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1、1.A pressure of 20 psi is measured at a depth of 20 ft. Calculate the density of the liquid if p = 0 on the surface. 解：1psi=6.895kPa 而1ft=304.8mm 根据P=gh 得：=P/gh=6.895103/（9.80.3048）=2.308103kg/m32A vacuum of 31 kPa is measured in an airflow at sea level. Find the absolute pressure in: (a) kPa (b) mm
2、 Hg (c) psi (d) ft H20 (e) in. Hg 解：根据 P=P-Pa 得：P=P+PaP=101.325-31=70.325 kPaA因为 1mmHg=133.33Pa 所以得：P=70.325 kPa=70325/133.33=527.45mmHg b因为 1psi=6.895kPa ， 所以 70.325/6.895psi=10.20psic因为 1mmH2o=9.807Pa 所以 P=70.325/(304.89.807)=22.97ftH2Od70kpa=20.77in.Hge3.The water level is same in the closure tan
3、k, tube 1, pipe 3, and 4. Pipe 1 can fluctuate, so as to adjust the water pressure in the bank. If (1) tube 1 decline in a certain height, (2) tube 1 rise in a certain height. Try to illustrate which is the highest level, which is the lowest level, and which are the same level at each time.解：在升降前有：P
4、2=P4=Pa 在变化前后都有：h1=h3 h2=h4 筒一下降一定的高度，此时水箱中的水流向筒中，以保持 h1=h3.有液面 2.3.4 均下降，1而液面 1 下降的高度与液面 3 下降高度相同，但小于筒下降的高度。由于筒下降，水箱中的空气被扩充，则有：P2Pa=0又PA=P2+ga 得 PA=0+gb=gb 推出PA/g=P2/g+a PA/g=b 因为 P20 所以 ba h3h2所以 h2=h4h1=h3筒一上升一定的高度。同理可得：相反的 ba h3h2 有2h1=h3h2=h44. In order to accurately measure the pressure differ
5、ence between A and B in the liquid with density of , a micromanometer was designed specially as shown in the figure. Try to determine the relationship between and, and the pressure difference between A and B.依题意有：从1-1面的静压相等可知PA-gH=PB-gH 1可得：PA-PB=gH-gH=gH(-) 由于1面和3面中间是空气，则P1=P22由2-2平面的静压相等可得：PB-gb=P
6、B-g(h1+h2)又b=a+h2+h1 所以推出 gb=g（b-a）进一步得：=(b-a)/b将=(b-a)/b 代入 PA-PB=gH-gH得：PA-PB= agH/b5. There was a flat rectangular metal gate supported with 3 beams on the opposite surface to water, the height of gate and the depth of water are same as h=3m. Calculate the locations of beams which are uniformity
7、in force.解：作闸门的压强分布图如图右，将其面积分为、三部分。各梁的位 置分别为闸门各部分AE、EG、GB受力的压力中心D1、D2和D3AABC=1/2ghh=1/21039.80732=44.13103N/mAAEF=1/3AABC=1/2gyEyE =1/344.13103=1/21039.807y2E得：yE=1.73m 由AAGH=2/3AABC=1/2gyGyG=1/21039.807y2G 得yG=2.449m yGE=yG-yE=2.449-1.73=0.719m yBG=yB-yG=3-2.449=0.551my1=2/3yE=2/31.73=1.15m对梯形EFHG分成矩形和三角形，根据合力距定理有：1/2yGE（yGEgyE）+2/3yGE(1/2yGEgyGE)=AEFHG(y2-yE)化简代入数据得：y2=2.11m 同理解得:y3=2.73m
