Basics of Agricultural Equipment Technology PDF

[Type text] Faculty of Industrial and Energy Technology Program: Technology of Tractors and Agricultural Equipment. Level: 1 Basics of Agricultural Equipment Technology Faculty of Industrial and Energy Technology Program: Technology of...

[Type text] Faculty of Industrial and Energy Technology Program: Technology of Tractors and Agricultural Equipment. Level: 1 Basics of Agricultural Equipment Technology Faculty of Industrial and Energy Technology Program: Technology of Tractors and Agricultural Equipment. Level: 1 Course title: Basics of Agricultural Equipment Technology Section (1) Agricultural machinery is classified according to its use into the following categories:  Land preparation for agriculture machines ‫آالت تهيئة األرض للزراعة‬  Crop planting and seeding machines ‫آال ت الزراعة ووضع البذرة‬  Crop maintenance ‫آالت خدمة المحصول النامي‬  Harvesting machines ‫آالت الحصاد والدراس‬  Animal production machinery ‫آالت مزارع االنتاج الحيواني‬ (1)Land preparation for agriculture machines  Plows and their various types.  Harrows and their various types.  Leveling machines  Furrowing machines ‫ سالح‬9 ‫المكونات األساسية لمحراث حفار معلق‬ ‫ بدن‬2 ‫المكونات األساسية لمحراث قالب مطرحي‬ 1 Faculty of Industrial and Energy Technology Program: Technology of Tractors and Agricultural Equipment. Level: 1 Course title: Basics of Agricultural Equipment Technology ‫ بدن‬3 ‫المكونات األساسية لمحراث قرصى‬ ‫األشكال المختلفة للمطارح واستخدامها‬ ‫معلق‬ Ex(1): A 40-horsepower (Hp) tractor shaft pulls a seven Chisel Plows shanks. Find the tensile force if the plowing speed (S) is 7.2 km/h? ‫ أوجد قوه الشد اذا‬.‫ حصان يقوم بجر محراث حفار ذي سبعه أسلحة‬40 ‫ جرار قدرته على عمود جرار‬: )1( ‫مثال‬ ‫ ساعة ؟‬/‫ كيلو متر‬7.2 ‫كانت سرعه الحرث‬ Answer 𝐦 𝐦 𝐏𝐨𝐰𝐞𝐫 𝐤𝐠. = 𝐓𝐞𝐧𝐬𝐢𝐥𝐞 𝐅𝐨𝐫𝐜𝐞 𝐤𝐠 × 𝐒𝐩𝐞𝐞𝐝 1hp = 75 kg.m/ sec 𝐬 𝐬 𝟕.𝟐∗𝟏𝟎𝟎𝟎 Tensile Force = 40* 75 = Tensile Force × Plowing Resistance 𝟔𝟎∗𝟔𝟎 3000 = Tensile strength × 2 𝟑𝟎𝟎𝟎 𝐓𝐞𝐧𝐬𝐢𝐥𝐞 𝐬𝐭𝐫𝐞𝐧𝐠𝐭𝐡 = = 𝟏𝟓𝟎𝟎 𝐤𝐠 𝟐 2 Faculty of Industrial and Energy Technology Program: Technology of Tractors and Agricultural Equipment. Level: 1 Course title: Basics of Agricultural Equipment Technology Ex(2): What is the number of shanks in the Chisel