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As described in Sec. PT3.1.2,
Where
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Chapter 11 Solutions
EBK NUMERICAL METHODS FOR ENGINEERS
- A closed thermodynamic system consists of a fixed amount of substance (i.e. mass) in which no substance can flow across the boundary, but energy can. For a closed themodynamic system we cannot add energy to the system, via substance (E ) (1.e. matter which contains energy is not allowed across the boundary) Across the Boundaries E° = No Q = = Yes W mass NO CLOSED = Yes SY STEM m = constant | energy YES Figure 1.1. If the substance inside the thermodynamic system shown in figure 1.1. (i.e. piston cylinder device) is air, is the system a Fixed closed system Moveable closed system A. В.arrow_forward3. Find the two-dimensional temperature distribution T(r,z) under steady state condition. Where, To=20 °C, TL=5200 °C. The density, conductivity and specific heat of the material are p 800 kg/m³, k=200 W/m.K, and cp=2500 J/kg.K, respectively. Also, r= 520 cm and L=1040cm. To TL L Toarrow_forwardExamples 2 PPL.pdf Homework Solve the following Liner Programming Problems (LPP) which to maximize profit. Use Graphical Method Ql: Max Z= 4X1 + 3X2 S.T 5X1+3X2arrow_forward
- First Order Differential Equations are inherent in almost all aspects of engineering, e.g., electronics (RC/RL circuits or charge/discharge of capacitors), thermodynamics (i.e., Newton’s Law of Cooling), mechanical systems (stress/strain) etc. In fact, virtually anywhere there are time varying dynamics. You need to demonstrate how different engineering systems models are used to solve them using first-order differential equations.arrow_forward23:38 Fri 22 Jul Q6. 4 of 6 Q6. Cont. library.qol.qub.ac.uk - Private Q7. An engineer is designing a pressure vessel, and selects an initial concept design consisting of a cylinder or length L, and radius r, both measured in metres, as shown in figure Q6. The volume of the vessel should be 1200 litres. The wall of the vessel will be a constant thickness throughout. The engineer wants to minimise the amount of material used in order to minimise the cost. 4% 1 [1 marks] Figure Q6 (a) Determine expressions for the surface area and the volume of the tank in L. Hence find an expression for surface area in terms of r only. Q6. Cont./ MEE1001/2021 rms of r and [4 marks] (b) Determine the combination of radius and length that will require the minimum amount of material to be used. You should use an appropriate method to verify that it is a minimum. [10 marks]arrow_forward*3.2 PQ1 A 100 mm diameter sphere contains an ideal gas at 20°C. Apply the grid method (p. 9) to calculate the density in units of kg/m³. a. Gas is helium. Gage pressure is 20 in H,O. b. Gas is methane. Vacuum pressure is 3 psi.arrow_forward
- 3. The relationship between arterial blood flow and blood pressure in a single artery satisfies the following first-order differential equation: dP(t) + dt RC mmHg (cm³/s) P(t) = where Qin is the volumetric blood flow, R is the peripheral resistance, and C is arterial compliance (all constant). Qin-60 cm³/s and the initial arterial pressure is 6 mmHg. Also, assume R = 4 and C= 0.4- Oin cm³ mmHg (a) Find the transient solution Ptran(t) for the arterial pressure. The unit for P(t) is mmHg. (b) Determine the steady-state solution Pss(t) for the arterial pressure. (c) Determine the total solution P(t) assuming that the initial arterial pressure is 0.arrow_forwardA generic property is being transported through a fluid of constant cross-sectional area at steady-state. The concentration of the property, IT, at point 1, is 0.015 / unit volume. The concentration of the property, T, at point 2 is 0.0075 / unit volume. Points 1 and 2 are 2 m apart. The constituitive property, ô, however, is not constant.... Rather it is a function of I according to: 8 = 0.15 + 1.65 I a.) Derive an integrated form of the equation for steady-state flux, yarrow_forwardis a mass hanging by a spring under the influence of gravity. The force due to gravity, Fg, is acting in the negative-y direction. The dynamic variable is y. On the left, the system is shown without spring deflection. On the right, at the beginning of an experiment, the mass is pushed upward (positive-y direction) by an amount y₁. The gravitational constant g, is 9.81 m/s². DO C.D Frontly у Your tasks: No Deflection m k Fg = mg Initial Condition y m k Write down an expression for the total energy If as the sum Write down an expression for the total energy H Fg = mg Figure 3: System schematic for Problem 4. Yi & X Write down, in terms of the variables given, the total potential energy stored in the system when it is held in the initial condition, relative to the system with no deflection. as the sum of potential and kinetic energy in terms of y, y, yi C After the system is released, it will start to move. Write down an expression for the kinetic energy of the system, T, in terms of…arrow_forward
- A hot water leak in one of the faucets of your apartment can be very wasteful. A continuous leak of one quart per hour (a "slow" leak) at 155 degrees°F causes a loss of 1.81 million Btu per year. Suppose your water is heated with electricity. a. How many pounds of coal delivered to your electric utility does this leak equate to if one pound of coal contains 12,000 Btu and the boiler combustion process and water distribution system have an overall efficiency of 28%? b. If a pound of coal produces 1.81 pounds of CO2 during the combustion process, how much extra carbon dioxide does the leaky faucet produce in a yeararrow_forwardCakulate the time rate of change of air density during expiration Assume that the lung (Fig. 3.11) has a total volume of 6000 ml, the diameter of the trachea is 18 mm, the airflow velocity out of the trachea is 20 cm/s, and the density of air is 1.225 kg/m. Also assume that lung volume is decreasing at a rate of 100 mL/s. Hello sir, I want the same solution, but in a detailed way and mention his data, a question, and a solution in detailing mathematics without words. Solution We will start from Eq. (3.24) because we are asked for the time rate of change of density. We are asked to find the time rate of change of air density; this suggests that Example 3.5 condis tions are representing a nonsteady flow scenario. In addition, we were told what the rate of change in the lung volume is during this procedure, further supporting the use of Eq. (3.24). pdV+ (3.24 ams Assume that at the instant in time that we are measuring the system, density is uniform within the volume of interest. This…arrow_forwardfluid mechanics A pipe 200 (mm) diameter carries Oil at a flow rate of 0.030 (m³/s). The pipe diameter reduces from 200 (mm) to 150 (mm). Point 1 is located at the beginning of the pipe and point 2 is located at the end of the pipe. Elevation of point 1 is 165 (m) lower than elevation of point 2. Water pressure at point 2 is atmospheric pressure. Water flow in the pipe ascending from point 1 to point 2. Total head losses of flow in the pipe equals to 15 (m). 1- Find the value of pressure head of Oil at point 1. 2- Draw the H.G.L. of flow in the pipe.arrow_forward
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