Question

这段话怎么翻译，不要用无脑有道翻译。如果满意我会另加悬赏。

energy of ethylene. The third factor, which is contributory, is that the oxidation of one mole of ethylene requires a smaller amount of oxygen (3 versus 3.5 moles), and hence there is less nitrogen in the combustion product that needs to be heated up. Thus the dependence of Tad on the C/H ratio is the net effect of the bond energy, the number of H available per C atom to form H2O, and the collateral need to heat up the N2 for the amount of O2 reacted.
The above dependence is the basis for the development of high-energy-density fuels as explosives or for tactical propulsion. These fuel molecules are usually highly strained,which imply large bond energies and C/H ratios.Nitrogen is also frequently present as an elemental constituent in explosives because of the associated large bond energy. Furthermore, since fuel and oxidizer are present together in these explosives as well as the monopropellants for rocket propulsion, the resulting Tad can be quite high because there is no atmospheric nitrogen that needs to be heated up.
Figures 1.4.2 and 1.4.3 also show that Tad behaves asymmetrically with respect to φ in three major aspects, namely the slope is gentler on the rich side of the maximum than that on the lean side, that there are sharp “bends” in the curves for some fuels at very rich compositions, and that it peaks slightly on the rich side of stoichiometry. These behaviors are explained in the following.

Answer 1

能源的乙烯。第三个因素,这是分摊,是氧化一摩尔乙烯需要少量的氧气(3和3.5摩尔数),并因此减少氮的燃烧产物,需要加热。这样的依赖在C / H有点比率的净效应结合能,可用的数量每C H原子形成水,和抵押品需要加热N2金额O2反应。
上面的依赖是发展的基础高能量密度燃料作为炸药或战术推进。这些燃料分子通常是高度紧张,这意味着大型债券能量和C / H比率。氮也是经常出现作为一个元素组成在炸药由于相关大型债券能源。此外,由于燃料和氧化剂存在一起在这些炸药以及monopropellants火箭推进,结果小孩子会相当高,因为没有大气氮需要加热。
数据还表明,1.4.2和地被有点不对称地表现对φ在三个主要方面,即斜率是温和的对富人一边的最大比精益方面,有锋利的“弯曲”的曲线对于一些燃料在非常丰富的成分,它对富人的山峰略侧的化学计量学。这些行为是解释在下面。

Answer 2

能源乙烯。第三个因素，这是促成，是一摩尔的乙烯氧化需要更小的氧气量（3与3.5摩尔），因此有较少的氮在燃烧的产品，需要加热。因此，TAD对C / H比的依赖的键能的净效应，H每C原子形成的水的数量，和抵押品需要加热的氧气量的N2反应。上面的依赖的高能量密度燃料爆炸或战术推进发展的基础。这些燃料分子通常是高度紧张的，这意味着大的键能和C / H比值。氮也常表现为爆炸物的元素组成因相关大键能。此外，由于燃料和氧化剂是目前在这些炸药以及火箭发动机的燃料，产生的点可以相当高，因为没有大气氮需要加热。数字图像和1.4.3还表明，有些行为不对称相对于φ等三大方面，即坡平缓的最大丰富侧比在精益的一面，有鲜明的“弯曲”的曲线的一些燃料非常丰富的组合物，以及它在化学计量比峰丰富的旁边。这些行为的解释如下。