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Skyler MacDougall

Homework 5: due 2/19/2020

  1. The primary of a transformer has twice as many turns as the secondary. (a=2) The primary voltage is 220V and a 5\Omega load is connected across the secondary. Calculate the power delivered by the transformer, as well as the primary and secondary current. V_p=220V;\ Z_s=5\Omega\ V_s={V_p\over a}={220V\over2}=110V\ P={V_s^2\over Z_s}={110V^2\over5\Omega}\ P=2420W\[16pt] I_s={V_s\over Z_s}={110V\over 5\Omega}\ I_s=22A\[16pt] I_p={I_s\over a}={22A\over2}\ I_p=11A\[20pt] \overline{\underline{|P=2420W;\ I_s=22A;\ I_p=11A|}}

  2. Explain why the secondary voltage of a practical transformer decreases with increasing resistive loads.

The voltage decreases, because there is more voltage going to the internal losses.

  1. A 66.7MVA transformer has an efficiency of 99.3% when it delivers full power to a load having a power factor of 100%.

    1. Calculate the losses in the transformer under these conditions. efficiency={P_s\over P_p}\ P_s=66.7MW*0.993=66.2331MW\ losses=P_p-P_s=66.7MW-66.2331MW\ \overline{\underline{|losses=466.9kW|}}

    2. Calculate the losses and efficiency when the transformer delivers 66.7MVA to a load having a power factor of 80%.

losses\ are\ consistent\ \therefore\ losses=466.9kW\[16pt] P_s=66.2MW;\ S={P\over pf}={66.2MW\over0.8}=82.75MVA\ S_p={66.7MW\over0.8}=83.375MVA\ efficiency = {S_s\over S_p}={83.375MVA\over82.75MVA}\ \overline{\underline{|losses = 466.9kW;\ efficiency=98.7%|}}

  1. During a short-circuit test on a 10MVA, 66kV-7.6kV transformer, the following results were obtained. E_g=2640V\ I_{sc}=72A\ P_{sc}=9.85kW Calculate:

    1. The total resistance and total leakage reactance referred to the 66kV primary side. 10MVA>> 100kVA\ \therefore \ Z\approx X\[16pt] Z={V\over I}={2640V\over 72A}\approx37\Omega\ \overline{\underline{|R\approx0\Omega;\ X\approx37\Omega|}}

    2. The nominal impedance of the transformer referred to the primary side. Z_n={E^2\over S_n}={(66kV)^2\over10MVA}\ \overline{\underline{|Z_n=435.6\Omega|}}

    3. The percent impedance of the transformer. Z_p (pu)={Z_p\over Z_{n_p}}={37\Omega\over435.6\Omega}\ \overline{\underline{|Z_p(pu)=8.42%|}}

  2. In the above problem, if the iron losses at rated voltage are 35kW, calculate the full-load efficiency of the transformer if the power factor of the load is 85%.

35kW@pf=0.85;\ S={P\over pf}={35kW\over0.85}=41.2kVA\ efficiency={S_s\over S_p}={10MVA-41.2kVA\over10MVA}\ \underline{\overline{|efficiency=99.5%|}}

  1. If a transformer were actually built according to the below diagram, it would have very poor voltage regulation. Explain why, and propose a method for improving it.

When the load is reflected across the transformer, it becomes significantly smaller, where the internal losses are significant.

One can change this in as simple a step as changing the sides that the power and the load are connected to, although if current or voltage requirements exist, it may be more difficult.