Question

clear handwriting , no typing , show all work please!

A steam engine runs at 150 r.p.m. Its turning moment diagram gave the following area measurements in mm? taken in order above and below the mean torque line: 500,- 250, 270, – 390, 190, – 340, 270, - 250 The scale for the turning moment is 1 mm 500 N-m, and for crank angle is 1mm = 5º. The fluctuation of speed is not to exceed + 1.5% of the mean, determine the cross-section of the rim of the flywheel assumed rectangular with axial dimension equal to 1.5 times the radial dimension. The hoop stress is limited to 3 MPa and the density of the material of the flywheel is 7500 kg/m3. [Ans. 222 mm; 148 mm]

Answer 1

Solution Nazan =N=150rpm W= 20N 60 => w = 15.70 radis T * 500 270 let us assume 290 190 d bc/ e Ea= E f 5 Trrean - 250 -250 -390 - 340 -> Ebs Et500 = E+500 Ec = E +500-250 = E +250 Ed= E +500-250 +270= 520+E Eę = E +500 - 250 7270 -390 = E+130 Ef = Et 500 -250 +270 -390 +190= E+320 Eg = E + 500-250 + 270-3907190-340 = E-20 En= E + 500 - 250 +270-390 +190-340 +270= E +250 Ei = E +500 -250 +270-390+190-340 7270-250=E Emax. = Eb = E +500 Emin. = Eg = E-20 DE= SE-Eman-Emin Eb-Eg = (E+500)–(E-20) DE=520 mm2 Now moment scale => imm=500N-m given Turning and Crank angle o=3 Imm=5° 1 mm= sxa = I rad 180° 36 Imm? - 500X1 O X7 36 1 mm² mm = 43.63 N-m=J DE = 520X43.63 J DE- 22689.28 J

change in energy of flywheel DER I w2Cs for Cs → coefficient of fluctuatim{ given £1.5% + 1.5% - 1,5 difference is 37 ( magnitude] 0.03 30% Cs= = 3 100 I = MR? for rim-flywheel. m = mass R- Radicy of rim R? we know that on v=W.R 3x10 N/m2 Oj = Hoop stress = 3 MPa = 8 = density of flywheel material = 7500 kg/m3 3x106 15.70 x R = 7500 R=1.273 m =) I = MR? = m(1.27372 I = 1.62 m Now putting all values in egn ☺ DE = I w2Cs => 22 689.28 = 1.62 m x(15.70)’x0.03 m= 1894.02 kg

mass = volume x density =) 1894.02 = Volume x 7500 Volume = 0.2525 m3 volume of the Rim-flywheel. = (29R)x thickness x wid tol height) Given that - axial dimension = 1.5X Radial dimension along the anis along the Radial direction width" height) thickness (t) width 1,5 x thickness (+) 0.2525 = 27X1,273 x 115 x 72 t=0.145 m t= 145 mm width = 1.58+ Width = 0.218 m (height) width= 218 mm dimensions found out to be 218 mm and 145 mm * So Given answer below the question is 222 mm and 148 mm whey not same t Reason Because of the spokes. Rim-type flywheel have Spokes in their design - with given data spokes mass can not be calculated , so mass what we found is not including spokes mass (less than Actual) thats why found dimensions are slighly lesser than given Answcy.