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Stolb23 [73]
4 years ago
15

A gasoline engine takes in air at 290 K, 90 kPa and then compresses it. The combustion adds 1000 kJ/kg to the air after which th

e temperature is 2050 K.
Using a cold air-standard analysis, i.e., constant specific heats at 300 K, find the compression ratio, compression specific work, and the highest pressure in the cycle. Repeat your analysis assuming temperature-dependent specific heats and comment on your results.

Engineering
1 answer:
Inessa [10]4 years ago
4 0

Answer:

attached below

Explanation:

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1. A wood board is one of a dozen different parts in a homemade robot kit. The width, depth, and height dimensions of the board
otez555 [7]

Answer:

183.75 cubic inches.

Explanation:

The volume of the wood board is determine by means of this expression:

V = w \cdot h \cdot l

By replacing variables:

V = (7.5 in) \cdot (14 in) \cdot (1.75 in)\\V = 183.75 in^{3}

3 0
3 years ago
Can someone explain it to me!
Darina [25.2K]

Answer:

It will be B

Explanation:

Since resisitors in series are added together, 1 + 1 + 1 would = 3kilo ohms. But with resistors in parallel would be (1/1+1/1)^-1. That would equal 0.5 Now you have two resistors in series for B, and because now that they are in series you add them together, so 0.5 + 1 = 1.5 kilo ohms which is what is needed.

7 0
3 years ago
Technician A says that TSBs are typically updates to the owner's manual. Technician B says that TSBs are generally
vichka [17]

Answer:

Technician A says that TSBs are typically updates to the owner's manual. Technician B says that TSBs are generally

updated information on model changes that do not affect the technician. Who is correct? the answer is c

4 0
3 years ago
A parallel circuit with two branches and an 18 volt battery. Resistor #1 on the first branch has a value of 220 ohms and resisto
likoan [24]

Answer:

  2.455 W

Explanation:

The power dissipated in each branch is ...

  P = V^2/R

So, the branch powers are ...

  branch 1: 18^2/220 ≈ 1.473 W

  branch 2: 18^2/330 ≈ 0.982 W

Total power is ...

  1.473 W + 0.982 W = 2.455 W

8 0
3 years ago
Consider a cylinder of height h, diameter d, and wall thickness t pressurized to an internal pressure P_0 (gauge pressure, relat
Serggg [28]

Explanation:

Note: For equations refer the attached document!

The net upward pressure force per unit height p*D must be balanced by the downward tensile force per unit height 2T, a force that can also be expressed as a stress, σhoop, times area 2t. Equating and solving for σh gives:

 Eq 1

Similarly, the axial stress σaxial can be calculated by dividing the total force on the end of the can, pA=pπ(D/2)2 by the cross sectional area of the wall, πDt, giving:

Eq 2

For a flat sheet in biaxial tension, the strain in a given direction such as the ‘hoop’ tangential direction is given by the following constitutive relation - with Young’s modulus E and Poisson’s ratio ν:

 Eq 3

 

Finally, solving for unknown pressure as a function of hoop strain:

 Eq 4

 

Resistance of a conductor of length L, cross-sectional area A, and resistivity ρ is

 Eq 5

Consequently, a small differential change in ΔR/R can be expressed as

 Eq 6

Where ΔL/L is longitudinal strain ε, and ΔA/A is –2νε where ν is the Poisson’s ratio of the resistive material. Substitution and factoring out ε from the right hand side leaves

 Eq 7

Where Δρ/ρε can be considered nearly constant, and thus the parenthetical term effectively becomes a single constant, the gage factor, GF

 Eq 8

For Wheat stone bridge:

 Eq 9

Given that R1=R3=R4=Ro, and R2 (the strain gage) = Ro + ΔR, substituting into equation above:

Eq9

Substituting e with respective stress-strain relation

Eq 10

 

part b

Since, axial strain(1-2v) < hoop strain (2-v). V out axial < V out hoop.

Hence, dV hoop < dV axial.

part c  

Given data:

P = 253313 Pa

D = d + 2t = 0.09013 m

t = 65 um

GF = 2

E = 75 GPa

v = 0.33

Use the data above and compute Vout using Eq3

Eq 11  

Download docx
3 0
3 years ago
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