The mass flow rate of air through the cycle

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35. The double-reduction, helical gear reducer shown in Figure P12-35 transmits 5.0 hp. Shaft 1 is the input, rotating at 1800 rpm and receiving power directly from an electric motor through a flexible coupling. Shaft 2 rotates at 900 rpm. Shaft 3 is the output, rotating at 300 rpm. A chain sprocket is mounted on the output shaft as shown and delivers the power upward. The data for the gears are given in Table 12-5. Each gear has a 1412° normal pressure angle and a 45° helix angle. The combinations of left- and right-hand helixes are arranged so that the axial forces oppose each other on shaft 2 as shown. Use SAE 4140 OQT 1200 for the shafts. Figure P12-35: P35-Double-reduction helical drive Each problem includes the following details: ■Design the complete shaft, including the specification of the overall geometry and the consideration of stress con- centration factors. The analysis would show the minimum acceptable diameter at each point on the shaft to be safe from the standpoint of…

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Chapter 9.12, Problem 82P

To determine

The mass flow rate of air through the cycle

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35. The double-reduction, helical gear reducer shown in Figure P12-35 transmits 5.0 hp. Shaft 1 is the input, rotating at 1800 rpm and receiving power directly from an electric motor through a flexible coupling. Shaft 2 rotates at 900 rpm. Shaft 3 is the output, rotating at 300 rpm. A chain sprocket is mounted on the output shaft as shown and delivers the power upward. The data for the gears are given in Table 12-5. Each gear has a 1412° normal pressure angle and a 45° helix angle. The combinations of left- and right-hand helixes are arranged so that the axial forces oppose each other on shaft 2 as shown. Use SAE 4140 OQT 1200 for the shafts. Figure P12-35: P35-Double-reduction helical drive Each problem includes the following details: ■Design the complete shaft, including the specification of the overall geometry and the consideration of stress con- centration factors. The analysis would show the minimum acceptable diameter at each point on the shaft to be safe from the standpoint of…

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Bài 1. Cho cơ hệ như hình 1. Hình biểu diễn lược đổ cơ hệ tại vị trí cân bằng tĩnh. Trục tọa độ Oy hướng theo phương chuyển động của vật 1, gốc O đặt tại vị trí cân bằng của vật 1(tức khi lò xo biến dạng tĩnh). Bỏ qua khối lượng của thanh số 3. Vật rắn 2 là pulley 2 tầng đồng chất có bán kính ngoài 21, bán kính trong I, bán kính quán tính đối với trục qua tâm P-1.5, khối lượng m:. Vật rắn 4 là thanh thắng đồng chất có khối lượng m, chiều dài 1. Cho các số liệu: m = 2kg, m= = 5kg, m = 4kg, k=40(N/cm), ! – 0.8(m),r=0.1(m). Điều kiện đầu y; =0.5 cm );j = 10 cm/s) . Giả sử hệ dao động bé, Vật rắn 2 chuyển động lăn không trượt trên mặt phẳng ngang. 1. Viết phương trình chuyển động của hệ. 2. Xác định tần số dao động tự do của hệ. 3. Xác định đáp ứng dao động tự do của hệ. dây dây 1 2r Hình 1 y 3 -2 I k www. -2