高等傳熱學課程作業(yè)1

1、高等傳熱學 課程作業(yè)Homework 5.21. Slug flow in a tube(u(r) = V) with a fully developed temperature profile:- Constant heat flux- What is T(r)?·Remember dTm/dx = constant·Use dT/dr = 0 at r = 0 and Tw at r = R to get ingration constants- What is Nu?A: 根據(jù)能量方程， (1)對于充分發(fā)展的管內(nèi)彈狀流，u=0, 且由于 = constant，因此(1

2、)式可化簡為： (2)解之可得， (3)根據(jù)邊界條件，代入(3)中，得到C1=0, 平均溫度因此，得到2. Fully developed Poiseuille flow between parallel plates- Constant heat flux- Top and bottom at the same temperature- Neglect viscous dissipation- What is T(y)?·Remember dTm/dx = constant·Use Tw at y = 0 and y = h to get integration cons

3、tants- What is Nu?A: 根據(jù)能量方程， (1)對于x方向充分發(fā)展的Poiseuille流，v=w=0由于 = constant，(1)式可化為： (2)根據(jù)，且，代入得到解得溫度分布為 (3)根據(jù)Poiseuille流邊界條件：帶入(3)式中解得，, 因此，得到溫度分布為根據(jù) Nu Homework 5.31. Find the equation for the boundary layer thickness and for Cf for:Compare to the exact values and the fourth order equation solutions.

4、A: 對于，根據(jù)邊界條件， a = 0， 1 = b + c + d對于平板，有， c = 0根據(jù)， b + 3d = 0由此可得，a = 0, b = 1.5, c = 0, d = -0.5.對于U = constant，積分后得到，與精確解相差7.2%，與4級近似解相差20.5%，且與精確解相差2.6%，與4級近似解相差5.5%2. Derive the ordinary differential equation and the boundary conditions for the Blasius solution energy equation for flow over a fl

5、at plate. (See pages 29-30)A: 根據(jù)能量方程，常壁溫邊界條件：根據(jù)Blasius solution 代入能量方程中得到，即 邊界條件為3. Start from the local friction coefficient for flow over a flat plate, Cf, on slide 31 and derive the average friction coefficient over the entire plate, CL. Show your work.A: 4. A projectile in the form of a bluff-en

6、ded cylinder 20 cm in diameter and 60 cm long, moves through the air in the direction of its long axis at a velocity U of 100 m/s. The drag coefficient, C, is equal to 1.0 for this object. Frontal area=pD2/4, FD= total dragAssuming that the boundary layer thickness over the cylindrical surface of th

7、e projectile at a distance x from the leading edge is given by:and that the momentum thickness of the boundary layer is given by: Find what proportion of the total drag on the projectile is attributable to skin friction over the curved surface, assuming no pressure gradient in the boundary layer in

8、the streamline direction. Data. Air kinematic viscosity is 0.15 cm2/sAir density is 1.15 Kg/m3Assumptions: Assume this is like a flat plate, then use the equation on slide 15. A: 若將圓柱表面看成平板，U = constant, ，則因為12.16N180.64N6.7%Homework 5.41. Derive the equation relating q” to the temperature differenc

9、e for natural convection driven flow in a round tube (like on slide 19 for parallel plates).- water properties at 25 C- tube length = 100 cm- tube diameter = 1 mm- constant heat flux- fully developed flow in a tube (from the first homework)a) Plot DT and Q (m3/s) versus heat flux for fluxes of 500 t

10、o 5000 W/m2 b) Is the assumption of fully developed flow valid?- hint: is L/(D Re) > 0.05c) Is the Boussinesq approximation valid here?A: 圓管內(nèi)充分發(fā)展的流體，其速度分布為平均速度體積流量為根據(jù)能量平衡，即 將代入，得到，a) 根據(jù)以上推導過程可以得到，分別作出當時與Q隨q變化的曲線，如下圖所示。b) 125.4 > 0.05 滿足充分發(fā)展的假設(shè)c) 熱膨脹系數(shù)0.085 < 1滿足Boussinesq假設(shè)2. a) Derive the similarity solution equation for