangerous software errors

Due: Monday, 6 May 2024, 3:00 PM The requirements for assessment 1: Too many developers are prioritising functionality and performance over security. Either that, or they just don’t come from a security background, so they don’t have security in mind when they are developing the application, therefore leaving the business

Read More Β»

LNDN08003 DATA ANALYTICS FINAL PROJECT

Business School                                                                 London campus Session 2023-24                                                                   Trimester 2 Module Code: LNDN08003 DATA ANALYTICS FINAL PROJECT Due Date: 12th APRIL 2024 Answer ALL questions. LNDN08003–Data Analytics Group Empirical Research Project Question 2-The project (2500 maximum word limit) The datasets for this assignment should be downloaded from the World Development Indicators (WDI)

Read More Β»

Imagine you are an IT professional and your manager asked you to give a presentation about various financial tools used to help with decisions for investing in IT and/or security

Part 1, scenario: Imagine you are an IT professional and your manager asked you to give a presentation about various financial tools used to help with decisions for investing in IT and/or security. The presentation will be given to entry-level IT and security employees to understand financial investing. To simulate

Read More Β»

DX5600 Digital Artefact and Research Report

COLLEGE OF ENGINEERING, DESIGN AND PHYSICAL SCIENCES BRUNEL DESIGN SCHOOL DIGITAL MEDIA MSC DIGITAL DESIGN AND BRANDING MSC DIGITAL DESIGN (3D ANIMTION) MSC DIGITAL DESIGN (MOTION GRAPHICS) MSC DIGITAL DESIGN (IMMERSIVE MIXED REALITY) DIGITAL ARTEFACT AND RESEARCH REPORTΒ Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β  Module Code: DX5600 Module Title: MSc Dissertation Module Leader: XXXXXXXXXXXXXXXXX Assessment Title:

Read More Β»

Bsc Public Health and Health Promotion (Top up) LSC LONDON

Health and Work Assignment Brief.Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β  Assessment brief: A case study of 4,000 words (weighted at 100%) Students will present a series of complementary pieces of written work that:   a) analyse the key workplace issues; b) evaluate current or proposed strategies for managing them from a public health/health promotion perspective

Read More Β»

6HW109 Environmental Management and Sustainable Health

ASSESSMENT BRIEF MODULE CODE: 6HW109 MODULE TITLE: Environmental Management and Sustainable Health MODULE LEADER: XXXXXXXXX ACADEMIC YEAR: 2022-23 1        Demonstrate a critical awareness of the concept of Environmental Management linked to Health 2        Critically analyse climate change and health public policies. 3        Demonstrate a critical awareness of the concept of

Read More Β»

PROFESSIONAL SECURE NETWORKS COCS71196

PROFESSIONAL SECURE NETWORKS– Case Study Assessment Information Module Title: PROFESSIONAL SECURE NETWORKS   Module Code: COCS71196 Submission Deadline: 10th May 2024 by 3:30pm Instructions to candidates This assignment is one of two parts of the formal assessment for COCS71196 and is therefore compulsory. The assignment is weighted at 50% of

Read More Β»

CYBERCRIME FORENSIC ANALYSIS – COCS71193

CYBERCRIME FORENSIC ANALYSIS – COCS71193 Assignment Specification Weighted at 100% of the module mark. Learning Outcomes being assessed by this portfolio. Submission Deadline: Monday 6th May 2024, 1600Hrs. Requirements & Marking Scheme General Guidelines: This is an individual assessment comprised of four parts and is weighted at 100% of the

Read More Β»

Social Media Campaigns (SMC) Spring 2024 – Winter 2024

Unit: Dynamic Websites Assignment title: Social Media Campaigns (SMC) Spring 2024 – Winter 2024 Students must not use templates that they have not designed or created in this module assessment. This includes website building applications, free HTML5 website templates, or any software that is available to them to help with

Read More Β»

ABCJ3103 NEWS WRITING AND REPORTING Assignment

ASSIGNMENT/ TUGASAN _________________________________________________________________________ ABCJ3103 NEWS WRITING AND REPORTING PENULISAN DAN PELAPORAN BERITA JANUARY 2024 SEMESTER SPECIFIC INSTRUCTION / ARAHAN KHUSUS Jawab dalam bahasa Melayu atau bahasa Inggeris. Jumlah patah perkataan: 2500 – 3000 patah perkataan tidak termasuk rujukan. Hantar tugasan SEKALI sahaja dalam PELBAGAIfail. Tugasan ini dihantar secara ONLINE. Tarikh

