che649-spring2010-exam1 - ChE 649 Spring 2010 Exam I 6 questions for a total of 120 points 1(10 points This is a problem of conduction through a

che649-spring2010-exam1 - ChE 649 Spring 2010 Exam I 6...

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Unformatted text preview: ChE 649, Spring 2010 Exam I 6 questions for a total of 120 points 1. (10 points) This is a problem of conduction through a composite cylinder. 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 R1 11111111111111111 00000000000000000 11111111111111111 00000000000000000 Ri 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 R2 11111111111111111 00000000000000000 11111111111111111 00000000000000000 1 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 2 11111111111111111 00000000000000000 11111111111111111 00000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 The temperature at Ri (inside the cylinder) is maintained at Ti while the temperature on the outside, radius R2 is maintained at T2 . As you can see, R1 is the radius of the first layer and R2 that of the second layer. At the interface between the first layer and the second layer you can assume that the temperatures and the flux are continuous. Let the length of the cylinder be L, k1 and k2 the thermal conductivities of layers 1 and 2 respectively. Show that the flux multiplied by the area for heat transfer from the inside to the outside is given by 2πL [Ti − T2 ] ln( R2 ) ln( R1 ) Ri [ R1 + k2 k1 2. (10 points) A pipe with an outer diameter of 4 cm, carrying steam is to be covered with two layers of insulation, each of which has a thickness of 2 cm. The thermal conductivity of one insulation material is 5 times that of the other insulation material. Assuming that the inner and the outer temperatures of the composite insulation are fixed, calculate the percent reduction in heat transfer when the better insulating material is next to the pipe than when it is the outer layer. I am aware that I have not given you the thermal conductivities or the temperatures 1 3. (10 points) Consider the insulation around a house. It has four layers, the first layer is sheet rock (one that contacts the inside of the house), the second layer is an air space, the third is celotex board and the fourth (facing the outside, the elements of weather) is brick. Information about these four layers is given in the table below. Name Sheet Rock Air Space Celotex Brick Thickness Thermal Conductivity (kg )(m) (in cm) (sec)3 (K ) 1.27 0.74421 8.89 1.99032 1.91 0.04673 10.2 0.6923 The inside wall surface temperature is 22.2 o C (72 o F)and the outside wall temperature is 1.7 o C (35 o F, cold day!) 1. Determine the rate of heat loss from the house per square meter. 2. Determine the temperature in the wall at a point 12 cm from the outside wall surface. 4. (30 points) Consider a binary flowing stream which consists of a water phase and an oil phase that is moving between two parallel plates. The flow rates at the inlet and the absolute viscosities of oil and water are 0.1 ft3 /sec/ft and .0.1 ft3 /sec/ft of width of the plate and 100 and .89 centipoises, respectively. If the distance between the plates is 2 in., find 1. the elevation of the oil-water interface, 2. the pressure gradient and 3. the velocity distribution. Assume the flow is steady, uniform and laminar. 2 5. (30 points) A system consists of two porous concentric spherical shells and a gas is blown outward radially from the inner shell to the outside through the outer shell. Express the rate of heat removal from the inner sphere as a function of the mass flow rate of gas for steady state laminar flow and low gas velocity. The following data is given. The radius of the outer shell: R The radius of inner shell: kR The temperature of inner surface of the outer shell: T1 The temperature of outer surface of the inner shell: Tk The temperature of gas: Tk (Tk < T1 ) 3 6. (30 points) A stenosis is a narrowing of a blood vessel or valve. Stenosis of a blood vessel arises during atherosclerosis and could occlude an artery, depriving the tissue downstream of oxygen. Further, the fluid shear stresses acting on the endothelial cells lining blood vessels may affect the expression of genes that regulate endothelial cell function. Such gene expression can influence whether the stenosis grows or not. Consider a symmetric stenosis as shown in the Figure. Assume that the velocity profile within the stenosis of radius Ri (z ) has the same shape as the profile outside the stenosis and is represented as vz (r) = vmax r2 1− 2 R Outside the stenosis, the radius equals Ro and the maximum velocity is constant. Within the stenosis, the radius of the fluid channel R(z) equals R(z ) = Ro 1 − 0.5 1 − 4 z L 2 1/2 The origin of the z axis is the midpoint of the stenosis. 1. Develop an expression for vmax in a stenosis in terms of the volumetric flow rate Q, z cylindrical tube of radius Ro , and distance along the stenosis L 2. Compute the shear stress acting on the surface of the stenosis (r = Ri ) at z = 0 relative to the value outside the stenosis. 4 ...
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