Plows suitable for use with a 30- horsepower (hp) tractor, since the distance between the shanks (d) is 50 cm, the specific resistance (r) is 0.7 kg/cm2, the plowing depth (pd) is 15 cm, and the plowing speed (s) 3 km/h. ‫ حيث أن المسافة بين‬، ‫ حصان‬30 ‫ ما عدد األسلحة في المحراث الحفار المناسب لالستخدام مع جرار‬: )2( ‫مثال‬. ‫ ساعة‬/‫ كم‬3 = ‫ وسرعة الحرث‬، ‫ سم‬15 ‫ وعمق الحرث‬2‫سم‬/‫ كجم‬0.7 ‫ والمقاومة النوعية‬، ‫ سم‬50 ‫األسلحة‬ Answer 𝐓𝐞𝐧𝐬𝐢𝐥𝐞 𝐅𝐨𝐫𝐜𝐞 𝐤𝐠 × 𝐒𝐩𝐞𝐞𝐝 𝐦/𝐬 𝐇𝐨𝐫𝐬𝐞𝐩𝐨𝐰𝐞𝐫 𝐡𝐩 = 𝟕𝟓 𝐓𝐞𝐧𝐬𝐢𝐥𝐞 𝐅𝐨𝐫𝐜𝐞 × 𝟑 × 𝟏𝟎𝟎𝟎 𝟑𝟎 𝐡𝐩 = 𝟔𝟎 × 𝟔𝟎 × 𝟕𝟓 𝟑𝟎∗ 𝟔𝟎 ∗𝟔𝟎∗𝟕𝟓 𝐓𝐞𝐧𝐬𝐢𝐥𝐞 𝐅𝐨𝐫𝐜𝐞 𝐏𝐥𝐨𝐰𝐢𝐧𝐠 𝐫𝐞𝐬𝐢𝐬𝐭𝐚𝐧𝐜𝐞 = = 2700 kg 𝟑∗𝟏𝟎𝟎𝟎 Plowing resistance = plowing cross-sectional area × specific resistance 2700 kg = plowing cross-sectional area × 0.7 kg / cm2 𝟐𝟕𝟎𝟎 𝐏𝐥𝐨𝐰𝐢𝐧𝐠 𝐜𝐫𝐨𝐬𝐬 − 𝐬𝐞𝐜𝐭𝐢𝐨𝐧𝐚𝐥 𝐚𝐫𝐞𝐚 = = 𝟑𝟖𝟓𝟕 𝐜𝐦𝟐 𝟎. 𝟕 Plowing cross-sectional area = Plowing width × Plowing depth 3857 cm2 = plowing width × 15 cm 𝟑𝟖𝟓𝟕 𝐏𝐥𝐨𝐰𝐢𝐧𝐠 𝐰𝐢𝐝𝐭𝐡 = = 𝟐𝟓𝟕 𝐜𝐦 𝟏𝟓 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐬𝐡𝐚𝐧𝐤𝐬 𝐏𝐥𝐨𝐰𝐢𝐧𝐠 𝐰𝐢𝐝𝐭𝐡 = × distance between 𝐬𝐡𝐚𝐧𝐤𝐬 𝟐 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐬𝐡𝐚𝐧𝐤𝐬 𝟐𝟓𝟕 = × 50 𝟐 𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐬𝐡𝐚𝐧𝐤𝐬 = 𝟏𝟎. 𝟐𝟖 So, it selected to be 9 OR 11 shanks 3 Faculty of Industrial and Energy Technology Program: Technology of Tractors and Agricultural Equipment. Level: 1 Course title: Basics of Agricultural Equipment Technology Ex(3): Find the performance rate ( P) of a Chisel Plows that works with a 50 hp (HP) tractor, the plowing depth was (pd) 25 cm, the tractor speed was (S) 7 km / h, the specific soil resistance (r) was 0.7 kg / cm2, and the field efficiency (E) was 75%. ‫ سم‬25 ‫ حصان علما بأن عمق الحرث‬50 ‫ أوجد معدل أداء المحراث المطرحي الذي يعمل مع جرار‬: )3( ‫مثال‬.%75 ‫ والكفاءة الحقلية‬2‫سم‬/‫ كجم‬0.7 ‫ ساعه والمقاومة النوعية للتربة‬/‫ كم‬7 ‫وسرعه الجرار‬ Answer 𝐓𝐞𝐧𝐬𝐢𝐥𝐞 𝐅𝐨𝐫𝐜𝐞 𝐤𝐠 × 𝐒𝐩𝐞𝐞𝐝 𝐦/𝐬 𝐇𝐨𝐫𝐬𝐞𝐩𝐨𝐰𝐞𝐫 𝐡𝐩 = 𝟕𝟓 𝐓𝐞𝐧𝐬𝐢𝐥𝐞 𝐅𝐨𝐫𝐜𝐞 × 𝟕 × 𝟏𝟎𝟎𝟎 𝟓𝟎 𝐡𝐩 = 𝟔𝟎 × 𝟔𝟎 × 𝟕𝟓 𝟓𝟎 × 𝟔𝟎 × 𝟔𝟎 × 𝟕𝟓 𝐓𝐞𝐧𝐬𝐢𝐥𝐞 𝐅𝐨𝐫𝐜𝐞 = 𝟕 × 𝟏𝟎𝟎𝟎 𝐓𝐞𝐧𝐬𝐢𝐥𝐞 𝐅𝐨𝐫𝐜𝐞 = 𝐩𝐥𝐨𝐰𝐢𝐧𝐠 𝐫𝐞𝐬𝐢𝐬𝐭𝐚𝐧𝐜𝐞 = 𝟏𝟗𝟐𝟖. 𝟓 𝐤𝐠 Plowing resistance = Plowing cross-sectional area × specific resistance 1928.5 = Plowing cross-sectional area × 0.7 Plowing cross-sectional area = 1928.5 / 0.7 = 2755 cm2 Plowing cross-sectional area = Plowing width × Plowing depth 2755 = plowing width × 25 Plowing width = 2755 / 25 = 110.2 cm = 1.1 m 𝐬𝐩𝐞𝐞𝐝 × 𝐰𝐢𝐝𝐭𝐡 𝐏𝐥𝐨𝐰 𝐩𝐞𝐫𝐟𝐨𝐫𝐦𝐚𝐧𝐜𝐞 𝐫𝐚𝐭𝐞 = × field efficiency 𝟒𝟐𝟎𝟎 𝟕 × 𝟏𝟎𝟎𝟎 × 𝟏.