Answer:
W=-109.12 kJ/kg
Q=-76.34 kJ/kg
Explanation:
The needed work W we will calculate by using the work equation for polytropic process and the heat transfer Q we will calculate by using the energy balance equation.
Before the calculations we first need to determine the final temperature T2. We will do that by using the given initial temperature T1 = 10°C, the given initial p_1 = 120 kPa and final p_2 = 800 kPa pressure and the polytropic index n = 1.2. Before the calculation we need to express the temperature in K units.
T1 = 10°C + 273 K = 283 K
T2 = ((p_2/p_1)^(n-1)/n)* T1
T2 = 388 K
Now we can use the heat capacity C_v, = 0.3122 kJ /kg K and the temperatures T1 and T2 to determine the change in internal energy ΔU.
ΔU = C_v*(T2-T1)
ΔU = 32.78 kJ/kg
to determine the work we will also need the initial v1 and final v2 specific volume. The initial specific volume v1 we can determine from the ideal gas equation. For the calculation we will need the initial pressure p_1, temperature T1 and the specific gas constant R = 0.2081 kJ /kg K.
v1=R*T1/p_1
v1=0.4908 m^3/kg
For the final specific volume we need to replace the initial temperature and pressure with the final.
v2=R*T2/p_2
v2=0.1009 m^3/kg
The work W is then:
W=p_2*v2-p_1*v1/n-1
W=-109.12 kJ/kg
The heat transfer Q we can calculate form the energy balance equation. For the calculation we will need the calculated work W and the change in internal energy ΔU.
Q=W+ΔU
Q=-76.34 kJ/kg
Answer:
The heat absorbed portion that is transformed into work is therefore helpful and accessible for our use, hence it is termed as available energy. But the energy or heat which is of no use is termed as unavailable energy.
Explanation:
A engine requires heat from the reservoir and transforms portion of it into function. The extra heat is expelled (generally atmosphere) to cold reservoir. The refused heat can't be equal to zero, as per the 2nd law of thermodynamics. The heat absorbed portion that is transformed into work is therefore helpful and accessible for our use, hence it is termed as available energy. But the energy or heat which is of no use is termed as unavailable energy.
Given Information:
Balanced Y-Y three phase circuit
Phase Voltage = Van = 120 < 10° V
Load Impedance = Zy = 20 +j15 Ω = 25 < 36.86° Ω
Required Information:
Load Voltages = Vab, Vbc, Vca = ?
Line currents = Ia, Ib, Ic = ?
Answer:
Vab = 208 < 40°
Vbc = 208 < -80°
Vca = 208 < 160°
Ia = 4.8 < -26.86°
Ib = 4.8 < -146.86°
Ic = 4.8 < 93.14°
Explanation:
Since it is balanced 3 phase system, all phases have equal magnitude and 120° phase shift
Van = 120 < 10° V
Vbn = 120 < -110° V
Vcn = 120 < 130° V
In a Y connected system, the phase voltage and line voltage are related as
Vab = 
(Van) <+30°
Vab = *120<10°+30°
Vab = 208 < 40°
Vbc = 208 < -80°
Vca = 208 < 160°
In a Y connected circuit, Iphase = Iline
Ia = Van/Zy
Ia = 120 < 10°/25 < 36.86°
Ia = 4.8 < -26.86°
Ib = 4.8 < -146.86°
Ic = 4.8 < 93.14°
