zadanie z prędkością ciężaru

$$V = 2,6\frac{m}{s}$$

$$V = w*r = > w_{2} = \frac{V}{r} = \frac{2,6}{0,5} = 5,2\frac{\text{rad}}{s}$$

$$z = \frac{w_{1}}{w_{2}},\ w_{1} = z*w_{2} = 5,2*20 = 104\frac{\text{rad}}{s}$$

T_b = G * r = 1000 * 0, 5 = 500Nm

$$T_{s} = \frac{T_{B}}{z*r} = 27,8\ Nm$$

Y = I_x * L_x = 2 * 0, 071 = 1, 42

$$w = \frac{U - I*R_{z}}{Y},\ wY = I - R_{z}$$

$$U_{s} = w_{s}Y + I_{s}R = wY + \frac{T}{Y}*R = 157,47$$

T = I * Y

$$I = \frac{T}{Y}$$

zadanie z ciężarem

P_N = T_N * w_N

$$w_{N} = 152,9\frac{\text{rad}}{s}$$

$$T_{N} = \frac{P_{N}}{w_{N}} = \frac{12000}{152,9} = 78,5\ Nm$$

$$z = \frac{w_{1}}{w_{2}},\ \ z*w_{2} = w_{1},\ \ w_{2} = \frac{w_{1}}{z} = \frac{152,9}{15} = 10,193\frac{\text{rad}}{s}$$

$$n = \frac{P_{2}}{P_{1}},\ P_{2} = P_{1}*n = 0,9*12000 = 10800W$$

$$P_{2} = T_{2}*w_{2},\ \ T_{2} = \frac{P_{2}}{w_{2}} = \frac{10820}{10,192} = 1059,65\ Nm$$

1059, 65 − 1000 = 59, 65 Nm

zadanie ze zmniejszeniem I 80% i U z 440V na 300V

schemat postepowania : ∖n1) T_N    ,    Y_N

2) T_S

3) I_A

4) V

5) n

$$1\mathbf{)\ }T_{N} = \frac{T_{c}}{z*n} = \frac{P}{w} = \frac{55000}{120} = 458,33\ Nm$$

$$Y_{N} = \frac{T_{N}}{I} = \frac{458,33}{130} = 3,53$$

Y^′_N = 4 * 0, 75 = 3

2) G = m * g = 1500 * 10 = 15000 N

$$T_{c} = G*\frac{D}{2} = \frac{15000*0,3}{z*n} = \frac{4500}{15*0,95} = 315,79\ Nm$$

$$w = \frac{U^{'} - I'R}{Y'} \rightarrow w_{B} \rightarrow V$$

$$w = \frac{300 - 104*0,083}{3} = 97,12\frac{\text{rad}}{s}$$

$z = \frac{w_{1}}{w_{2}} \rightarrow w_{2}z = w_{1},\ w_{2} = \frac{w_{1}}{z} = \frac{97,12}{15} = 6,475\frac{\text{rad}}{s}$_,

$$4)\ V = 1,95\frac{m}{s}$$

$$5)\ n = \frac{P_{m}}{P_{e}} = \frac{T*w_{s}}{U^{'}*{I'}_{u}} = 0,97$$

zadanie obliczyć predkosc obrotowa silnika obcowzbudnego(z rysunkiem)

$$w^{'} = 150,5\frac{\text{rad}}{s},\ \ U_{N} = U_{r} = 500V$$

Y = I_t * L_t

$$w = \frac{U_{N} - I_{N}(R_{N} + R_{d})}{Y}$$

T = I_N * Y

$$I_{N} = \frac{T}{Y} = 213$$

$$R_{N} + R_{d} = \frac{w*Y - V_{N}}{I_{N}}$$

$$R_{d} = - \left( \frac{150,5*2,6 - 5000}{213} \right) - 0,04 = 0,47oma$$

silnik obcowzbudny wzory i zadanie jeśli U=500V a U'=400V

$w = \frac{U - IR_{z}}{Y_{N}} = \frac{U_{t}}{C*Y} - \frac{R_{z}}{{(C*Y)}^{2}}*T_{e}\ \ \ ,\ \ I = \frac{T'}{Y_{n}}\ \ \ ,\ $ M_e = c_e * Y * w_m

