Answer:
2jz - ge
Explanation:
it has the non turbo 2jz. its the engine the supra has
Answer:
The three types of lubrication occurred in the bearing are:
When the lubrication occur in the bearing it play an important role in the rolling life of bearing element. Basically, lubricant task is to reduced the friction.
1) Grease lubricant: Grease contain various type of additives which help in enhance the performance. It consist of oil where thickness are added as, this thickness improved the characteristics of grease.
2) Oil lubricant: This is used to reduced or minimized the friction and the lubricant oil is used to in those vehicles when they are motorized. Bearing lubricant oil is used to higher the speed capability.
3) Solid films: These are non fluid coating surface that are applied in the friction surface for prevention. It is used in very extreme situation where oil and grease types of lubricant does not work.
Answer:
P=3.31 hp (2.47 kW).
Explanation:
Solution
Curve A in Fig1. applies under the conditions of this problem.
S1 = Da / Dt ; S2 = E / Dt ; S3 = L / Da ; S4 = W / Da ; S5 = J / Dt and S6 = H / Dt
The above notations are with reference to the diagram below against the dimensions noted. The notations are valid for other examples following also.
32.2
Fig. 32.2 Dimension of turbine agitator
The Reynolds number is calculated. The quantities for substitution are, in consistent units,
D a =2⋅ft
n= 90/ 60 =1.5 r/s
μ = 12 x 6.72 x 10-4 = 8.06 x 10-3 lb/ft-s
ρ = 93.5 lb/ft3 g= 32.17 ft/s2
NRc = (( D a) 2 n ρ)/ μ = 2 2 ×1.5×93.5 8.06× 10 −3 =69,600
From curve A (Fig.1) , for NRc = 69,600 , N P = 5.8, and from Eq. P= N P × (n) 3 × ( D a )5 × ρ g c
The power P= 5.8×93.5× (1.5) 3 × (2) 5 / 32.17 =1821⋅ft−lb f/s requirement is 1821/550 = 3.31 hp (2.47 kW).
