Reactive power would put no extra load on a generator shaft if everything were perfect. However, real generators have real losses, with some of those proportional to the square of the current. The reactive load causes more current in the wires than there would be with a purely resistive load of the same real power. The extra current causes additional real power to be lost.
So the answer is that the engine will see a somewhat higher load and therefore use a bit more fuel. This is because of more inefficiencies and losses in the system, not be reactive power itself makes the generator harder to turn.
Added:
I should have mentioned this before, but somehow it slipped thru my mind at the time.
A reactive load on a perfect generator does not require more shaft power averaged over a cycle, but it does add "bumps" to the torque. One attribute of a 3 phase AC generator is that the torque is constant over a cycle with a resistive load. However, with a reactive load parts of the cycle will require more power and other parts less. The average power is still the same, but constant pushing forwards and backwards relative to the average torque can cause undesirable mechanical stresses and vibrations.
You can think of this a bit like moving two magnets past each other. Let's say they are oriented to repell. At a distance there is little force. You have to apply force to move them close together, meaning you put energy into the system. The magnets push in the direction of motion as they move away, thereby giving you back the energy you put in earlier. The net energy spent is 0, but there was definitely energy flow back and forth. There is always some loss as energy is moved around or converted back and forth in real systems.
Again, the reactive power itself doesn't cause the problem, but real power is lost because energy can't be moved around and converted with perfect efficiency. This real power loss has to be made up with more real power input. In addition, the extra mechanical forces can decrease the life of the generator and engine driving it.
Any help would be appreciated
You can measure phase-neutral voltage and if the supply and load are balanced then you can infer line voltage by: -
line volts = phase volts\$\times\sqrt3\$
This should be one of the easier measurements yet you say you can calculate "real power, reactive power and apparent with ease". This does make me think that you have used one of these measurements, and the RMS measurement of current and back-calculated phase/line voltage and that's when you are seeing a discrepency.
If this is so then I suspect your current measurement may be at fault either through incorrect use of a current transformer or some ratio being incorrect. Or, it could be the \$\sqrt3\$ thing mentioned above.
If you need any further help with this you should consider detailing how you make the other measurements and why you believe voltage to be incorrectly calculated/measured.
Best Answer
Intuitively, if all the load power factors are lagging, the overall power factor is lagging too.
$$% outer vertical array of arrays \begin{array}{c} % inner horizontal array of arrays \begin{array}{c|c} % inner array of minimum values \begin{align} S_1&=20\; \;\text{kVA}\\ V&=600 \angle{0}^{\circ} \\ I_1&=\frac{S}{V}=\frac{20k}{600}=33.3 \;\text{A}\\ \therefore {\overline{I}_{1}}&=33.3\angle{-36.87^{\circ}} \end{align} & % inner array of maximum values \overline{I}_{2}=\frac{V}{Z}=\frac{600}{15+30j}=8\sqrt{5}\angle{-63.43^{\circ}}\\ \end{array} \\ % inner array of delta values \hline \begin{matrix} &\overline{I}_{source}=\overline{I}=I\angle{-\arccos(0.707)}=I\angle{-45^{\circ}} \Longleftarrow\\ &\overline{I}_3=I_3\angle{-\arccos(0.6)}=I_3\angle{-53.13^{\circ}}\\ &{\overline{I}}={\overline{I}_{1}}+{\overline{I}_{2}}+{\overline{I}_{3}}\\ &\Im\{{\overline{I}_{1}}+{\overline{I}_{2}}\}=-35.98j\\ &\therefore I_3\sin(-53.13^{\circ})=-45^{\circ}-(-35.98^{\circ})=-4.02^{\circ}\\ &\text{So}\quad \overline{I}_{3}=5.03\angle{-53.13^{\circ}}\Longleftarrow\\ & \therefore I=54.94\angle{-45^{\circ}}\Longleftarrow\\ \end{matrix} \end{array}$$
$$\begin{array} &S_{L_3}={V}\times{I}_3=600\times5.03=3018\; \text{VA}\\ P_{L_3}=3018\times 0.6=1810.8 \;\text{W}\\ Q_{L_3}=\sqrt{3018^2-1810.8^2}=2414.4 \;\text{VAr}\\ P_{loss}=I^2R=54.94^2\times 0.3=905.5 \;\text{W}\\ V_i=\overline{V}+\overline{I}*\overline{Z}_{\text{line}}=600+(54.94\angle{-45^{\circ}})\cdot(0.3+j0.8)=643.0\angle1.73^{\circ} \end{array}$$