$$w = \frac{U - IR_{z}}{Y_{N}},\ \ \ \ \ Y_{n} = \frac{U - IR_{z}}{w} = \frac{500 - 440*0,04}{180} = 2,68\left\lbrack \frac{\text{Vs}}{\text{rad}} \right\rbrack$$

$$w_{N} = 1720\frac{\text{obr}}{\min}*\frac{2\pi}{60} = 180\frac{\text{rad}}{s}$$

moment zanamionowy $T_{n} = \frac{P_{n}}{w_{n}} = \frac{200}{180} = 1111Nm$

P_N = T_N * w_N

$$I^{'} = \frac{T_{n}*0,5}{Y_{N}} = \frac{1111*0,5}{2,68} = 207,28\ A$$

$$w^{'} = \frac{U^{'} - I'R_{z}}{Y_{n}} = \frac{400 - 207,28*0,04}{2,68} = 146,16\frac{\text{rad}}{s}$$

Jak sie zmieni moc zarówki

P = U * I

$R = \frac{U}{I}$

IR = U

$$I = \frac{U}{R}$$

$$P = U*\frac{U}{R}$$

$$P = \frac{U^{2}}{R}$$

$$R = \frac{U^{2}}{P}$$

$$\frac{{U^{2}}_{N}}{P_{n}} = \frac{U^{2}}{P}$$

PU² = P_nU²

$$P = \frac{P_{n}U^{2}}{{U^{2}}_{N}} = 95,20W$$

zadanie z prędkością ciężaru

$$V = 2,6\frac{m}{s}$$

$$V = w*r = > w_{2} = \frac{V}{r} = \frac{2,6}{0,5} = 5,2\frac{\text{rad}}{s}$$

$$z = \frac{w_{1}}{w_{2}},\ w_{1} = z*w_{2} = 5,2*20 = 104\frac{\text{rad}}{s}$$

T_b = G * r = 1000 * 0, 5 = 500Nm

$$T_{s} = \frac{T_{B}}{z*r} = 27,8\ Nm$$

Y = I_x * L_x = 2 * 0, 071 = 1, 42

$$w = \frac{U - I*R_{z}}{Y},\ wY = I - R_{z}$$

$$U_{s} = w_{s}Y + I_{s}R = wY + \frac{T}{Y}*R = 157,47$$

T = I * Y

$$I = \frac{T}{Y}$$

zadanie z ciężarem

P_N = T_N * w_N

$$w_{N} = 152,9\frac{\text{rad}}{s}$$

$$T_{N} = \frac{P_{N}}{w_{N}} = \frac{12000}{152,9} = 78,5\ Nm$$

$$z = \frac{w_{1}}{w_{2}},\ \ z*w_{2} = w_{1},\ \ w_{2} = \frac{w_{1}}{z} = \frac{152,9}{15} = 10,193\frac{\text{rad}}{s}$$

$$n = \frac{P_{2}}{P_{1}},\ P_{2} = P_{1}*n = 0,9*12000 = 10800W$$

$$P_{2} = T_{2}*w_{2},\ \ T_{2} = \frac{P_{2}}{w_{2}} = \frac{10820}{10,192} = 1059,65\ Nm$$

1059, 65 − 1000 = 59, 65 Nm

zadanie ze zmniejszeniem I 80% i U z 440V na 300V

schemat postepowania : ∖n1) T_N    ,    Y_N

2) T_S

3) I_A

4) V

5) n

$$1\mathbf{)\ }T_{N} = \frac{T_{c}}{z*n} = \frac{P}{w} = \frac{55000}{120} = 458,33\ Nm$$

$$Y_{N} = \frac{T_{N}}{I} = \frac{458,33}{130} = 3,53$$

Y^′_N = 4 * 0, 75 = 3

2) G = m * g = 1500 * 10 = 15000 N

$$T_{c} = G*\frac{D}{2} = \frac{15000*0,3}{z*n} = \frac{4500}{15*0,95} = 315,79\ Nm$$

$$w = \frac{U^{'} - I'R}{Y'} \rightarrow w_{B} \rightarrow V$$

$$w = \frac{300 - 104*0,083}{3} = 97,12\frac{\text{rad}}{s}$$

$z = \frac{w_{1}}{w_{2}} \rightarrow w_{2}z = w_{1},\ w_{2} = \frac{w_{1}}{z} = \frac{97,12}{15} = 6,475\frac{\text{rad}}{s}$_,

