1 year ago

9. An automobile's oxygen sensor output needs

Engineering

1 answer:

Maslowich1 year ago

5 0

Answer:

the answer is Letter C

Explanation:

DON'T TRULY TRUST ME PLEASE..

You might be interested in

2. The device that varies incoming line pressure using centrif-

Kryger [21]

Answer:

I believe reverse boost valve.

3 0

3 years ago

2. A mild steel wire of radius 0.5mm and length 3m is stretched by a force of 49 N. Calculate:

damaskus [11]

I don’t know how to answer :’(

6 0

2 years ago

Two balls are chosen randomly from an urn containing 8 white 4 black, and orange balls. Suppose that we win $ 2 for each black b

Scorpion4ik [409]

(-2,-10,-1,-2,-3,-4)

8 0

3 years ago

If the channel-Length modulation effect is neglected, ID in the saturation region is considered to be independent of VDS

djverab [1.8K]

The answer is true because if the effect is neglected, the saturation id region is considered true

4 0

2 years ago

A rigid tank with a total volume of 0.05 m3 initially contains a two-phase liquid-vapor mixture of water at a pressure of 15 bar

Westkost [7]

Answer:

a) $m_{2} = 0.753\,kg$ , b) $Q_{in} = 2122.963\,kJ$

Explanation:

A rigid tank means a storage whose volume is constant. Process is entirely isobaric. Initial and final properties of water are included below:

State 1 - Gas-Vapor Mixture

$P = 1500\,kPa$

$T = 198.29^{\textdegree}C$

$\nu = 0.02726\,\frac{m^{3}}{kg}$

$u = 1192.94\,\frac{kJ}{kg}$

$h_{g} = 2791.0\,\frac{kJ}{kg}$

$x = 0.2$

State 2 - Gas-Vapor Mixture

$P = 1500\,kPa$

$T = 198.29^{\textdegree}C$

$\nu = 0.06643\,\frac{m^{3}}{kg}$

$u = 1718.12\,\frac{kJ}{kg}$

$h_{g} = 2791.0\,\frac{kJ}{kg}$

$x = 0.5$

The model for the rigid tank is created by using the First Law of Thermodynamics:

$Q_{in} - (m_{1}-m_{2})\cdot h_{g} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}$

Initial and final masses are:

$m_{1} = \frac{V_{1}}{\nu_{1}}$

$m_{1} = \frac{0.05\,m^{3}}{0.02726\,\frac{m^{3}}{kg} }$

$m_{1} = 1.834\,kg$

$m_{2} = \frac{V_{2}}{\nu_{2}}$

$m_{2} = \frac{0.05\,m^{3}}{0.06643\,\frac{m^{3}}{kg} }$

$m_{2} = 0.753\,kg$

a) The final mass within the tank is:

$m_{2} = 0.753\,kg$

b) The total amount of heat transfer is:

$Q_{in} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}+ (m_{1}-m_{2})\cdot h_{g}$

$Q_{in} = (0.753\,kg)\cdot (1718.12\,\frac{kJ}{kg} )- (1.834\,kg)\cdot (1192.94\,\frac{kJ}{kg} ) + (1.081\,kg)\cdot (2791.0\,\frac{kJ}{kg} )$

$Q_{in} = 2122.963\,kJ$

5 0

3 years ago