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shtirl [24]
3 years ago
8

Which function has an inverse that is also a function?

Mathematics
1 answer:
UNO [17]3 years ago
5 0
The answer is C or the third one
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The standard gasoline engine for the 2017 Toyota Corolla includes the following data: Configuration: In-line four-stroke, four-c
aleksandrvk [35]

Answer:

a) The total engine volume is 30.3265 in³

b) the mass of air contained in the engine is 0.00129 lbm

c) The efficiency of the engine is 55.6%

Step-by-step explanation:

Given the data in the question;

The configuration is in line with four-stroke, four-cylinder engine.

Bore ( d ) = 3.17 in

Stroke ( L ) = 3.48 in

compression ratio ( r ) = 10.6 : 1

First, we calculate the swept Volume V₁ = π/4 × d²L

we substitute

V₁ = π/4 × (3.17 )² × (3.48) = 27.4655 in³

Now, Compression Ration; r = Vs + Vc / Vc

r = (V₁ + V₂) / V₂

we substitute

10.6 = (27.4655 + V₂) / V₂

10.6V₂ = 27.4655 + V₂

10.6V₂ - V₂ = 27.4655

9.6V₂ = 27.4655

V₂ = 2.861 in³

So

a) total engine volume

V_{BDC = V₁ + V₂

we substitute

V_{BDC = 27.4655 in³ + 2.861 in³

V_{BDC = 27.4655 in³ + 2.861 in³

V_{BDC = 30.3265 in³

Therefore, The total engine volume is 30.3265 in³

b) The mass of air

given that; T = 540°R, P = 14.7 psia

we know that; PV = mRT   { R is constant }

we solve for m

m = PV / RT

we substitute

m = (14.7 × 30.3265)  / ( 0.3704 × 1728 )540

m = 445.79955 / 345627.648

m = 0.00129 lbm

Therefore, the mass of air contained in the engine is 0.00129 lbm

c) The efficiency

from table { properties of air}

At 540°R; u₁ = 92.04 BTu/lbm, u_r₁ = 144.32

Now process 1 - 2 ( Isentropic )   u_r₁ / u_r₂  = V₁/V₁ = r

so,  u_r₁ / u_r₂ = r

144.32  / u_r₂ = 10.6

u_r₂ = 144.32/10.6

u_r₂ = 13.61

Thus, at u_r₂ = 13.61, u₂ = 235 BTU/lbm, T₂ = 1340°R

Now, we know that;

P₁V₁/T₁ = P₂V₂/T₂  ⇒ (P₁/V₁)r = P₂/T₂

⇒ ( 14.7 / 540 )10.6 = P₂/1340

P₂ = 386.664 psia  

process 2 - 3; constant volume heat addition;

Now, At T₃ = 3600°R, u₃ = 721.44 BTU/lbm, u_r₃ = 0.6449

so, q_{in =  u₃ -  u₂

we substitute

so, q_{in =  721.44  -  235  = 486.44 BTU/lbm

Next, process 3 - 4; Isentropic  

u_r₄/u_r₃ = V₄/V₃ = r

u_r₄ / 0.6449  = 10.6

u_r₄ = 6.8359

At u_r₄ = 6.8359; u₄ = 308 BTU/lbm

∴

Heat Rejection q_{out = u₄ - u₁ = 308 - 92.04 = 215.96 BTU/lbm

so, W_{net = q_{in - q_{out

W_{net = 486.44 - 215.96

W_{net = 270.48 BTU/lbm

so the efficiency; η_{thermal = (W_{net / q_{in ) × 100%

η_{thermal = ( 270.48 / 486.44  ) × 100%

η_{thermal = ( 0.55604  ) × 100%    

η_{thermal = 55.6%

Therefore, The efficiency of the engine is 55.6%

4 0
3 years ago
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