The pressure at which a liquid will boil is called its vapor pressure. This pressure is a function of temperature (vapor pressure increases with temperature). In this context we usually think about the temperature at which boiling occurs. For example, water boils at 100oC at sea-level atmospheric pressure (1 atm abs). However, in terms of vapor pressure, we can say that by increasing the temperature of water at sea level to 100 oC, we increase the vapor pressure to the point at which it is equal to the atmospheric pressure (1 atm abs), so that boiling occurs. It is easy to visualize that boiling can also occur in water at temperatures much below 100oC if the pressure in the water is reduced to its vapor pressure. For example, the vapor pressure of water at 10oC is 0.01 atm. Therefore, if the pressure within water at that temperature is reduced to that value, the water boils. Such boiling often occurs in flowing liquids, such as on the suction side of a pump. When such boiling does occur in the flowing liquids, vapor bubbles start growing in local regions of very low pressre and then collapse in regions of high downstream pressure. This phenomenon is called as cavitation
Compressibility and the Bulk modulus:
All materials, whether solids, liquids or gases, are compressible, i.e. the volume V of a given mass will be reduced to V - dV when a force is exerted uniformly all over its surface. If the force per unit area of surface increases from p to p + dp, the relationship between change of pressure and change of volume depends on the bulk modulus of the material.
Bulk modulus (K) = (change in pressure) / (volumetric strain)
Volumetric strain is the change in volume divided by the original volume. Therefore,
(change in volume) / (original volume) = (change in pressure) / (bulk modulus)
i.e., -dV/V = dp/K
Negative sign for dV indicates the volume decreases as pressure increases.
In the limit, as dp tends to 0,
K = -V dp/dV à 1
Considering unit mass of substance, V = 1/r à 2
Differentiating,
Vdr + rdV = 0
dV = - (V/r)dr à3
putting the value of dV from equn.3 to equn.1,
K = - V dp / (-(V/r)dr)
i.e. K = rdp/dr
The concept of the bulk modulus is mainly applied to liquids, since for gases the compressibility is so great that the value of K is not a constant.
The relationship between pressure and mass density is more conveniently found from the characteristic equation of gas.
For liquids, the changes in pressure occurring in many fluid mechanics problems are not sufficiently great to cause appreciable changes in density. It is therefore usual to ignore such changes and consider liquids as incompressible.
Gases may also be treated as incompressible if the pressure changes are very small, but usually compressibility cannot be ignored. In general, compressibility becomes important when the velocity of the fluid exceeds about one-fifth of the velocity of a pressure wave (velocity of sound) in the fluid.
參考答案:液体将煮沸的压力叫做它的蒸气压力。这压力是温度的作用(蒸气压力增加以温度) 。在这上下文我们通常考虑煮沸发生的温度。例如, 水煮沸在100oC 以海平面大气压(1 atm 吸收) 。但是, 根据蒸气压力, 我们能说, 由增加温度水在海平面对100 oC, 我们增加蒸气压力对它与大气压的点(1 atm 吸收) 是相等的, 以便煮沸发生。它容易形象化煮沸可能并且发生在水中在温度在100oC 之下如果压力在水中被减少到它的蒸气压力。例如, 水蒸气压力在10oC 是0.01 atm 。所以, 如果压力在水之内在那个温度被减少到那价值, 水煮沸。这样煮沸经常发生在流动的液体, 譬如在泵浦的吸边。当这样煮沸发生在流动的液体, 蒸气泡影开始生长在非常低pressre 的地方地区和然后崩溃在高顺流压力的地区。这种现象叫当气蚀压缩性和大块模数: 所有材料, 是否固体, 液体或气体, 是可压缩的, 即容量v 的被测量的大量将被减少到V - dV 当力量一致地被施加到处它的表面。如果每单位面积作用力表面增加从p 到p + dp, 关系在压力的容量之间的变动和变动依靠材料的大块模数。大块模数(k) = (变化在压力)/(上容量张力) 容量张力是体积变化由原始的容量划分。所以, (体积变化)/(原始的容量) = (变化在压力)/(上大块模数) 即, - dV/V = dp/K 负号为dV 表明容量减退当压力增加。在极限, 作为dp 趋向到0, K = - V dp/dV 2a 物质, V = 1/r 2a 2 区分, Vdr + rdV 1 考虑的单位大量= 0 dV = - (V/r)dr 2a3 投入dV 的价值从equn.3 到equn.1, K = - V dp/(- (V/r)dr) 即K = rdp/dr 大块模数的概念向液体主要被应用, 因为为气体压缩性是很伟大的K 的价值不是常数。关系在压力和大量密度之间更加方便地被发现从气体的典型等式。为液体, 变化在压力上发生在许多流动力学问题不是充足地伟大导致看得出的变化在密度上。它是因此通常的忽略这样变动和液体把不可压缩视为。气体不可压缩也许并且对待如果压力变动非常小, 但压缩性无法通常被忽略。总之, 压缩性变得重要当流体的速度超出压力波浪(速度的速度的大约五分之一声音) 在流体。