$$4)\ V = 1,95\frac{m}{s}$$

$$5)\ n = \frac{P_{m}}{P_{e}} = \frac{T*w_{s}}{U^{'}*{I'}_{u}} = 0,97$$

zadanie obliczyć predkosc obrotowa silnika obcowzbudnego(z rysunkiem)

$$w^{'} = 150,5\frac{\text{rad}}{s},\ \ U_{N} = U_{r} = 500V$$

Y = I_t * L_t

$$w = \frac{U_{N} - I_{N}(R_{N} + R_{d})}{Y}$$

T = I_N * Y

$$I_{N} = \frac{T}{Y} = 213$$

$$R_{N} + R_{d} = \frac{w*Y - V_{N}}{I_{N}}$$

$$R_{d} = - \left( \frac{150,5*2,6 - 5000}{213} \right) - 0,04 = 0,47oma$$

silnik obcowzbudny wzory i zadanie jeśli U=500V a U'=400V

$w = \frac{U - IR_{z}}{Y_{N}} = \frac{U_{t}}{C*Y} - \frac{R_{z}}{{(C*Y)}^{2}}*T_{e}\ \ \ ,\ \ I = \frac{T'}{Y_{n}}\ \ \ ,\ $ M_e = c_e * Y * w_m

$$w = \frac{U - IR_{z}}{Y_{N}},\ \ \ \ \ Y_{n} = \frac{U - IR_{z}}{w} = \frac{500 - 440*0,04}{180} = 2,68\left\lbrack \frac{\text{Vs}}{\text{rad}} \right\rbrack$$

$$w_{N} = 1720\frac{\text{obr}}{\min}*\frac{2\pi}{60} = 180\frac{\text{rad}}{s}$$

moment zanamionowy $T_{n} = \frac{P_{n}}{w_{n}} = \frac{200}{180} = 1111Nm$

P_N = T_N * w_N

$$I^{'} = \frac{T_{n}*0,5}{Y_{N}} = \frac{1111*0,5}{2,68} = 207,28\ A$$

$$w^{'} = \frac{U^{'} - I'R_{z}}{Y_{n}} = \frac{400 - 207,28*0,04}{2,68} = 146,16\frac{\text{rad}}{s}$$

Jak sie zmieni moc zarówki

P = U * I

$R = \frac{U}{I}$

IR = U

$$I = \frac{U}{R}$$

$$P = U*\frac{U}{R}$$

$$P = \frac{U^{2}}{R}$$

$$R = \frac{U^{2}}{P}$$

$$\frac{{U^{2}}_{N}}{P_{n}} = \frac{U^{2}}{P}$$

PU² = P_nU²

$$P = \frac{P_{n}U^{2}}{{U^{2}}_{N}} = 95,20W$$

zadanie z prędkością ciężaru

$$V = 2,6\frac{m}{s}$$

$$V = w*r = > w_{2} = \frac{V}{r} = \frac{2,6}{0,5} = 5,2\frac{\text{rad}}{s}$$

$$z = \frac{w_{1}}{w_{2}},\ w_{1} = z*w_{2} = 5,2*20 = 104\frac{\text{rad}}{s}$$

T_b = G * r = 1000 * 0, 5 = 500Nm

$$T_{s} = \frac{T_{B}}{z*r} = 27,8\ Nm$$

Y = I_x * L_x = 2 * 0, 071 = 1, 42

$$w = \frac{U - I*R_{z}}{Y},\ wY = I - R_{z}$$

$$U_{s} = w_{s}Y + I_{s}R = wY + \frac{T}{Y}*R = 157,47$$

T = I * Y

$$I = \frac{T}{Y}$$

zadanie z ciężarem

P_N = T_N * w_N

$$w_{N} = 152,9\frac{\text{rad}}{s}$$

$$T_{N} = \frac{P_{N}}{w_{N}} = \frac{12000}{152,9} = 78,5\ Nm$$

$$z = \frac{w_{1}}{w_{2}},\ \ z*w_{2} = w_{1},\ \ w_{2} = \frac{w_{1}}{z} = \frac{152,9}{15} = 10,193\frac{\text{rad}}{s}$$

$$n = \frac{P_{2}}{P_{1}},\ P_{2} = P_{1}*n = 0,9*12000 = 10800W$$

$$P_{2} = T_{2}*w_{2},\ \ T_{2} = \frac{P_{2}}{w_{2}} = \frac{10820}{10,192} = 1059,65\ Nm$$

