Ask question

Ask question

All categories

3 years ago

8

An individual is planning to take an 800-mile trip between two large cities. Three pos-sibilities exist: air, rail, or auto. The

person is willing to pay $25 for every hour savedin making the trip. The trip by air costs $600 and travel time is 8 hours, by rail the costis $450 and travel time is 16 hours, and by auto the cost is $200 and travel time is 20 hours.
(a) Which mode is the best choice?
(b) What factors other than cost might influence the decision regarding which mode to use?

1 answer:

vagabundo [1.1K]3 years ago

8 0

Answer:

1. Auto is the best mode since it is the cheapest

2. Safety, reliability, convenience should be considered

Explanation:

1.

This person is willing to pay $25 per hour.

To get the mode, which is the best choice I used this formula

Cost of trip + 25T

T = time or number of hours

By air travel

Cost = $600, T = 8 hours

600+25(8)

= 600+200

= 800 dollars

By rail,

Cost = $450

T = 16 hours

450+25(16)

= 450+400

= 850 dollars

By auto,

Cost = $200

T = 20 hours

200+25(20)

= 200+500

= 700 dollars

From these calculations the cost of Traveling by auto is the cheapest so it is the best mode.

2. Other factors to consider when choosing an alternative mode of transport

A. Safety: this transportation material should be able to get to it's destination without any form of damages happening

B. Reliability: the material should be able to fulfill it's requirements

C. Convenience: in terms of availability per day and also the frequency of traveling time

You might be interested in

9. A vehicle is having routine maintenance performed.

dexar [7]

Answer: b

Explanation:

8 0

3 years ago

Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th

Aleksandr-060686 [28]

Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine = $1 - \frac{Tl}{Th} \\$

where Th =temperature at which the heat enters the engine

Tl is the temperature of the environment

since both engines have the same thermal capacities <em> $n_{th}$ </em> therefore $n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\$

We have now that

$\frac{-1400}{T}+\frac{T}{300}=0\\$

multiplying through by T

$-1400 + \frac{T^{2} }{300}=0\\$

multiplying through by 300

- $420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000} \\T=648.07k$

The temperature T= 648.07k

5 0

3 years ago

Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat trans

aksik [14]

Answer:

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet, $T_{i} = 20^{\circ}C$ = 273 + 20 = 293 K

Temperature at the outlet, $T_{o} = 200{\circ}C$ = 273 + 200 = 473 K

Pressure at inlet, $P_{i} = 80 kPa = 80\times 10^{3} Pa$

Pressure at outlet, $P_{o} = 800 kPa = 800\times 10^{3} Pa$

Speed at the outlet, $v_{o} = 20 m/s$

Diameter of the tube, $D = 10 cm = 10\times 10^{- 2} m = 0.1 m$

Input power, $P_{i} = 400 kW = 400\times 10^{3} W$

Now,

To calculate the heat transfer, $Q$ , we make use of the steady flow eqn:

$h_{i} + \frac{v_{i}^{2}}{2} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}$

where

$h_{i}$ = specific enthalpy at inlet

$h_{o}$ = specific enthalpy at outlet

$v_{i}$ = air speed at inlet

$p_{s}$ = specific power input

H and H' = Elevation of inlet and outlet

Now, if

$v_{i} = 0$ and H = H'

Then the above eqn reduces to:

$h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}$

$Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}$ (1)

Also,

$p_{s} = \frac{P_{i}}{ mass, m}$

Area of cross-section, A = $\frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}$

Specific Volume at outlet, $V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s$

From the eqn:

$P_{o}V_{o} = mRT_{o}$

$m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s$

Now,

$p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg$

Also,

$\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg$

Now, using these values in eqn (1):

$Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW$

Now, rate of heat transfer, q:

q = mQ = $0.925\times 813.33 = 752.33 kW$

4 0

2 years ago

What the answer fast

anygoal [31]

Viscosity isT=u(U/y) where T is shear stress & u is velocity and y is thr length
The answer is =2.57

7 0

3 years ago

Ben leads a team of a few engineers at a robotics firm. A couple of them would like to improve their skills by taking additional

Anit [1.1K]

Answer:

i dont know

Explanation:

4 0

2 years ago