𝟏 Plow performance rate = × 75% 𝟒𝟐𝟎𝟎 𝐏𝐥𝐨𝐰 𝐩𝐞𝐫𝐟𝐨𝐫𝐦𝐚𝐧𝐜𝐞 𝐫𝐚𝐭𝐞 = 𝟏. 𝟑 𝐟𝐞𝐝 / 𝐡𝐨𝐮𝐫 4 Faculty of Industrial and Energy Technology Program: Technology of Tractors and Agricultural Equipment. Level: 1 Course title: Basics of Agricultural Equipment Technology Ex(4): What is the plowing resistance and horse power required for a 9 Chisel plow, the distance between the shanks (d) is 50 cm, the plowing depth (pd) is 15 cm, the speed (s) is 8 km/h, and the specific resistance of the soil (r) is 0.8 kg/cm2. ‫ ما هي مقاومة الحرث والقدرة المطلوبة بالحصان لمحراث حفار زي تسعة اسلحة في أرض‬: )4( ‫مثال‬ ‫ ساعة والمقاومة النوعية‬/‫ كم‬8 ‫ وسرعة‬، ‫ سم‬15 ‫ سم وعمق الحرث‬50 ‫متوسطة والمسافة بين األسلحة‬. 2‫سم‬/ ‫ كجم‬0.8 ‫للتربة‬ Answer 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐬𝐡𝐚𝐧𝐤𝐬 𝐏𝐥𝐨𝐰𝐢𝐧𝐠 𝐰𝐢𝐝𝐭𝐡 = × distance between 𝐬𝐡𝐚𝐧𝐤𝐬 𝟐 𝟗 𝐏𝐥𝐨𝐰𝐢𝐧𝐠 𝐰𝐢𝐝𝐭𝐡 = × 50 = 225 cm 𝟐 Plowing cross-sectional area = Plowing width * Plowing depth Plowing cross-sectional area = 225 × 15 = 3375 cm2 Plowing resistance = Plowing cross-sectional area × specific resistance Plowing resistance = 3375 cm2 × 0.8 kg/cm2 = 2700 kg 𝐓𝐞𝐧𝐬𝐢𝐥𝐞 𝐅𝐨𝐫𝐜𝐞 𝐤𝐠 × 𝐒𝐩𝐞𝐞𝐝 𝐦/𝐬 𝐇𝐨𝐫𝐬𝐞𝐩𝐨𝐰𝐞𝐫 𝐡𝐩 = 𝟕𝟓 𝟐𝟕𝟎𝟎 × 𝟖 × 𝟏𝟎𝟎𝟎 𝐇𝐨𝐫𝐬𝐞𝐩𝐨𝐰𝐞𝐫 = = 𝟖𝟎 𝐡𝐩 𝟔𝟎 × 𝟔𝟎 × 𝟕𝟓 5 Faculty of Industrial and Energy Technology Program: Technology of Tractors and Agricultural Equipment. Level: 1 Course title: Basics of Agricultural Equipment Technology Ex(5): Find the power needed to pull a 3 Moldboard in moldboard plows, with width (w) 40 cm, plowing depth (pd) 15 cm, at a speed (s) of 5 km/h and specific resistance (r) 0.8 kg/cm2 and what is the average performance (P) of the plow per hour if the field efficiency (E) is 80%, then calculate the final needed power for the tractor if transmission coefficient 70%. ‫ سم وعمق الحرث‬40 ‫عرض البدن‬، ‫ أوجد القدرة الالزمة لجر محراث مطروحي ذو ثالثة ابدان‬: )5( ‫مثال‬ ‫ وما هو معدل اداء المحراث في‬، 2‫سم‬/‫ كجم‬0.8 ‫ ساعه و المقاومة النوعية‬/‫ كم‬5 ‫ سم على سرعه‬15 = ‫ وايضا احسب القدرة النهائية مع اخذ في االعتبار معامل النقل‬، %80 ‫الساعة اذا كانت الكفاءة الحقلية‬. 