ĐP P2 R₂ 92 (+) P₁ 71(+) A 24 Ps ل A ₂ 73(+) P₁₁ = ATM
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Get the electrical diagram (Voltage and current) from this pressure system
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- Taking U to be a function of P and V, derive the following equations: (), [), (a) đ = dP+ + P|dV. V Сук (b) V Cp Р. VB (c) - avThe shaft is supported by a smooth thrust bearing at A and a smooth journal bearing at B. (Figure 1) -0.8 m 600 N-m -0.8 m -0.8 m 900 N Draw the shear diagram for the shaft. Draw the moment diagram for the shaft.10 / 36 110% LQ23]( problem 지25's p37) Exepe20-원 At What angle &' must the (400N) force be applied in order to that the resultant `R of the two 400 N forces has a of (1000 N)? magnitude What will be the angle o between 700 N R and the horizontal (x-axis) ? Ane ∞ = 51.3° 0 = 18 19° P Type here to search
- Anaerobic treatment is a common treatment process in environmental engineering to degrade organic contaminants. During this process, methane (CHA) is produced which is an explosive gas. So, its pressure in the treatment tank has to be maintained below explosion limit. Question: You want to keep the methane pressure in the tank below 400 kPa, so what should be the maximum pressure you measured at point A? Methane Mercury (sg = 13,6) 8 cm 1 cm A 3 cm 11 cm water | 4 cm Oil (sg = 0,8)M-(1zy.z- ly.y) + My,(1;.z – Izy -y) Pr My Formüller; a =- I.ly - lzy Pr Ox + Oy o - Gạ0p + o < ož , Jort =: %3D •Imaks = 2t 2 Oz + Øy. + t, , εαΔΤ, τ, Tc d²y M(x) Omaks.min 2 I dx² EI MalanıcoXe, Opc EI Malanıpc El Tmaks = ƏM -dx. Pgr ӘР VQ It y = L/2- -L/2 - AB beam is a W section drawn profile, P= 50 kN, L= 1,25 m ve E= 200 GPa since it is ( Iw= 26,9.106 mmª) b) Find the slope at A. a) Find the deflection at C. 0,0015 rad 0,886 mm 0,0012 rad 1,859 mm 0,0007 rad 0,554 mm 0,0009 rad 1,246 mm 0,0005 rad 0,378 mmCp = Cv + R Prove?? %3D
- kv² 1. The discharge flowing down the pipe is 0.03 m³/s and the pressure at 2 is 0.14 kg/cm² greater than at 1. Assuming the losses in the pipe between 1 and 2 is find the value of k. Liquid flowing in the pipe is water. } 2g 20 cm dia 10 cm dia mercury 3 mThere is a heated continuous stirred tank reactor shown below where the density is constant through out the system. On the other hand the volume of reaction mixture may vary. The thermal capacitance of the coil used to heat the tank is assumed to be negligible. There is a 2nd order elementary chemical reaction 2A → C taking place in the reactor. Cai Tai qa 2A ->C v „ro Tc Ac In the figure shown above Cai, Tai,qa is the inlet concentration of component A , inlet stream temperature and inlet volumetric flow rate, V is the volume of the reaction mixture, ro is the density through out the system. Tc, Ac and U are the coil temperature, surface area of the coil and heat transfer coefficient of the coil. Cc, T and Ca are the concentration of product C, temperature of the reactor and exit concentration of component A, qc is the outlet volumetric flowrate. The rate constant is assumed to be independent of temperature denoted as ko. The heat of reaction is denoted as AHrxn. Cp is the heat…What is the surface tension of benzene at 537-degree Rankine, if Y = 31.58 – 0.137T + 0.0001T^2? * O 28.2175 dynes/cm O 27.8224 dynes/cm O 21.6239 dynes/cm O No answer from the choices
- Q1:360 liters per second of water is flowing in a pipe. The pipe is bent by 120°.The pipe bend measures 360 mm x 240 mm and volume of the bend is 0.14 m3. The pressure at the entrance is 73 kN/m and the exit is 2.4 m above the entrance section. Find the force exerted on the bend. O 360 mm 240 mm 120 だい →バ 240 inin diaExhaust gas leaves the exhaust pipe of a motor car with a pressure of 1.2bar and a temperature of 400 degree C. If the exhaust gas specific was 1.6653 m^3per kg, determine the identity of the exhaust gas.The horizontal force due to water acting on the bolt (Fh): Fh=ρ×A1×V1×V1--V2 Fh=1000×π4×0.0752×1.254×1.254+11.286( Fh)=69.472 This is the only part I did not understand how to get. Can you please explain more? I arrived with this equation, -P1A1 - P2A2 (cancelled since atmos. Pres.) +Fh = rho A1v1 (v2-v1) Why did P1A1 disappear in your solution? How did you come up with (v1 - - v2) ? What is the control volume? Why did you not do a summation of forces is equal to change in momentum but instead go immediately to getting the horizontal force acting on the bolt?