1059, 65 − 1000 = 59, 65 Nm

zadanie ze zmniejszeniem I 80% i U z 440V na 300V

schemat postepowania : ∖n1) T_N    ,    Y_N

2) T_S

3) I_A

4) V

5) n

$$1\mathbf{)\ }T_{N} = \frac{T_{c}}{z*n} = \frac{P}{w} = \frac{55000}{120} = 458,33\ Nm$$

$$Y_{N} = \frac{T_{N}}{I} = \frac{458,33}{130} = 3,53$$

Y^′_N = 4 * 0, 75 = 3

2) G = m * g = 1500 * 10 = 15000 N

$$T_{c} = G*\frac{D}{2} = \frac{15000*0,3}{z*n} = \frac{4500}{15*0,95} = 315,79\ Nm$$

$$w = \frac{U^{'} - I'R}{Y'} \rightarrow w_{B} \rightarrow V$$

$$w = \frac{300 - 104*0,083}{3} = 97,12\frac{\text{rad}}{s}$$

$z = \frac{w_{1}}{w_{2}} \rightarrow w_{2}z = w_{1},\ w_{2} = \frac{w_{1}}{z} = \frac{97,12}{15} = 6,475\frac{\text{rad}}{s}$_,

$$4)\ V = 1,95\frac{m}{s}$$

$$5)\ n = \frac{P_{m}}{P_{e}} = \frac{T*w_{s}}{U^{'}*{I'}_{u}} = 0,97$$

zadanie obliczyć predkosc obrotowa silnika obcowzbudnego(z rysunkiem)

$$w^{'} = 150,5\frac{\text{rad}}{s},\ \ U_{N} = U_{r} = 500V$$

Y = I_t * L_t

$$w = \frac{U_{N} - I_{N}(R_{N} + R_{d})}{Y}$$

T = I_N * Y

$$I_{N} = \frac{T}{Y} = 213$$

$$R_{N} + R_{d} = \frac{w*Y - V_{N}}{I_{N}}$$

$$R_{d} = - \left( \frac{150,5*2,6 - 5000}{213} \right) - 0,04 = 0,47oma$$

silnik obcowzbudny wzory i zadanie jeśli U=500V a U'=400V

$w = \frac{U - IR_{z}}{Y_{N}} = \frac{U_{t}}{C*Y} - \frac{R_{z}}{{(C*Y)}^{2}}*T_{e}\ \ \ ,\ \ I = \frac{T'}{Y_{n}}\ \ \ ,\ $ M_e = c_e * Y * w_m

$$w = \frac{U - IR_{z}}{Y_{N}},\ \ \ \ \ Y_{n} = \frac{U - IR_{z}}{w} = \frac{500 - 440*0,04}{180} = 2,68\left\lbrack \frac{\text{Vs}}{\text{rad}} \right\rbrack$$

$$w_{N} = 1720\frac{\text{obr}}{\min}*\frac{2\pi}{60} = 180\frac{\text{rad}}{s}$$

moment zanamionowy $T_{n} = \frac{P_{n}}{w_{n}} = \frac{200}{180} = 1111Nm$

P_N = T_N * w_N

$$I^{'} = \frac{T_{n}*0,5}{Y_{N}} = \frac{1111*0,5}{2,68} = 207,28\ A$$

$$w^{'} = \frac{U^{'} - I'R_{z}}{Y_{n}} = \frac{400 - 207,28*0,04}{2,68} = 146,16\frac{\text{rad}}{s}$$

Jak sie zmieni moc zarówki

P = U * I

$R = \frac{U}{I}$

IR = U

$$I = \frac{U}{R}$$

$$P = U*\frac{U}{R}$$

$$P = \frac{U^{2}}{R}$$

$$R = \frac{U^{2}}{P}$$

$$\frac{{U^{2}}_{N}}{P_{n}} = \frac{U^{2}}{P}$$

PU² = P_nU²

$$P = \frac{P_{n}U^{2}}{{U^{2}}_{N}} = 95,20W$$