0.7 Answer Plowing width = number of moldboards × width of it Plowing width = 3 * 40 = 120 cm Plowing cross-sectional area = Plowing width × Plowing depth Plowing cross-sectional area = 120 × 15 = 1800 cm2 Plowing resistance = Plowing cross-sectional area × specific resistance Plowing resistance = 1800 × 0.8 = 1440 kg 𝐓𝐞𝐧𝐬𝐢𝐥𝐞 𝐅𝐨𝐫𝐜𝐞 𝐤𝐠 × 𝐒𝐩𝐞𝐞𝐝 𝐦/𝐬 𝐇𝐨𝐫𝐬𝐞𝐩𝐨𝐰𝐞𝐫 𝐡𝐩 = 𝟕𝟓 𝟏𝟒𝟒𝟎 × 𝟓 × 𝟏𝟎𝟎𝟎 𝐏𝐨𝐰𝐞𝐫 = = 𝟐𝟔. 𝟔 𝐡𝐩 𝟔𝟎 × 𝟔𝟎 × 𝟕𝟓 𝐬𝐩𝐞𝐞𝐝 × 𝐰𝐢𝐝𝐭𝐡 𝐏𝐥𝐨𝐰 𝐩𝐞𝐫𝐟𝐨𝐫𝐦𝐚𝐧𝐜𝐞 𝐫𝐚𝐭𝐞 = × field efficiency 𝟒𝟐𝟎𝟎 𝟓 × 𝟏𝟎𝟎𝟎 × 𝟏.𝟐 𝐏𝐥𝐨𝐰 𝐩𝐞𝐫𝐟𝐨𝐫𝐦𝐚𝐧𝐜𝐞 𝐫𝐚𝐭𝐞 = × 80% = 𝟏. 𝟏 𝐟𝐞𝐝 / 𝐡𝐨𝐮𝐫 𝟒𝟐𝟎𝟎 Actual Power = Calculated Power / Transmission Factor Actual Power = 26.6 / 0.7 = 38 hp 6 ‫‪Faculty of Industrial and Energy Technology‬‬ ‫‪Program: Technology of Tractors and Agricultural Equipment.‬‬ ‫‪Level: 1‬‬ ‫‪Course title: Basics of Agricultural Equipment Technology‬‬ ‫الخطوات المتبعة لحساب القدرة الالزمة لعملية الحرث‬ ‫‪.1‬حساب عرض الحرث (سم)‬ ‫=‬ ‫‪× distance between‬‬ ‫عدد األسلح ة‬ ‫* المسافة بين األسلحة‬ ‫عرض الحرث ( للمحراث الحفار) =‬ ‫𝟐‬ ‫‪Plowing width = number of moldboards × width of it‬‬ ‫عرض الحرث ( المطرحى) = عدد األبدان * عرض البدن‬ ‫‪.2‬حساب مساحة مقطع الحرث‬ ‫‪Plowing cross-sectional area = Plowing width × Plowing depth‬‬ ‫مساحة مقطع الحرث = عرض الحرث * عمق الحرث‬ ‫‪.3‬حساب المقاومة التي يواجهها المحراث من التربة (كجم)‬ ‫‪Plowing resistance = plowing cross-sectional area × specific resistance‬‬ ‫مقاومة الحرث = مساحة مقطع الحرث* المقاومة النوعية‬ ‫‪.4‬حساب القدرة االزمة لعملية الحرث ( بالحصان)‬ ‫×‬ ‫‪/‬‬ ‫=‬ ‫مقاومة الحرث∗ السرعة‬ ‫القدرة المطلوبة =‬ ‫𝐡𝐭𝐝𝐢𝐰 × 𝐝𝐞𝐞𝐩𝐬‬ ‫= 𝐞𝐭𝐚𝐫 𝐞𝐜𝐧𝐚𝐦𝐫𝐨𝐟𝐫𝐞𝐩 𝐰𝐨𝐥𝐏‬ ‫‪× field efficiency‬‬ ‫𝟎𝟎𝟐𝟒‬ ‫العرض ∗ السرعة‬ ‫* الكفاءة الحقلية‬ ‫معدل أداء المحراث =‬ ‫𝟎𝟎𝟐𝟒‬ ‫‪7‬‬ Faculty of Industrial and Energy Technology Program: Technology of Tractors and Agricultural Equipment. Level: 1 Course title: Basics of Agricultural Equipment Technology Tensile Force ( F) ‫قوة الشد‬ Plowing Resistance (Pr) ‫مقاومة الحرث‬ plowing speed (s) ‫سرعة الحرث‬ number of shanks (n) ‫عدد األسلحة‬ specific resistance (r) ‫المقاومة النوعية‬ plowing depth (pd) ‫عمق الحرث‬ distance between the shanks (d) ‫المسافة بين األسلحة‬ Plowing width (w) ‫عرض الحرث‬ Plowing cross-sectional area (A) ‫مساحة مقطع الحرث‬ field efficiency (E) ‫الكفاءة الحقلية‬ performance rate ( P) ‫معدل األداء‬ 8