Example: Step Down Transformer - Revision history

Amy.crosby at 09:55, 29 January 2010

2010-01-29T09:55:09Z

← Older revision		Revision as of 02:55, 29 January 2010
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	From the given data of the step down transformer we can see that		From the given data of the step down transformer we can see that
	<math>~~~a=2~and~s=48000~\angle \cos^{-1}(0.7)</math>		<math>~~~a=2~and~s=48000~\angle \cos^{-1}(0.7)</math>


	<math>I_2~=~\frac{s^}{v_2^}~=~200\angle -46^\circ </math>		<math>I_2~=~\frac{s^}{v_2^}~=~200\angle -46^\circ </math>


	<math> R_H~+~a^2~R_l~=~1.6</math>		<math> R_H~+~a^2~R_l~=~1.6</math>
	<math>j~(X_H~+A^2~X_L)~=~J~4~=~\frac{1}{a}~\vec I_2~\left((R_H~+~a^2~R_L)~+~j(X_H~+~a^2X_L))+a\~~vecV_2~~\right)</math>
	<math>\vec V_1~=~893\angle 10.7\circ ~~/text~~{instead of} 480 V </math>
	<math>~~P_core~~~=~ P_c~=~250 w = \frac{v_1^2/R_{cH}=\frac{893^2}{3200}</math>		<math>j~(X_H~+A^2~X_L)~=~J~4~=~\frac{1}{a}~\vec I_2~\left((R_H~+~a^2~R_L)~+~j(X_H~+~a^2X_L))+a\vec V_2\right)</math>
	<math>P_{cu}=16000 w </math>
	<math> n~=~\frac{Re(s)}{Re(s)~+~P_{cu}~+~P_c}</math>
			<math>\vec V_1~=~893\angle 10.7\circ ~~{instead~ of}~~ 480 V </math>


			<math>P_{core}~=~ P_c~=~250 w = \frac{v_1^2}{R_{cH}}=\frac{893^2}{3200}</math>


			<math> P_{cu}~=~16000 w </math>


			<math> n~=~\frac {Re(s)}{Re(s)~+~P_{cu}~+~P_c}</math>


			<math>VR~=~\frac{V_1-a~V_2}{a~V_2}</math>



			Conclusion:

			<math>n~=~67~% </math>

			<math> VR~=~86~%</math>

Amy.crosby at 09:38, 29 January 2010

2010-01-29T09:38:57Z

← Older revision		Revision as of 02:38, 29 January 2010
Line 9:		Line 9:

	From the given data of the step down transformer we can see that		From the given data of the step down transformer we can see that
	<math>~~\text~~ a=2~and~s=48000~\angle cos^{-1} \</math>		<math>~~~a=2~and~s=48000~\angle \cos^{-1}(0.7)</math>

			<math>I_2~=~\frac{s^}{v_2^}~=~200\angle -46^\circ </math>

			<math> R_H~+~a^2~R_l~=~1.6</math>
			<math>j~(X_H~+A^2~X_L)~=~J~4~=~\frac{1}{a}~\vec I_2~\left((R_H~+~a^2~R_L)~+~j(X_H~+~a^2X_L))+a\vecV_2\right)</math>
			<math>\vec V_1~=~893\angle 10.7\circ /text{instead of} 480 V </math>
			<math>P_core~=~ P_c~=~250 w = \frac{v_1^2/R_{cH}=\frac{893^2}{3200}</math>
			<math>P_{cu}=16000 w </math>
			<math> n~=~\frac{Re(s)}{Re(s)~+~P_{cu}~+~P_c}</math>

Amy.crosby at 09:06, 29 January 2010

2010-01-29T09:06:59Z

← Older revision		Revision as of 02:06, 29 January 2010
Line 2:		Line 2:
	Given a 48 kVA, 480/240 V/V step down transformer operating at rated load with a power factor of 0.7 (lag), determine the efficiency and voltage regulation.		Given a 48 kVA, 480/240 V/V step down transformer operating at rated load with a power factor of 0.7 (lag), determine the efficiency and voltage regulation.

	Given: <math> R_L~=~0.2~\Omega,~~X_L~=~0.6~\Omega,~~ R_H~=~0.8~\Omega,~~X_H~=~1.6~\Omega,~~ R_{cH}~=~3.2~k\Omega,~~H_{cH}~=~1.2~k\Omega</math>		Given: <math> R_L~=~0.2~\Omega,~~X_L~=~0.6~\Omega,~~ R_H~=~0.8~\Omega,~~X_H~=~1.6~\Omega,~~ R_{cH}~=~3.2~k\Omega,~~X_{mH}~=~1.2~k\Omega</math>

	*this problem is from EMEC EXAM 1, Winter 08 by legendary Dr. Cross		*this problem is from EMEC EXAM 1, Winter 08 by legendary Dr. Cross

	==Solution==		==Solution==

			From the given data of the step down transformer we can see that
			<math>\text a=2~and~s=48000~\angle cos^{-1} \</math>

Amy.crosby: New page: ==Problem Statement== Given a 48 kVA, 480/240 V/V step down transformer operating at rated load with a power factor of 0.7 (lag), determine the efficiency and voltage regulation. Given: <...

2010-01-29T07:42:01Z

New page: ==Problem Statement== Given a 48 kVA, 480/240 V/V step down transformer operating at rated load with a power factor of 0.7 (lag), determine the efficiency and voltage regulation. Given: <...

New page

==Problem Statement==
Given a 48 kVA, 480/240 V/V step down transformer operating at rated load with a power factor of 0.7 (lag), determine the efficiency and voltage regulation.

Given: <math> R_L~=~0.2~\Omega,~~X_L~=~0.6~\Omega,~~ R_H~=~0.8~\Omega,~~X_H~=~1.6~\Omega,~~ R_{cH}~=~3.2~k\Omega,~~H_{cH}~=~1.2~k\Omega</math>

*this problem is from EMEC EXAM 1, Winter 08 by legendary Dr. Cross

==Solution==