<jupyter_start><jupyter_text>Excercises Electric Machinery Fundamentals Chapter 2  Problem 2-9<jupyter_code>%pylab notebook<jupyter_output>Populating the interactive namespace from numpy and matplotlib<jupyter_text>Description A 5000-kVA 230/13.8-kV single-phase power transformer has a per-unit resistance of 1 percent and a per-unit reactance of 5 percent (data taken from the transformer’s nameplate).<jupyter_code>Sbase = 5000e3  # [VA]
Vp0 = 230e3     # [V]
Vs0 = 13.8e3    # [V]
Req_pu = 0.01
Xeq_pu = 0.05<jupyter_output><empty_output><jupyter_text>The open-circuit test performed on the low-voltage side of the transformer yielded the following data:$$V_{OC} = 13.8kV \qquad I_{OC} = 21.1A \qquad P_{OC} = 90.8kW$$<jupyter_code>Voc = 13.8e3  # [V]
Ioc = 21.1    # [A]
Poc = 90.8e3  # [W]<jupyter_output><empty_output><jupyter_text>(a) * Find the equivalent circuit referred to the low-voltage side of this transformer.  (b)If the voltage on the secondary side is 13.8 kV and the power supplied is 4000 kW at 0.8 PF lagging. * Find the voltage regulation of the transformer.  * Find its efficiency.  SOLUTION  (a)The open-circuit test was performed on the low-voltage side of the transformer, so it can be used to directly find the components of the excitation branch relative to the low-voltage side.$$|Y_{EX}| = |G_{C} - jB_{M}| = \frac{I_{OC}}{V_{OC}}$$<jupyter_code>yex = Ioc / Voc
print('yex = {:.6f}'.format(yex))<jupyter_output>yex = 0.001529<jupyter_text>$\theta = \arccos\frac{P_{OC}}{V_{OC}I_{OC}}$<jupyter_code>theta = arccos(Poc/(Voc*Ioc))
print('theta = {:.2f}°'.format(theta/pi*180))  # print the angle in degrees
Yex = yex * (cos(theta) - sin(theta)*1j)
print('Yex = {:.7f} S'.format(Yex))<jupyter_output>Yex = 0.0004768-0.0014527j S<jupyter_text>$$Y_{EX} = G_C - jB_M$$<jupyter_code>Gc = Yex.real
Bm = -Yex.imag
Rc = 1/Gc
Xm = 1/Bm
print('Rc = {:4.0f} Ω'.format(Rc))
print('Xm = {:4.0f} Ω'.format(Xm))<jupyter_output>Rc = 2097 Ω
Xm =  688 Ω<jupyter_text>The base impedance of this transformer referred to the secondary side is:$$Z_\text{base} = \frac{V^2_\text{base}}{S_\text{base}}$$<jupyter_code>Vbase = Vs0
Zbase = Vbase**2 / Sbase
print('Zbase = {:.2f} Ω'.format(Zbase))<jupyter_output>Zbase = 38.09 Ω<jupyter_text>Thus:<jupyter_code>Req = Req_pu * Zbase
Xeq = Xeq_pu * Zbase
print('Req = {:4.2f} Ω'.format(Req))
print('Xeq = {:4.2f} Ω'.format(Xeq))<jupyter_output>Req = 0.38 Ω
Xeq = 1.90 Ω<jupyter_text>**The resulting equivalent circuit is shown below:**<jupyter_code>print('''
Req,s = {:.2f} Ω   Xeq,s = {:.1f} Ω 
Rc,s  = {:.0f} Ω    Xm,s = {:.0f} Ω 
==============================
'''.format(Req, Xeq, Rc, Xm))<jupyter_output>Req,s = 0.38 Ω   Xeq,s = 1.9 Ω 
Rc,s  = 2097 Ω    Xm,s = 688 Ω 
==============================<jupyter_text>(b)If the load on the secondary side of the transformer is 4000 kW at 0.8 PF lagging and the secondary voltage is 13.8 kV, the secondary current is:$$I_S = \frac{P_{LOAD}}{V_SPF}$$<jupyter_code>Pload_b = 4000e3 # [W]
VS_b    = 13.8e3 # [V]
PF      = 0.8

Is_b = Pload_b / (abs(VS_b) * PF)
print('Is_b = {:.1f} A'.format(Is_b))
Is_b_angle = -arccos(PF)    # [rad]
IS_b = Is_b * (cos(Is_b_angle) + sin(Is_b_angle)*1j)
print('IS_b = {:.1f} A ∠{:.2f}°'.format(
            abs(IS_b), Is_b_angle/pi*180))<jupyter_output>IS_b = 362.3 A ∠-36.87°<jupyter_text>The voltage on the primary side of the transformer (referred to the secondary side) is:$$\vec{V}'_P = \vec{V}_S + \vec{I}_SZ_{EQ}$$<jupyter_code>Zeq = Req + Xeq*1j
V_P = VS_b + IS_b*Zeq
V_P_angle = arctan(V_P.imag/V_P.real)
print('V_P = {:.0f} V ∠{:.1f}°'.format(
            abs(V_P), V_P_angle/pi*180))<jupyter_output>V_P = 14332 V ∠1.9°<jupyter_text>There is a voltage drop under these load conditions. **Therefore the voltage regulation of the transformer is:**$$VR = \frac{V'_P-V_S}{V_S} \cdot 100\%$$<jupyter_code>VR = (abs(V_P)-abs(VS_b)) / abs(VS_b) *100
print('VR = {:.2f} %'.format(VR))
print('===========')<jupyter_output>VR = 3.86 %
===========<jupyter_text>The transformer copper losses and core losses are:$$P_{CU} = I^2_S R_{EQ,S}$$$$P_{core} = \frac{(V'_P)^2}{R_C}$$<jupyter_code>Pcu = abs(IS_b)**2 * Req
Pcore = abs(V_P)**2 / Rc
print('Pcu   = {:.1f} kW'.format(Pcu/1000))
print('Pcore = {:.1f} kW'.format(Pcore/1000))<jupyter_output>Pcu   = 50.0 kW
Pcore = 97.9 kW<jupyter_text>**Therefore the efficiency of this transformer at these conditions is:**$$\eta = \frac{P_{OUT}}{P_{OUT}+P_{CU}+P_{core}} \cdot 100\%$$<jupyter_code>Pout = Pload_b
eta = Pout / (Pout + Pcu +Pcore) * 100
print('η = {:.1f} %'.format(eta))
print('==========')<jupyter_output>η = 96.4 %
==========
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