Who made the frist ever car

Independent auto lots usually have ____ finance rates than dealerships

wel

Independent auto lots usually have <u>higher</u> finance rates than dealerships

<u>Explanation:</u>

The finance rates that are charged by the dealers are lower than the finance charges that are charged by the independent auto. In case if you are getting financed through dealerships, you can also negotiate with them to charge finance rates and lower the charges of the finance.

But this negotiation and lowering of the finance rates is not possible with the independent auto lots and thus they charge higher rates compared to the dealerships.

8 0

3 years ago

A Rankine steam power plant is considered. Saturated water vapor enters a turbine at 8 MPa and exits at condenser at 10 kPa. The

Ray Of Light [21]

Answer:

0.31

126.23 kg/s

Explanation:

Given:-

- Fluid: Water

- Turbine: P3 = 8MPa , P4 = 10 KPa , nt = 85%

- Pump: Isentropic

- Net cycle-work output, Wnet = 100 MW

Find:-

- The thermal efficiency of the cycle

- The mass flow rate of steam

Solution:-

- The best way to deal with questions related to power cycles is to determine the process and write down the requisite properties of the fluid at each state.

First process: Isentropic compression by pump

P1 = P4 = 10 KPa ( condenser and pump inlet is usually equal )

h1 = h-P1 = 191.81 KJ/kg ( saturated liquid assumption )

s1 = s-P1 = 0.6492 KJ/kg.K

v1 = v-P1 = 0.001010 m^3 / kg

P2 = P3 = 8 MPa( Boiler pressure - Turbine inlet )

s2 = s1 = 0.6492 KJ/kg.K .... ( compressed liquid )

- To determine the ( h2 ) at state point 2 : Pump exit. We need to determine the wok-done by pump on the water ( Wp ). So from work-done principle we have:

$w_p = v_1*( P_2 - P_1 )\\\\w_p = 0.001010*( 8000 - 10 )\\\\w_p = 8.0699 \frac{KJ}{kg}$

- From the following relation we can determine ( h2 ) as follows:

h2 = h1 + wp

h2 = 191.81 + 8.0699

h2 = 199.88 KJ/kg

Second Process: Boiler supplies heat to the fluid and vaporize

- We have already evaluated the inlet fluid properties to the boiler ( pump exit property ).

- To determine the exit property of the fluid when the fluid is vaporized to steam in boiler ( super-heated phase ).

P3 = 8 MPa

T3 = ? ( assume fluid exist in the saturated vapor phase )

h3 = hg-P3 = 2758.7 KJ/kg

s3 = sg-P3 = 5.7450 KJ/kg.K

- The amount of heat supplied by the boiler per kg of fluid to the water stream. ( qs ) is determined using the state points 2 and 3 as follows:

$q_s = h_3 - h_2\\\\q_s = 2758.7 -199.88\\\\q_s = 2558.82 \frac{KJ}{kg}$

Third Process: The expansion ( actual case ). Turbine isentropic efficiency ( nt ).

- The saturated vapor steam is expanded by the turbine to the condenser pressure. The turbine inlet pressure conditions are similar to the boiler conditions.

- Under the isentropic conditions the steam exits the turbine at the following conditions:

P4 = 10 KPa

s4 = s3 = 5.7450 KJ/kg.K ... ( liquid - vapor mixture phase )

- Compute the quality of the mixture at condenser inlet by the following relation:

$x = \frac{s_4 - s_f}{s_f_g} \\\\x = \frac{5.745- 0.6492}{7.4996} \\\\x = 0.67947$

- Determine the isentropic ( h4s ) at this state as follows:

$h_4_s = h_f + x*h_f_g\\\\h_4_s = 191.81 + 0.67947*2392.1\\\\h_4_s = 1817.170187 \frac{KJ}{kg}$

- Since, we know that the turbine is not 100% isentropic. We will use the working efficiency and determine the actual ( h4 ) at the condenser inlet state:

$h4 = h_3 - n_t*(h_3 - h_4_s ) \\\\h4 = 2758.7 - 0.85*(2758.7 - 181.170187 ) \\\\h4 = 1958.39965 \frac{KJ}{kg} \\$

- We can now compute the work-produced ( wt ) due to the expansion of steam in turbine.

$w_t = h_3 - h_4\\\\w_t = 2758.7-1958.39965\\\\w_t = 800.30034 \frac{KJ}{kg}$

- The net power out-put from the plant is derived from the net work produced by the compression and expansion process in pump and turbine, respectively.

$W_n_e_t = flow(m) * ( w_t - w_p )\\\\flow ( m ) = \frac{W_n_e_t}{w_t - w_p} \\\\flow ( m ) = \frac{100000}{800.30034-8.0699} \\\\flow ( m ) = 126.23 \frac{kg}{s}$

Answer: The mass flow rate of the steam would be 126.23 kg/s

- The thermal efficiency of the cycle ( nth ) is defined as the ratio of net work produced by the cycle ( Wnet ) and the heat supplied by the boiler to the water ( Qs ):

$n_t_h = \frac{W_n_e_t}{flow(m)*q_s} \\\\n_t_h = \frac{100000}{126.23*2558.82} \\\\n_t_h = 0.31$

Answer: The thermal efficiency of the cycle is 0.31

7 0

3 years ago

A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic

DiKsa [7]

Answer:

The answer is below

Explanation:

1) The synchronous speed of an induction motor is the speed of the magnetic field of the stator. It is given by:

$n_s=\frac{120f_s}{p}\\ Where\ p\ is \ the \ number\ of\ machine\ pole, f_s\ is\ the\ supply \ frequency\\and\ n_s\ is \ the \ synchronous\ speed(speed \ of\ stator\ magnetic \ field)\\Given: f_s=60\ Hz, p=4. Therefore\\\\n_s=\frac{120*60}{4}=1800\ rpm$

2) The speed of the rotor is the motor speed. The slip is given by:

$Slip=\frac{n_s-n_m}{n_s}. \\ n_m\ is\ the \ motor\ speed(rotor\ speed)\\Slip = 0.05, n_s= 1800\ rpm\\ \\0.05=\frac{1800-n_m}{1800}\\\\ 1800-n_m=90\\\\n_m=1800-90=1710\ rpm$

3) The frequency of the rotor is given as:

$f_r=slip*f_s\\f_r=0.04*60=2.4\ Hz$

4) At standstill, the speed of the motor is 0, therefore the slip is 1.

The frequency of the rotor is given as:

$f_r=slip*f_s\\f_r=1*60=60\ Hz$

6 0

3 years ago

(a) For a given material, would you expect the surface energy to be greater than, the same as, or less than the grain boundary e

aksik [14]

Answer:

(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.

(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.

Explanation:

(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.

(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.

5 0

3 years ago

In DC electrode positive, how much power is at the work clamp?

Korolek [52]

Answer:

1/3 power

Explanation:

I'm just a smart guy

7 0

3 years ago