### Reynolds Number Calculations - HydroFlo Tech

A Newtonian fluid with a dynamic or absolute viscosity of 0.38 Ns/m 2 and a specific gravity of 0.91 flows through a 25 mm diameter pipe with a velocity of 2.6 m/s. The density can be calculated using the specific gravity like

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4) (243.40 82.90) For the condenser: kg kJ 5.78 kgK kJ 294 K(0.3113 1.0135) kgK kJ e e f (h h ) T o (s 3 s 2) (82.90 295.13) Comment: Although there is heat transfer to the refrigerant passing through the evaporator, the specific flow availability decreases. This can be explained by noting that

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7/26/2013· Leeward Roof -0.6 Foundation systems are considered rigid, therefore, G = 0.85. Design Pressure (p) The basic pressure equation (ASCE 7 6-17), which includes the internal pressure coefficient is as follows: p = qGC p – q i (GC pi) However, this would only .

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12.6 How does increasing temperature affect the parameters in the flow curve equation? Answer. Increasing temperature decreases both K and n in the flow curve equation. 12.7 Indicate some of the advantages of cold working relative to warm and hot working. Answer. Advantages of cold working are (1) better accuracy, (2) better surface finish, (3 ...

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6.02×1023 molecules mole ¶µ mole 18 g ¶ =3.34×1012 molecules g The density of water is 1 g/cm 3,so the number of molecules in a cm is 3.34× 1012.The volume of water that contains 104 molecules is Volume = 104 molecules 3.34×1012 molecules cm3 =3.0×10−19 cm3 Since the volume of a cube is L3,whereLis the length of a side L= 3 p 3.0×10 ...

Get price### CHAPTER 10 EXAMPLES & SOLUTIONS

4) (243.40 82.90) For the condenser: kg kJ 5.78 kgK kJ 294 K(0.3113 1.0135) kgK kJ e e f (h h ) T o (s 3 s 2) (82.90 295.13) Comment: Although there is heat transfer to the refrigerant passing through the evaporator, the specific flow availability decreases. This can be explained by noting that

Get price### Solution 12

Uo = 1621 Wm-2 °C-1 Ao = 3482.5 x 10 3 = 24.80 m2 1621 x 86.6 So the exchanger should be capable of fulfilling the duty required, providing the water in put through the shell. Note; the viscosity correction factor has been neglected when estimating the heat transfer

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Chapter 2 • Pressure Distribution in a Fluid 2.1 For the two-dimensional stress field in Fig. P2.1, let σxx yy==3000 psf 2000 psfσ σxy =500 psf Find the shear and normal stresses on plane AA cutting through at 30°. Solution: Make cut "AA" so that it just hits the bottom right corner of the element.

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5 P. 3) IMMEDIATE SETTLEMENT Question: A foundation 4 m 2 m, carrying a net uniform pressure of 200 kN/m2, is located at a depth of 1.5 m in a layer of clay 5 m thick for which the value of Eu is 45 MN/m 2.The layer is underlain by a second layer, 10 m thick, for which the value of Eu is 80 MN/m 2.

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The probability is .95 that demand during lead time will exactly equal the amount on hand at the beginning of lead time. Approximately 95 percent of demand during lead time will be satisfied. The probability is .95 that the order will arrive after the on-hand inventory is exhausted. A stockout only occurs if demand during lead time exceeds the ROP.

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• Floor live load = 40 psf • Floor dead load = 20 psf During construction, a 20-piece stack of 5/8 in. by 4'x 12' gypsum board is stored near the mid-span of two of the trusses with the long dimension parallel to the trusses. At the time, the trusses are sheathed with ¾ in. T&G plywood and

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CEE 345 Spring 2002 Problem set #3 Solutions Problem 4.7 A concrete-lined trapezoidal channel with bottom width of 10 ft and side slopes of 1 vertical to 2 horizontal is designed to carry a o w of 3000 cfs.

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Calculate the density of solid Ne (atomic mass = 20.18 g/mol). Solution a Let E = potential energy and x = distance variable. The energy E is given by Ex xx . .=− − 2 1445 1213 612 ε σσ The force F on each atom is given by Fx dE x dx x x x x () =− = .. − 2 145 56 86 7 11 2 5 ε σ σ σ σ ∴ Fx xx . .=− 2 145 56 86 7 12 13 6 .

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Ck10 steel cans of inner diameter 70 mm, wall thickness 5 mm, base thickness 6 mm and height of 120 mm (including the base thickness) will be manufactured using an axisymmetrical backward extrusion process. Annealed slugs used for the process have a diameter of 80 mm. The friction coefficient at the punch-workpiece and workpiece-die

Get price### Solution 12

Uo = 1621 Wm-2 °C-1 Ao = 3482.5 x 10 3 = 24.80 m2 1621 x 86.6 So the exchanger should be capable of fulfilling the duty required, providing the water in put through the shell. Note; the viscosity correction factor has been neglected when estimating the heat transfer

Get price### Student Solutions Manual to accompany - Prgmea

6.02×1023 molecules mole ¶µ mole 18 g ¶ =3.34×1012 molecules g The density of water is 1 g/cm 3,so the number of molecules in a cm is 3.34× 1012.The volume of water that contains 104 molecules is Volume = 104 molecules 3.34×1012 molecules cm3 =3.0×10−19 cm3 Since the volume of a cube is L3,whereLis the length of a side L= 3 p 3.0×10 ...

