An alternator with internal voltage of 1∠δ₁ p.u. and synchronous reactance of 0.4 p.u. is connected by a transmission line of reactance 0.1 p.u. to a synchronous motor having synchronous reactance 0.35 p.u. and internal voltage of 0.85 ∠δ₂ p.u. If the real power supplied by the alternator is 0.866 p.u., then (δ₁ - δ₂) is _____ degrees. (Round off to 2 decimal places.) (Machines are of non-salient type. Neglect resistances)

The correct answer is 60.

To find the angle difference (${\delta }_1-{\delta }_2$) between the alternator and the synchronous motor, we can represent the entire system using its equivalent reactance network.

Step 1: Calculate the total equivalent reactance of the system
Since the alternator, transmission line, and synchronous motor are connected in series, their reactances add up directly:
$X_{\mathrm{total}}=X_{s1}+X_{\mathrm{line}}+X_{s2}$
Given parameters:
- Synchronous reactance of the alternator, $X_{s1}=0.4\text{p.u.}$
- Transmission line reactance, $X_{\mathrm{line}}=0.1\text{p.u.}$
- Synchronous reactance of the motor, $X_{s2}=0.35\text{p.u.}$

Substituting these values:
$X_{\mathrm{total}}=0.4+0.1+0.35=0.85\text{p.u.}$

Step 2: Use the real power transfer equation
For a non-salient pole machine network where resistances are neglected, the real power ($P$) transferred between the two internal voltage sources is given by:
$P=\frac{E_1{E}_2}{X_{\mathrm{total}}}sin({\delta }_1-{\delta }_2)$
Where:
- Internal voltage of the alternator, $E_1=1\text{p.u.}$
- Internal voltage of the motor, $E_2=0.85\text{p.u.}$
- Real power supplied by the alternator, $P=0.866\text{p.u.}$

Substituting the known values into the equation:
$0.866=\frac{1\times 0.85}{0.85}sin({\delta }_1-{\delta }_2)$
Simplifying the expression:
$0.866=1\times sin({\delta }_1-{\delta }_2)$
$sin({\delta }_1-{\delta }_2)=0.866$

Step 3: Calculate the angle difference
Taking the inverse sine:
${\delta }_1-{\delta }_2={sin}^{-1}\left(0.866\right)$
Since $0.866\approx \frac{\sqrt{3}}{2}$:
${\delta }_1-{\delta }_2=60°$