Seminar Ski Transportna Sredstva I Uredji

V = 15[ m / s ] f = 0,015 u = 4,5% cos ≈ 1 sin = u Vw = 7[ m / s ]

M u = 4600[ kg ]

b p = 1400[ mm]

h = 2000[ mm] η m.tr . = 0,90

[

K w = 0,55 Ns 2 m − 4

[

]

]

g = 10 m / s 2 Pe. max = 170[ KW ] 1.∑ R = ?

2.R f , Ru , Rw = ?

3. p e. max = 170[ kw]; Pot = ?;η m = ?

Ukupan otpor koji vozilo može da savlada:

∑R = R

f

+ Ru + Rw + Ri + R pot

R f -otpot kretanja (kotrljanja); R f = m ⋅ g ⋅ f ⋅ cos α Ru − otpor uspona; Rw = m ⋅ g ⋅ sin α Rw -otpor vjetra;

Rw = k w ⋅ A ⋅(V ± Vw ) 2 Ri -otpor inercijalnih sila; G dv G Ri = δ =δ a g dt g R pot − otpor poteznice R pot = R 'f + Ru' + Ri' + Rw'

1.Otpori kod zadanog motorno teretnog vozila:

∑R = R

f

+ Ru + Rw

R = m ⋅ g ⋅ f ⋅ cos α cos α ≈ 1 R f = 10 ⋅ 4600 ⋅ 0,015 ⋅ 1 R f = 690[ N ];

R f = 690[ N ]

Ru = m ⋅ g ⋅ sin α sin α ≈ u u = 4,5% 4,5 tg = = 0,045 100 arctg 0,045 = 2,58  sin 2,58  = 0,045 Ru = 4600 ⋅ 10 ⋅ 0,045 Ru = 2070[ N ]

Rw = K w ⋅ A ⋅(V + Vw )

Ru = 2070[ N ] 2

A = 1,1 ⋅ b p ⋅ h A = 1,1 ⋅ 1.4 ⋅ 2

[ ]

A = 3,08 m 2

Rw = 0,55 ⋅ 3,08 ⋅(15 + 7 ) Rw = 0,55 ⋅ 3,08 ⋅ 484

[ ]

A = 3,08 m 2 2

Rw = 819,90[ N ]

Ukupni otpor kretanja vozila

Rw = 819,90[ N ]

∑R = R + R + R ∑ R = 690 + 2070 + 819,90 ∑ R = 3579,90[ N ] f

u

w

∑ R = 3579,90[ N ] 2.Procentualno učešće pojedinih otpora u ukupnom otporu. R f ⋅ 100

∑R

=

690 ⋅ 100 = 19,27% 3579,90 R f = 19,27%

Ru ⋅ 100 2070 ⋅ 100 = = 57,83% ∑ R 3579,90

Ru = 57,83%

Rw ⋅ 100 819,90 ⋅ 100 = = 22,90% 3579,90 ∑R Ru = 22,90% 3.Snaga koju razvija motor i procenat njenog iskorišćenja. Pot = ∑ R ⋅ V

Pot = 3579,90 ⋅ 15 Pot = 53698,5[W ]

Pot = 53,6985[ KW ] Pot = 53,699[ KW ]

Pef =

ηM =

Pot 53,6985 = = 59,665[ KW ] η m.tr . 0,90 Pef ⋅ 100 Pe. max

=

Pef = 59,665[ KW ]

59,665 ⋅ 100 = 35,09% 170

η M = 35,09%