diagramshorttransformer
For part a, I considered the following model of the transformer where v1 = 230 V, representing the low voltage side. For part b, I used the same model, with the right hand side short circuited rather than open circuited, except this time v1 = 4600 V because the low voltage side is short circuited. Is this the correct approach to draw the equivalent circuit diagram and calculate the required parameters?
What you may be getting confused about is the "ideal transformer" in the equivalent circuit. You should not regard it as having any magnetic qualities at all. Try and see it like this: -
Whatever voltage you have on the input to the ideal transformer, \$V_P\$ is converted to \$V_S\$ on the output of this "theoretical" and perfect device. It converts power in to power out without loss or degradation such that: -
\$V_P\cdot I_P = V_S\cdot I_S\$
The ratio of \$V_P\$ to \$V_S\$ happens to be also called the turns ratio and almost quite literally it is on an unloaded transformer because there will be no volt-drop across R3 and L3 and only the tiniest of volt-drops on R1 and L1.
This means you now have a relatively easy way of constructing scenarios of load effects and recognizing the volt drops across the leakage components that are present.
The equivalent circuit of the transformer is really quite good once you accept that the ideal power converter is "untouchable" and should just be regarded as a black box. For instance you can measure both R1 and R3 and, by shorting the secondary you can get a pretty good idea what L3 and L1 are. With open circuit secondary you can measure the current into the primary and get a pretty good idea what \$L_M\$ is too.
The nominal output voltage should be the rated voltage with nominal input voltage and the full load (resistive) as rated.
In other words, a 12V 300mA transformer should have 12V RMS output with nominal input and a 300mA resistive load. For loads less than the rated load, the voltage, of course, will be higher.
Edit:
The regulation of a transformer is typically defined as:
Regulation(%) = \$ \frac {V_{open} - V_{full.load}}{V_{full.load}}\$
Large high power transformers might have regulation of a few percent, cheap small transformers maybe 5 or 10x worse.
Best Answer
For the open circuit test, the general idea here is that the power dissipated is due to core losses (\$R_{lr}\$ in your diagram if I read the name correctly). The series winding losses will be negligible. So you have a power loss of 152 watts and that can be regarded as being exclusively dissipated in \$R_{lr}\$. Given that the applied voltage is 230 volts, you can calculate resistance by re-arranging the following equation for power: -
$$P = \dfrac{V^2}{R}$$
So, I calculate \$R_{lr}\$ to be 348 Ω.
And given that the apparent impedance is 230 volts ÷ 3.5 amps (65.7 Ω) you can work out what value of \$L_{lm}\$ (magnetization inductance is) using pythagoras. \$L_{lm}\$ is fairly dominant here (compared to \$R_{lr}\$) and so \$L_{lm}\$ is going to be a bit higher than 174 mH.
For the short circuit test, both \$L_{lm}\$ and \$R_{lr}\$ are much less dominant than the series losses (\$R_1\$ and \$R_2\$) and should be ignored.
Can you take it from here? This bit is the easier part but you do need to recognize that the shorted secondary projects its series loss resistance to the primary via the turns ratio squared.