习题答案201108
- 格式：doc
- 大小：1.06 MB
- 文档页数：28
PROBLEMS AND ANSWERSPROBLEMS 11.1 An object of 6.795kg has a volume of 0.5L. Determine its density, specific weight and relative density.1.1 一6.795kg 的物体体积为0.5L 。
确定其密度、重度和相对密度。
Solution:33kg/m1059.13L /13.59kg L500.0kg 795.6⨯====V m ρ)N/m (133264806.91059.1333=⨯⨯==g ργRD =ρ/ρwater =13.59⨯103/1000=13.591.2 A force acts upon a square area 4 by 4 cm in the xy plane. It expressed by F =8i +6j +9k (N). Resolve this force into a normal force and a shear-force component. What are the pressure and the shear stress? Repeat the calculations for F = -8i +6j -9k . Solution:(1) The normal force F n =9N and the shear force F τ=10(N) )68(2/122=+ The pressurePa)(5625m1016N 924=⨯==-A F p (upward)and the shear stressPa)(6250m1016N 1024=⨯==-A F ττ)(13.53N6N 8arctanarctan︒===xy F F α(2) The normal force F n =-9N and the shear force F τ=10(N) ]6)8[(2/122=+- The pressurePa)(5625m1016N 924-=⨯-==-A F p (downward)and the shear stressPa)(6250m1016N 1024=⨯==-A F ττ)(13.53N6N 8arctanarctan︒-=-==xy F F α1.2 (1) p =5625Pa(upward), τ=6250Pa,︒13.53; (2) p =5625Pa(downward ), τ=6250Pa,︒-13.531.3 The bulk modulus of elasticity of oil is K =2.2GPa, what pressure is required to reduce itsvolume by 1 percent? Solution:υυ/d dp K -=MPa22)(N/m102.2)01.0(102.2279=⨯=-⨯⨯-=-=υυd Kdp1.3 22MPa1.4 The upper plate, 1 mm distant from the lower plate, moves at 0.5m/s and requires a force per unit area of 4Pa (N/m 2) to maintain this speed as shown Fig. 1.4. Determine the fluid viscosity in SI units. Solution:hV μτ=, s)(Pa 1085.0101433⋅⨯=⨯⨯==--Vh τμ1.4 s Pa 1083⋅⨯=-μ1.5 An shaft of 200mm diam rotates in a sliding bearing filled with lube oil of μ=0.245Pa ⋅s. The width of bearing is b =300mm and radial clearance is h =0.8mm. Power loss in the bearing is N =50.7W. Determine the rotational speed of shaft n .1.5 一轴直径为200mm ，在充满μ=0.245Pa ⋅s 润滑油的滑动轴承中转动。
轴承宽度为b =300mm ，径向间隙为h =0.8mm 。
轴承中的功率损失为N =50.7W 。
确定轴的转速n 。
Solution:Dn n D V ππ==)2(2, hV DbhV AF μπμ==, FV N =minrev/5.89)rev/s (4915.12.03.02.0245.0108.07.503==⨯⨯⨯⨯⨯⨯==-πππμπDDb Nhn1.5 minrev/5.89rev/s 4915.1==n1.6 The diameter of a piston cylinder D is 120mm, piston diameter d is 119.6mm, and piston width L is 140mm. The viscosity of oil is μ=0.65Pa ⋅s. By neglecting the pressures at both ends of piston, determine the force F needed on pulling back the piston in a steady velocity of 0.5m/s.1.6 一活塞缸直径D 为120mm ，活塞直径d 为119.6mm ，活塞厚L 为140mm 。
油液粘度μ=0.65Pa ⋅s 。
忽略活塞两端压力，试求以0.5m/s 的稳定速度将活塞拉回时所需的力F 。
Solution:N)(5.85102.05.010140106.11965.02/)(333=⨯⨯⨯⨯⨯⨯⨯=-==---πμπμd D V dLdzdu AF1.6N5.85=F1.7 Compress air by a compressor, then the absolute pressure is from 1 atm to 6 atm, the temperature is from 20︒C up to 78︒C, how much volume of air will lose?1.7 压缩机将空气压缩，绝对压力从1大气压增大到6大气压，温度从20︒C 升高到 78︒C ，其体积损失多少？ Solution:T p R =υ, T m V p mTV p '''==R ,V p p T T V ''='V0.261)2015.273()7815.273(=++=The loss in volume∆V =0.8V1.7 ∆V =0.8V1.8 An airtight container of 200 L is filled with two kilogram of dry air at -10︒C. What is the pressure?1.8一200升的密封容器充满两公斤-10︒C 的干空气。
压力为多少？ Solution:T p R =υ, T m V p R =,)abs Pa (756109)1015.273(33.287102002R 3=-⨯⨯⨯==-T V m p1.8Pa(abs)756109=p1.9 The bulk modules of a gas in adiabatic process can be expressed asK =κpProve it.1.9 绝热过程的气体体积弹性模数可以表达为K =κp证明之。
SolutionFrom Eq. (1.4.7) the adiabatic equation isCp=κρi.e. 'C p =κυ,By making a differential,01=+-υυκυκκd p dp ,Simplifying,pd dp κυυ-=/From the definition of bulk modulesVV p K /d d -=For a given constant mass m , m =ρV , ρ=1/υ, V =m υ, so d V =md υpp p m m p VV p K κκυυυυ=--=-=-=-=)(/d d )/(d d /d d1.9 key: do a differential to the adiabatic equation1.10* There are two parallel disks of diameter d and the space between the two disks is δ, dynamic viscosity of fluid in the clearance is μ. The lower disk is fixed, and the upper disk rotates in a constant angular velocity ω, prove the torque M isδπμω324d M =Solution:From the Newton’s law of viscosityAdzdv T μ=in which r dv ω=, []22)(r dr r dA -+=π, and δ=dz In this problem,[]δπμωδπωμπδωμ322412)(44222/0d d rdr r r r M d =⎪⎭⎫ ⎝⎛=-+⋅⋅=⎰PROBLEMS 22.1已知作用在单位质量物体上的体积力分布为：f x =ax , f y =b , f z=cz 。
物体的密度ρ=lx 2+ry +nz π，坐标度量单位为m ；其中，a =0, b =0.1N/kg, c =0.5N/(kg ⋅m)；l =2.0kg/m 5, r =0, n =1.0kg/m 4。
试求：起点(0, 0, 0)至终点(3, 2, 4) 六面体区域的体积力F x 、F y 、F z 各为多少？ 2.1 The distribution of volume force acting on an object of unit mass is f x =ax , f y =b and f z=cz . The density of object is ρ=lx 2+ry +nz π. The measurement units of coordinates are m. In which a =0, b =0.1N/kg, c =0.5N/(kg ⋅m); l =2.0kg/m 5, r =0, n =1.0kg/m 4. If the coordinate of starting point is (0, 0, 0) and coordinate of ending point is (3, 2, 4) find the components of volume force on the hexahedron F x , F y and F z . 解：)N (0)(30204230204=++==⎰⎰⎰⎰⎰⎰dxdydz nz ry lx ax dxdydz f F x x πρ)N (472.29)(3020402302040=++==⎰⎰⎰⎰⎰⎰dxdydz nz ry lx b dxdydz f F y y πρ )N (96.488)(30204230204=++==⎰⎰⎰⎰⎰⎰dxdydz nz ry lx cz dxdydz f F z z πρAns 2.1 N 0=x F , N 472.29=y F ,N 96.488=z F2.2某地大气压为680mmH 2O ，A 点的相对压强为120mmH 2O 。
