# Need help with my writing homework on Basic Thermodynamics. Write a 1250 word paper answering;

Need help with my writing homework on Basic Thermodynamics. Write a 1250 word paper answering; The volume of the coolant system can be found by using the formula.

Increase in Volume = Original Volume X Co Efficient X Change in Temperature

Now in order to measure the radiator’s overflow we first find out the change in temperature by using the following equation.

Change in temperature = Temperature at the End – Temperature at the Start

Putting the values we get.

Change in temperature = 90℃ – 15℃ = 75℃

Step ‘’1’’ find Original Volume of Water

50 / 100 X 175 = Original Volume X 0.00045 X 75℃

Original Volume = (50 / 100 X 175) / (0.00045 X 75℃)

Original Volume = 87.5 / 0.03375

Original Volume = 2592.5 ml

Step “2” finds the original volume of Ethyl alcohol

30 / 100 X 175 = Original Volume X 0.001 X 75℃

Original Volume = (30 / 100 X 175) / (0.001 X 75)

Original Volume = 52.5 / 0.075

Original Volume = 700 ml

Step “3” finds the original volume of Glycerine

25 / 100 X 175 = Original Volume X 0.00053 X 75℃

Original Volume = (20 / 100 X 175) / (0.00053 X 75)

Original Volume = 35 / 0.0397

Original Volume = 881.6 ml

Total Volume of Coolant Expand

The total volume of coolant expand can be found by adding the volume of all solutions.

i.e. Total volume = original volume of water + original volume of Ethyl alcohol + original volume of Glycerine

Total volume = 2592.5 ml+700 ml+881.6 ml

Total volume = 4174.1 ml /1000

Total volume = 4.17 Litre.

Maximum primary force is given by the formula.

Primary Force=R x w^2 x r x Cos∅

We can find these variables by using the given data.

R = Piston + Con Rod = 0.6 + 0.4 = 1.0

r = 0.16/2 = 0.08

L= Length of Con Rod = 0.3 m

w= 2 x π x 2000/60

Therefore, w2 = (2 x π x 2000/60)²

Maximum Primary Forces

At 0^0 and 〖180〗^0 (T.D.C. & B.D.C)

R x w^2 x r x Cos∅

Primary Forces = 1.0x (2000/60 x 2 x π) ² x 0.08 x Cos 0

Primary Forces = 1.0x (2000/60 x 2 x π) ² x 0.08 x Cos 180

There is a Primary Force (0° + 180°) of 3509 KG m² (+&-) at 90° and 270°

Primary Forces = 1.0x (2000/60 x 2 x π) ² x 0.08 x Cos 90 Primary Forces = 1.0x (2000/60 x 2 x π) ² x 0.08 x Cos 270

There is a primary force (90° + 270°) of 0 kg m²

0 3509

15 3390

30 3039

45 2481

60 1755

75 908.2

90 0

105 -908.2

120 -1755

135 -2481

150 -3039

165 -3390

180 -3509

195 -3390

210 -3039

225 -2481

240 -1755

255 -908.2

270 0

285 908. 