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Given the following text description, write Python code to implement the functionality described below step by step
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Description:
 Excercises Electric Machinery Fundamentals
Chapter 2
Problem 2-11
Step1:  Description
A 100-MVA 230/115-kV $\Delta$-Y three-phase power transformer has a per-unit resistance of 0.015 pu and a per-unit reactance of 0.06 pu. The excitation branch elements are $R_C =100\,pu$ and $X_M = 20\,pu$.
Step2:  (a)
If this transformer supplies a load of 80 MVA at 0.8 PF lagging.
draw the phasor diagram of one phase of the transformer.
(b)
What is the voltage regulation of the transformer bank under these conditions?
(c)
Sketch the equivalent circuit referred to the low-voltage side of one phase of this transformer.
Calculate all the transformer impedances referred to the low-voltage side.
(d)
Determine the losses in the transformer and the efficiency of the transformer under the conditions of part (b).
SOLUTION
(a)
The transformer supplies a load of 80 MVA at 0.8 PF lagging. Therefore, the secondary line current of the transformer (referred to the secondary side) is
Step3:  The base apparent power is $S_\text{base} = 100\,MVA$ , and the base line voltage on the secondary side is $V_{LS_\text{base}} = 115\,kV$ , so the base value of the secondary line current is
Step4:  so the per-unit secondary current is
Step5:  The per-unit phasor diagram is shown below
Step6:  and the voltage regulation is
Step7:  (c)
The secondary side of this transfer is Y-connected, so the base phase voltage of the low voltage (secondary) side of this transformer is
Step8:  The base impedance of the transformer referred to the low-voltage side is
Step9:  Each per-unit impedance is converted to actual ohms referred to the low-voltage side by multiplying it by this base impedance. 
The resulting equivalent circuit is shown below
Step10:  (d)
The per-unit losses in the series resistance are
Step11:  and the actual losses in the series resistance are
Step12:  The per-unit losses in the excitation branch are
Step13:  and the actual losses in the excitation branch are
Step14:  The per-unit power supplied to the load
Step15:  Therefore, the transformer’s efficiency is
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Python Code:
%pylab notebook
Explanation: Excercises Electric Machinery Fundamentals
Chapter 2
Problem 2-11
End of explanation
Sbase = 100e6 # [VA]
Vp0   = 230e3 # [V]
Vs0   = 115e3 # [V]
Rc_pu  = 100.0
Xm_pu  =  20.0
Req_pu =   0.015
Xeq_pu =   0.06
Explanation: Description
A 100-MVA 230/115-kV $\Delta$-Y three-phase power transformer has a per-unit resistance of 0.015 pu and a per-unit reactance of 0.06 pu. The excitation branch elements are $R_C =100\,pu$ and $X_M = 20\,pu$.
End of explanation
Sload = 80e6  # [VA]
PF    = 0.8
Vls_a = Vs0
Ils_a = Sload / (sqrt(3)*Vls_a)
print('Ils_a = {:.0f} A'.format(Ils_a))
Explanation: (a)
If this transformer supplies a load of 80 MVA at 0.8 PF lagging.
draw the phasor diagram of one phase of the transformer.
(b)
What is the voltage regulation of the transformer bank under these conditions?
(c)
Sketch the equivalent circuit referred to the low-voltage side of one phase of this transformer.
Calculate all the transformer impedances referred to the low-voltage side.
(d)
Determine the losses in the transformer and the efficiency of the transformer under the conditions of part (b).