Read More Β»

ABCM2103 INFORMATION TECHNOLOGY, MEDIA AND SOCIETY Assignment

ASSIGNMENT/ TUGASAN _________________________________________________________________________ ABCM2103 INFORMATION TECHNOLOGY, MEDIA AND SOCIETY TEKNOLOGI MAKLUMAT, MEDIA DAN MASYARAKAT JANUARY 2021 SPECIFIC INSTRUCTION / ARAHAN KHUSUS Jawab dalam Bahasa Melayu atau Bahasa Inggeris. Jumlah patah perkataan : 2500 – 3000 patah perkataan tidak termasuk rujukan. Hantar tugasan SEKALI sahaja dalam SATU fail. Tugasan ini dihantar

Read More Β»

ABCR3203 COMMUNICATION LAW Assignment

ASSIGNMENT/ TUGASAN _________________________________________________________________________ ABCR3203 COMMUNICATION LAW UNDANG-UNDANG KOMUNIKASI JANUARY 2024 SEMESTER SPECIFIC INSTRUCTION / ARAHAN KHUSUS Jawab dalam Bahasa Melayu atau Bahasa Inggeris. Jumlah patah perkataan : 2500 – 3000 patah perkataan tidak termasuk rujukan. Hantar tugasan SEKALI sahaja dalam SATU fail. Tugasan ini dihantar secara ONLINE. Tarikh penghantaran        :

Read More Β»

ORGANISATIONAL STRATEGY PLANNING AND MANAGEMENT ASSIGNMENT

POSTGRADUATE DIPLOMA IN BUSINESS MANAGEMENT ORGANISATIONAL STRATEGY PLANNING AND MANAGEMENT ASSIGNMENT NOTE: At postgraduate level, you are expected to substantiate your answers with evidence from independent research. INTRODUCTION TO THE ASSIGNMENT β€’ This assignment consists of FOUR compulsory questions. Please answer all of them. β€’ When you answer, preferably use

Read More Β»

Solution: Scenario 1, Mirror therapy in patients post stroke

Title: Scenario 1, Mirror therapy in patients post stroke Part 1 : Summary Ramachandran and colleagues developed mirror therapy to treat amputees’ agony from phantom limbs. Patients were able to feel their amputated limb without experiencing any pain by presenting them a mirror image of their healthy arm. Since then,

Read More Β»

Solution: Exploring the Dominance of Silence

Slide 1: Title – Exploring the Dominance of Silence The title, “Exploring the Dominance of Silence,” sets the stage for a deep dive into the portrayal of silence in Philip K. Dick’s “Do Androids Dream of Electric Sheep?” Our presentation will dissect the literary techniques used by the author to

Read More Β»

Solution: Assessment: Critical Reflection S2 2023

The policies that hampered the cultural survival of Indigenous groups have a major effect on their health (Coffin, 2007). Cultural isolation can cause an identity crisis and a sense of loss, which can exacerbate mental health problems. Indigenous people have greater rates of chronic illness and impairment due to historical

Read More Β»

Solution: The Market – Product and Competition Analysis

Section 1: The Market – Product and Competition Analysis Industry and Competition Analysis: The baking mix market is very competitive, but My Better Batch is entering it anyhow. The prepackaged baking mixes sold in this market allow busy people to have bakery-quality products on the table quickly without sacrificing quality

Read More Β»

Solution: PDCA model for Riot

Student Name: Student ID: University Name: Date: Learning Outcome 1: Engage actively in recognizing a new product/service for Riot and detect the vital tasks required for its effective growth. In this comprehensive learning outcome, Riot’s progress towards innovation superiority is characterized by a deliberate scheme that draws on components from

Read More Β»

Solution: EDEN 100 – ASSIGNMENT 1

Part 1: Reflections on the Register Variables Use the questions in Column 1 and analyse the sample oral interactions provided under the assessment tile. The transcript for Viv’s conversation is provided on pages 4-5. Probe Questions  Link to readings and theory Interaction 1 Interaction 2 PART 1 – ANALYSING THE

Read More Β»

Can't Find Your Assignment?

Hire an Expert
Open chat
1
Free Assistance
Universal Assignment
Hello πŸ‘‹
How can we help you?