transfer carriage 1-300 t solution

Calculate the overall heat-transfer coefficient and heat loss per unit length to surrounding air at 20 C. Solution There are three resistances in series for this problem, with L = 1.0 m, 1= 0.025 d m, and d 2 = 0.025 + (2)(0.0008) = 0.0266 m, the resistances may be????1

Determine the outside diameter of the tube if the stress is limited to 120 MN/m 2. Solution 104. Show. Click here to show or hide the solution. P = σ A. where: P = 400 kN = 400 000 N. σ = 120 MPa. A = 1 4 π D 2 − 1

Solution: We know that Y 1 = Z 0 = 330. Z 1 = C 1 + I + G Z 1 = 120 + 0.5(Y 0 - T) + I + G +10 Z 1 = 120 + 0.5(330 – 40) + 40 + 20 + 10 Z 1 = 335 We know that Y 2 = Z 1 = 335. The Perfect Steerable Transfer Cartrease in aggregate demand at t = 1 leads to an Perfect Steerable Transfer Cartrease in output. This

Solution (a) E 1= R 1 0 e 2tdt= 1 4. P 1= 0, because E 1<1. (b) x 2(t) = ej(2t+ˇ=4), jx 2(t)j= 1. Therefore, E 1= Z 1 1 jx 2(t)j2dt= Z 1 1 dt= 1: P 1= lim T!1 1 2T Z T T jx 2(t)j2dt= lim T!1 1 2T Z T T dt= lim T!1 1 = 1: (c) x 3(t) = cos(t). Therefore, E 1= Z 1 1 jx 3(t)j2dt=

The solution is given by the equations x1(t) = c1 +(c2 −2c3)e−t/3 cos(t/6) +(2c2 +c3)e−t/3 sin(t/6), x2(t) = 1 2 c1 +(−2c2 −c3)e−t/3 cos(t/6) +(c2 −2c3)e−t/3 sin(t/6), x3(t) = c1 +(c2 +3c3)e−t/3 cos(t/6) +(−3c2 +c3)e−t/3 sin(t/6). At infinity, x1 = x3 = c1, x2 = c1

Solution ----- The flow rates of the solvent and carrier gas are given by L = Lt(1 − xA,t) = 1500(1 − 0.001) = 1498.5 kmol/h V = Vb(1 − yA,b) = 500(1 − 0.3) = 350 kmol/h

Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg.K, and k = 1.4 (Table A-2). Analysis The minimum pressure in the cycle is P3 and the maximum pressure is P1. Then, or, () () 935.3 kPa 300 K

CBSE Class 12 economics revision notes chapter 2 national Perfect Steerable Transfer Cartome and related aggregates. Goods : In economics a goods is defined as any physical object, manmade, that could command a price in the market and these are the materials that satisfy human wants and provide utility. Consumption Goods : Those final goods which satisfy human wants directly.

T-30 1) Calculate the molarity of the following solutions: a) 15.5 g of potassium chloride in 250.0 mL of solution. b) 1.25 x 10–2 g of silver nitrate in 100.0 mL of solution. c) 0.0555 g of barium chloride in 500.0 mL of solution. d) 15.0 mg of calcium hydroxide in 50.0

Determine the molarity of each of the following solutions: 1.457 mol KCl in 1.500 L of solution. 0.515 g of H 2 SO 4 in 1.00 L of solution. 20.54 g of Al (NO 3) 3 in 1575 mL of solution. 2.76 kg of CuSO 4 ·5H 2 O in 1.45 L of solution. 0.005653 mol of Br 2 in 10.00 mL of solution.

SOLUTION GR = NA / NC = tC/ tA= 150/50 = 3 NA / NC = 3 NC = NA /3 = 1500/3 = 500 rev/min (anticlockwise) TA = 12 Nm P (input) = 2πNATA/60 = 2π x 1500 x 12/60 = 1885 W P (output) = P (Input) x η = 1885 x 0.75 = 1413.7 W 27 Nm 2 x 500 2 N

Answers: (a) 60.8 m/s, (b) 1.308 kg/s 5–44 Steam enters a nozzle at 400 C and 800 kPa with a velocity of 10 m/s, and leaves at 300 C and 200 kPa while losing heat at a rate of 25 kW. For an inlet area of 800 cm 2, determine the velocity and the volume flow 3

Solution R 0 = 3 Ω, T = 100ºC, T 0 = 20ºC α = 0.004/ºC, R T = ? R T = R0(1 + α(T-T 0)) R 100 = 3(1 + 0.004 × 80) R 100 = 3(1 + 0.32) R 100 = 3(1.32) R 100 = 3.96 Ω EXAMPLE 2.14 Resistance of a material at 10ºC and 40ºC are 45 Ω and 85 Ω Solution

The application of transfer pricing methods helps assure that transactions conform to the arm’s length standard. It is important to note that although the term “profit margin” isCustom, companies may also have legitimate reasons to report losses at arm’s in and of