SOLUTION
(a)
The transformer supplies a load of 80 MVA at 0.8 PF lagging. Therefore, the secondary line current of the transformer (referred to the secondary side) is:
$$I_{LS} = \frac{S_\text{load}}{\sqrt{3}V_{LS}}$$
End of explanation
Vls_base = Vs0
Ils_base_a = Sbase / (sqrt(3)*Vls_base)
print('Ils_base_a = {:.0f} A'.format(Ils_base_a))
Explanation: The base apparent power is $S_\text{base} = 100\,MVA$ , and the base line voltage on the secondary side is $V_{LS_\text{base}} = 115\,kV$ , so the base value of the secondary line current is:
$$I_{LS_\text{base}} = \frac{S_\text{base}}{\sqrt{3}V_{LS_\text{base}}}$$
End of explanation
Ils_pu_a = Ils_a / Ils_base_a
ILS_pu_a_angle = -arccos(PF)
ILS_pu_a = Ils_pu_a * (cos(ILS_pu_a_angle) + sin(ILS_pu_a_angle)*1j)
print('ILS_pu_a = {:.1f} ∠{:.2f}°'.format(
            abs(ILS_pu_a), ILS_pu_a_angle/pi*180))
Explanation: so the per-unit secondary current is:
$$\vec{I}{LS{pu}} = \frac{\vec{I}{LS}}{I{LS_\text{base}}}$$
End of explanation
VS_pu = 1.0                                    
Zeq_pu = Req_pu + Xeq_pu*1j      
VP_pu = VS_pu + ILS_pu_a*Zeq_pu
VP_pu_angle = arctan(VP_pu.imag/VP_pu.real)
print('VS_pu = {:.3f} ∠{:.1f}°'.format(
           abs(VP_pu), VP_pu_angle/pi*180))
Explanation: The per-unit phasor diagram is shown below:
<img src="figs/Problem_2-11a.jpg" width="50%">
(b)
The per-unit primary voltage of this transformer is:
$$\vec{V}P = \vec{V}_S + \vec{I}Z\text{EQ}$$
End of explanation
VR = (abs(VP_pu) - abs(VS_pu)) / abs(VS_pu) * 100
print('''
VR = {:.1f}%
=========
'''.format(VR))
Explanation: and the voltage regulation is:
$$VR = \frac{V_P - V_S}{V_S} \cdot 100\%$$
End of explanation
Vphis_base = Vls_base / sqrt(3)
print('Vphi_base = {:.1f} kV'.format(Vphis_base/1000))
Explanation: (c)
The secondary side of this transfer is Y-connected, so the base phase voltage of the low voltage (secondary) side of this transformer is:
$$V_{\phi S_\text{base}} = \frac{V_{LS_\text{base}}}{\sqrt{3}}$$
End of explanation
Zbase = 3 * Vphis_base**2 / Sbase
print('Zbase = {:.0f} Ω'.format(Zbase))
Explanation: The base impedance of the transformer referred to the low-voltage side is:
$$Z_\text{base} = \frac{3V_{\phi S_\text{base}}^2}{S_\text{base}}$$
End of explanation
Req = Req_pu*Zbase
Xeq = Xeq_pu*Zbase
Rc = Rc_pu*Zbase
Xm = Xm_pu*Zbase
print('''
Req,s = {:.2f} Ω    Xeq,s = {:.2f}  Ω
Rc,s  = {:.1f} kΩ    Xm,s = {:.2f} kΩ
==================================
'''.format(Req, Xeq, Rc/1000, Xm/1000))
Explanation: Each per-unit impedance is converted to actual ohms referred to the low-voltage side by multiplying it by this base impedance. 
The resulting equivalent circuit is shown below:
<img src="figs/Problem_2-11b.jpg" width="60%">
End of explanation
Peq_pu = abs(ILS_pu_a)**2 *Req_pu
print('Peq_pu = {:.4f}'.format(Peq_pu))
Explanation: (d)
The per-unit losses in the series resistance are:
$$P_\text{EQ} = I^2R_\text{EQ}$$
End of explanation
Peq = Sbase * Peq_pu
print('''
Peq = {:.2f} MW
=============
'''.format(Peq/1e6))
Explanation: and the actual losses in the series resistance are:
$$P_\text{EQ} = S_\text{base}P_{\text{EQ}_\text{pu}}$$
End of explanation
Rex_pu = Rc_pu
Pex_pu = abs(VP_pu)**2 / Rex_pu
print('Pex_pu = {:.4f}'.format(Pex_pu))
Explanation: The per-unit losses in the excitation branch are:
$$P_{\text{EX}\text{pu}} = \frac{V^2}{R\text{EX}}$$
End of explanation
Pex = Sbase*Pex_pu
print('''
Pex = {:.2f} MW
=============
'''.format(Pex/1e6))
Explanation: and the actual losses in the excitation branch are:
$$P_{EX} = S_{base}P_{EX,pu}$$
End of explanation
Pload_pu = PF*Sload/Sbase
print('Pload_pu = {:.2f}'.format(Pload_pu))
Explanation: The per-unit power supplied to the load:
$$P_{\text{load}\text{pu}} = \frac{P\text{load}}{S_\text{base}}= \frac{PF  \cdot S \text{load}}{S\text{base}}$$
End of explanation
Pin_pu= Pload_pu + Peq_pu + Pex_pu
Pout_pu = Pload_pu
eta = Pout_pu/Pin_pu * 100
print('''
η = {:.1f} %
==========
'''.format(eta))
Explanation: Therefore, the transformer’s efficiency is:
$$\eta = \frac{P_{OUT}}{P_{IN}} \cdot 100\%$$
End of explanation
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