Get price### F. Example Calculations - FEMA.gov

7/26/2013· Leeward Roof -0.6 Foundation systems are considered rigid, therefore, G = 0.85. Design Pressure (p) The basic pressure equation (ASCE 7 6-17), which includes the internal pressure coefficient is as follows: p = qGC p – q i (GC pi) However, this would only .

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College of Engineering - Purdue University

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3/1/2006· Oil–water two-phase flow experiments were conducted in a 15 m long, 8.28 cm diameter, inclinable steel pipe using mineral oil (density of 830 kg/m 3 and viscosity of 7.5 mPa s) and brine (density of 1060 kg/m 3 and viscosity of 0.8 mPa s). Steady-state data on flow patterns, two-phase pressure gradient and holdup were obtained over the entire range of flow rates for pipe inclinations of .

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AREN 2110 SOLUTIONS FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8 SOLUTIONS: HOMEWORK #6 Chapter 5 Problems 5-45 A number of brass balls are to be quenched in a water bath at a specified rate. The rate at which heat needs to be removed from the water in order to keep its temperature constant is to be determined.

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23.6 What are the two basic methods of arc shielding? Answer. (1) Shielding gas, such as argon and helium; and (2) flux, which covers the welding operation and protects the molten pool from the atmosphere. 2 23.8 Describe the shielded metal arc-welding (SMAW) process. Answer. SMAW is an arc-welding process that uses a consumable electrode ...

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3–135 A rigid tank contains 6 kg of an ideal gas at 3 atm and 40°C. Now a valve is opened, and half of mass of the gas is allowed to escape. If the final pressure in the tank is 2.2 atm, the final temperature in the tank is (a) 186°C (b) 59° (c) -43°C (d) 20°C (e) 230°C Get 3.135 exercise solution

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40 225 125 60 As some concrete examples, first consider the cost of 192.5 in cell (1,2). This is the cost of ordering in period 1 the total demand for the first two periods, D1+D2. Hence the total cost is the ordering cost plus the inventory holding cost of carrying

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7/1/2008· High-strength Steel Bolt Capacity Calculator: Finds capacity of high-strength steel bolts stressed in shear. Systems: Reinforced Concrete Integrated T-beam, 1-way Slab, and Column Calculator : Finds required rebar area (or spacing) and selects bars for T-beams, 1-way slabs, and columns in a typical reinforced concrete building.

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This calc is mainly for pipes full with water at ambient temperature and under turbulent flow. If you know the slope rather than the pipe length and drop, then enter "1" in "Length" and enter the slope in "Drop". If the cunduit is not a full circular pipe, but you know the hydraulic radius, then enter (Rh×4) in "Diameter".

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Brainly is the knowledge-sharing community where 350 million students and experts put their heads together to crack their toughest homework questions.

Get price### Problem 4.2 Solution

CEE 345 Spring 2002 Problem set #3 Solutions Problem 4.7 A concrete-lined trapezoidal channel with bottom width of 10 ft and side slopes of 1 vertical to 2 horizontal is designed to carry a o w of 3000 cfs.

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0 40-= = m d ˜400 mm v. ... Solve Problem 4.9 if the 50-N load is replaced by an 80-N load. PROBLEM 4.9 The maximum allowable value of each of the reactions . is 180 N. Neglecting the weight of the beam, determine the range of the distance d for which the beam is safe. SOLUTION

Get price### Reynolds Number Calculations - HydroFlo Tech

A Newtonian fluid with a dynamic or absolute viscosity of 0.38 Ns/m 2 and a specific gravity of 0.91 flows through a 25 mm diameter pipe with a velocity of 2.6 m/s. The density can be calculated using the specific gravity like

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### Solution 12

Uo = 1621 Wm-2 °C-1 Ao = 3482.5 x 10 3 = 24.80 m2 1621 x 86.6 So the exchanger should be capable of fulfilling the duty required, providing the water in put through the shell. Note; the viscosity correction factor has been neglected when estimating the heat transfer

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associated network of piping systems. For example, a pipe standard might require that the leaching level of a heavy metal must not exceed 10% of the drinking-water standard based upon a standard test that simulates use and expo-sure conditions. Some authorities also insist that piping systems for soil and

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4/6/2013· P6.61 to deliver a flow rate of 0.015 ft3/sthrough the 1 -in commercial-steel pipe? 2 Fig. P6.61 35. Chapter 6 • Viscous Flow in Ducts 397Solution: For water at 20°C, take ρ = 1.94 slug/ft3 and µ = 2.09E−5 slug/ft⋅s.

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