Basic Thermodynamics - Assignment Example

Calculations Calculations Answer 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 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. Answer # 2 Maximum primary force is given by the formula; We can find these variables by using the given data; R = Piston + Con Rod = 0.6 + 0.4 = 1.0 r = = 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 (T.D.C. & B.D.C) 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 Answer # 3 In order to find the combine effect of Primary Force and Secondary we have to find the value of secondary Force. The value of Secondary Force can be found by using the following formula; Secondary Force = R x w² x r ((Cos (2x∅))/ (I/r)) Secondary Force At 0° and 180° (T.D.C & B.D.C) Secondary Force = R x w² x r ((Cos (2x∅))/ (I/r)) Secondary Force = 1.0x (2000/60 x 2 x π) ² x 0.08 x ((Cos (2x0))/ (0.3/0.08)) Secondary Force = 1.0x (2000/60 x 2 x π) ² x 0.08 x ((Cos (2x180))/ (0.3/0.08)) There is a secondary force of 935.78 kg m² At 90° and 270° there is also a negative force of - 935.78 kg m Value of secondary force at different angles are found and can be seen in the table given below, the graph representation of these values is also given below. Combined Primary and Secondary Forces When the primary and secondary forces are added together we are going to follow this Formula Force = R x w² x r (Cos+cos2/n), Where, n = (Con Rod length / Crank Radius) = L/r Maximum Combined Primary and Secondary Forces At T.D.C. there is 3509kg m² and 935.78kg m² = 4444.97kg m² At B.D.C. there is – 3509kg m² and 935.78kg m² = -2573.41kg m² Value of combined primary and secondary force at different angles is given in table Answer # 4 Graph of combined primary and secondary force is given below; Answer # 5 Thermal Efficiency For any system or device, thermal efficiency is defined as the ratio of desired output and required input. Mathematically it can be written as: For any heat engine thermal efficiency is defined as the total output work divided by the total input energy from the heat source (Venkanna & Swati, 2010: 177). Therefore thermal efficiency is defined in the most intuitive way as the ratio of work done by the engine to the heat energy provided to the engine. This can be expressed in mathematical way as, In the above equation the values of Qout and Qin are absolute. From the above equation it can be observed that the value of the thermal efficiency should always be less than 1. Thermal efficiency is a tool through which the measurement of efficient conversion of heat energy into work is observed. Through this thermal efficiency, the efficiency of the heat engine can be improved by the engineers, which cause less consumption of fuel and thus less pollution and lower fuel cost. Answer # 6 Answer # 7 Bearing Radius = 15 mm (0.015m) (Diameter 30 mm = 0.03) Clearance = 0.038 mm (0.000038 m) Length = 25 mm (0.025m) Bearing load = 4.5 kN (4500 N) SAE 30 @ 50°C = 0.045 Pa from the fig 1-1 RPM = 2500 RPM (41.7nrev / sec) Sommer Field number can be found by the following formula; (Fig 1-1) (SAE 30 @ 50° C) = 0.045 Pa The first step is finding “P” which is ratio of Bearing load and Projected Area Projected Area is the Area that the force is coming down on. Therefore, P= 3000000 Pa The second step is finding “S”. Therefore, S =155817 x 0.0000006255 = 0.09746 The third step is finding the length to Diameter, Ratio (L/D) Length of bearing = 0.025 m Diameter = 0.03 m Ratio = 0.8333333 . The fourth step is finding the temperature rise on chart, and Average Temperature S = 0.09746 L/D = 0.8333333 Temperature rises with oil going through bearing = 13 °C Change in Temp = temp rise 50 + (26/2) = 63° C Then the calculations need to be redone with the temperature at 63° C Figure 23-2a: Absolute viscosity versus temperature 155817 x 0.0000003197 = 0.0498 Therefore the value of is 0.0498 Answer # 8 Minimum Thickness of the Oil Film As found above, S = 0.0498 L/d = 0.833333 We go up the line 0.0498 to around 0.8333333 and then we can get /c and € value. € = 0.85 C (clearance) = 0.000038 0.00000646 m or 0.0646 mm Minimum thickness required is 0.0646 mm Actual clearance is 0.038 mm Answer # 9 S = 0.0724 L/D = 0.83333333 Flow Variable = 4.6 x Q / (r x c x N x I) Therefore oil required is given by using above equation, Q = 4.6 x 0.015 x 0.000038 x 41.7 x 0.025 = 2.7334×cumec / sec Q = 0.002733 Litters / second References Venkanna, B. K. & Swati B. V. (2010). Basic Thermodynamics. 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