Question

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Fills with the blank. (Each blank worth 2 score, together 24 score) 1. Write the definition equation of enthalpy H = . _(1) 2. A steady state flow in a throttle, the enthalpy change of the working fluid is = _(2) 3. Below functions, which one is right? _(3) 3. A container encloses some oxygen in it. The pressure meter tells us the pressure of the oxygen is 1.2MPa. the atmosphere pressure is 0.1MPa. The absolute pressure of oxygen is_(4). 4. When a system changes the state in such a way that at any instant during the process the state point can be located on the diagram, then the process is said to be _(5)_ s. In free expansion process, which state parameters keep constant _(6_? From the first and second law, we have Then *_(z)_? 7. One closed system with 2 kg working fluid undergoes a process and loses heat energy 25kJ. At the same time, outside does work to it with 100kJ.Howmany internal energy will change_(8)_? 8. In constant pressure process, system obtains 30kJ heat. Technic work is (9)__.Enthalpy changes _(10) 6. As a figure, a piston did the total, work is W. if p is the pressure of the surroundings, V, and V is the volume before and after system change, respectively. The exergy of the work Wis_(11) 8. Known and based on Tas-duu pod, Please express with T and y. (12)

Answer 1

(1). H = E + PV enthalpy is the sum of internal energy plus the product of pressure and volume of the system.

(2). In a steady-state throttling process the enthalpy change (dH) = constant and the work done (W) = 0.

(3). Where are the functions?

(4). From formula Pab = Pg + Patm , where Pab= absolute pressure, Pg= gauge pressure, Patm= atmospheric pressure

Pab = 1.2 + 0.1

Pab = 1.3 MPa is the absolute pressure of the oxygen.

(5). Reversible process

(6).  In free expansion process, there is no work exchanged to the surroundings, no heat exchange. Therefore as both are functions of temperature, In free expansion process temperature (T1 = T2 = constant), internal energy ($\Delta$U = 0) So (U) = constant, heat energy (Q) and Work done (W) are zero (Q = W = 0)
(7). $ds = \frac{1}{T}du + \frac{P}{T}dv$

du = Tds - Pdv

The internal energy U is the function of temperature and volume u = f(T, v)

Using Maxwell's third equation $\left ( \frac{\partial s}{\partial v} \right )_{T}=\left ( \frac{\partial P}{\partial T} \right )_{v}$

and

ди au = T ӘР ОТ — Р

So, $\frac{\partial s}{\partial u}=\frac{1}{T} \left [ \frac{1+Pdv}{du} \right ]$

(8). The change in internal energy ($\Delta$U) = Q - W, $\Delta$ U = (Qin - Qout) - (Wout - Win)

Given, Qout = 25KJ, Win = 100KJ and Qin = Wout = 0

Therefore, $\Delta$ U = -25+100 = 75KJ

(9). Technical workdone in a constant pressure process (W) = P.dV

(10). In a constant pressure process, change in enthalpy ($\Delta$H) is equal to the Heat flow (q)

Therefore, $\Delta$ H = q = 30KJ