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3 years ago

What's the difference between accuracy and percision in measuring and gaging?

Engineering

1 answer:

lidiya [134]3 years ago

4 0

Answer:

The term Accuracy means that how close our result to the original result.

Suppose we do any experiment in laboratory and we calculate mass = 7 kg but answer is mass = 15 kg then our answer is not accurate.

And the term Precision means how likely we get result like this.

Suppose we do any experiment in laboratory and we calculate mass five times and each time we get mass = 7 kg then our answer is precised but not accurate.

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Liquid water is fed to a boiler at 24°C and 10 bar is converted at a constant pressure to saturated steam.

zepelin [54]

We can find the change in the enthalpy through the tables A5 for Saturated water, pressure table.

For 1bar=1000kPa:

$T_{sat}=179.88\°c$

$H_{fg} = 2014.6kJ/kg$

$c_p=4.18 kJkg^{-1}{K^{-1}$

$\nu_g = 0.19436m^3/kg$

Replacing,

$\Delta h = h_{fg}+c_p(T_{sat}-T_{inlet})$

$\Delta h = 2014.6+4.18(179.88-24)$

$\Delta h=2666.17kJ/kg$

With the specific volume we know can calculate the mass flow, that is

$\dot{m}=\frac{\frac{15000}{3600}}{0.19436}$

$\dot{m} = 21.4378kg/s$

Then the heat required in input is,

$Q=\dot{m}\Delta h$

$Q=21.4378*2666.17$

$Q=57157.036kW$

With the same value required of 15000m^3/h, we can calculate the velocity of the water, that is given by,

$V= \frac{\dotV}{A}$

$V = \frac{\frac{15000}{3600}}{\pi /4 *(0.15)^2}$

$V=235.79m/s$

Finally we can apply the steady flow energy equation, that is

$\dot{m}(h_1+\frac{V^2}{2000})+Q = \dot{m}h_2$

Re-arrange for Q,

$Q=\dot{m}(h_2-h_1-\frac{V^2}{2000})$

$Q=\dot{m}(\Delta h-\frac{V^2}{2000})$

$Q= (21.4378)(2666.17-\frac{235.79^2}{2000})$

$Q= 56560.88kW$

We can note that consider the Kinetic Energy will decrease the heat input.

4 0

3 years ago

At an impaired driver checkpoint, the time required to conduct the impairment test varies (according to an exponential distribut

professor190 [17]

Answer:

Option (d) 2 min/veh

Explanation:

Data provided in the question:

Average time required = 60 seconds

Therefore,

The maximum capacity that can be accommodated on the system, μ = 60 veh/hr

Average Arrival rate, λ = 30 vehicles per hour

Now,

The average time spent by the vehicle is given as

⇒ $\frac{1}{\mu(1-\frac{\lambda}{\mu})}$

thus,

on substituting the respective values, we get

Average time spent by the vehicle = $\frac{1}{60(1-\frac{30}{60})}$

Average time spent by the vehicle = $\frac{1}{60(1-0.5)}$

Average time spent by the vehicle = $\frac{1}{60(0.5)}$

Average time spent by the vehicle = $\frac{1}{30}$ hr/veh

Average time spent by the vehicle = $\frac{1}{30}\times60$ min/veh

[ 1 hour = 60 minutes]

thus,

Average time spent by the vehicle = 2 min/veh

Hence,

Option (d) 2 min/veh

7 0

3 years ago

15- Vipsana's Gyros House sells gyros. The cost of ingredients (pita, meat, spices, etc.) to make a gyro is $2.00. Vipsana pays

sineoko [7]

D is a great answer

4 0

2 years ago

Water flows through a nozzle at the end of a fire hose. If the nozzle exit velocity must be 20 m/s and the exit diameter is 40 m

lara31 [8.8K]

Answer:

minimum flow rate provided by pump is 0.02513 m^3/s

Explanation:

Given data:

Exit velocity of nozzle = 20m/s

Exit diameter = 40 mm

We know that flow rate Q is given as

$Q = A \times V$

where A is Area

$A =\frac{\pi}{4} \times (40\times 10^{-3})^2 = 1.256\times 10^{-3} m^2$

$Q = 1.256\times 10^{-3} \times 20 = 0.02513 m^3/s$

minimum flow rate provided by pump is 0.02513 m^3/s

5 0

3 years ago

3. A 4-m × 5-m × 7-m room is heated by the radiator of a steam-heating system. The steam radiator transfers heat at a rate of 10

Natali [406]

Answer:

14.52 minutes

14 minutes and 31 seconds

Explanation:

Let's first start by mentioning the specific heat of air at constant volume. We consider constant volume and NOT constant pressure because the volume of the room remains constant while pressure may vary.

Specific heat at constant volume at 27°C = 0.718 kJ/kg*K

Initial temperature of room (in kelvin) = 283.15 K

Final temperature (required) of room = 293.15 K

Mass of air in room= volume * density= (4 * 5 * 7) * (1.204 kg/m3) = 168.56kg

Heat required at constant volume: 0.718 * (change in temp) * (mass of air)

Heat required = 0.718 * (293.15 - 283.15) * (168.56) = 1,210.26 kJ

Time taken for temperature rise: heat required / (rate of heat change)

Where rate of heat change = 10000 - 5000 = 5000 kJ/hr

Time taken = 1210.26 / 5000 = 0.24205 hours

Converted to minutes = 0.24205 * 60 = 14.52 minutes

4 0

3 years ago