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7nadin3 [17]
3 years ago
10

Which of the following is the typical carbon concentration range for cast irons?

Engineering
1 answer:
AveGali [126]3 years ago
8 0

Answer:

typical carbon concentration range for cast irons is

F. 3.0 wt% - 4.5wt% C

You might be interested in
Refrigerant-134a enters an adiabatic compressor as saturated vapor at -24°C and leaves at 0.8 MPa and 60°C. The mass flow rate o
Leni [432]

Answer:

(a) The power input to the compressor: \dot{W}=73.07 kJ/s = 73.07 kW

(b) The volume flow rate of the refrigerant at the compressor inlet: \dot{v}=0.209 m^{3}/s

Explanation:

(a)

We need to check the values of enthalpy (as we have an open system) for both states, being the inlet, state 1 and the outlet, state 2. We will know these values by checking the vapor charts of R134a, I used the ones found in Thermodynamics of Cengel, 7th edition.

Then, our values are:

h_{1}=235.92kJ/kg\\h_{2}=296.81kJkg

Now we proceed to know the work with the following expression:

\dot{W}=\dot{m}(h_{2}-h_{1})

Now we replace values and our result is:

\dot{W}=73.07 kJ/s = 73.07 kW

(b)

To know the volume rate at the compressor inlet, we need to know the specific volume in that phase, as we have that is saturated and at -24°C, we can read our table:

\nu=0.1739m^{3}/kg

With our specific volume and the mass rate, we can calculate the volume rate:

\dot{v}=\nu * \dot{m}\\\dot{v}=0.209 m^{3}/s

6 0
3 years ago
Assign numMatches with the number of elements in userValues that equal matchValue. userValues has NUM_VALS elements. Ex: If user
erica [24]

Answer:

import java.util.Scanner;

public class FindMatchValue {

  public static void main (String [] args) {

     Scanner scnr = new Scanner(System.in);

     final int NUM_VALS = 4;

     int[] userValues = new int[NUM_VALS];

     int i;

     int matchValue;

     int numMatches = -99; // Assign numMatches with 0 before your for loop

     matchValue = scnr.nextInt();

     for (i = 0; i < userValues.length; ++i) {

        userValues[i] = scnr.nextInt();

     }

     /* Your solution goes here */

         numMatches = 0;

     for (i = 0; i < userValues.length; ++i) {

        if(userValues[i] == matchValue) {

                       numMatches++;

                }

     }

     System.out.println("matchValue: " + matchValue + ", numMatches: " + numMatches);

  }

}

6 0
3 years ago
A motor vehicle engine operating with a diesel engine takes in atmospherics air at a temperature of 30°C and pressure of 1 bar.
nydimaria [60]

Answer:

η=0.568

Explanation:

At inlet condition

temperature = 30 C and pressure P=1 bar

The maximum temperature = 13456 C

Compression ratio r= 12

We know that for process 1-2

\dfrac{T_2}{T_1}=r^{\gamma -1}

\dfrac{T_2}{303}=12^{1.4 -1}

T_2=818.68 K

Now for process 2-3

\dfrac{T_3}{T_2}=\dfrac{V_3}{V_2}

\dfrac{273+1345}{818.86}=\dfrac{V_3}{V_2}

\dfrac{V_3}{V_2}=1.97

So the cut off ratio ρ=1.97

Efficiency of diesel engine

\eta =1-\dfrac{\rho ^{\gamma}-1}{r^{\gamma -1}\gamma \left (\rho -1\right )}

Now put the values

\eta =1-\dfrac{1.97 ^{1.4}-1}{12^{1.4-1}\times 1.4\times \left (1.97 -1\right )}

   ⇒η=0.568

So the efficiency is 56.8%.

4 0
3 years ago
Explain how does optimism and open-minded can help engineers succeed at their jobs
Aleonysh [2.5K]

Answer:There is an idea floating around that being an optimist is good for you. In her book, ‘Smile or Die’, Barbara Ehrenreich traces the origins of this idea, from early religious eccentrics to modern day psychology and motivational gurus. Related notions are that self belief is important to achievement, that confidence is needed, and you get what you expect. I freely admit that there was a time when I also thought that expectations influenced results, and sometimes they do.

However, since then I have examined this much more closely and changed my views. I now realise that there are downsides to optimism too, and that optimism and pessimism are not the only options. If you are interested in exploring this in more detail, especially if you have been uncomfortable with the self-delusion that positive thinking usually involves, then this article may provide a clearer understanding of a reasonable alternative that works and makes sense.

What is the choice?

Research has fairly consistently shown two biases in our predictions about the net benefits of actions we plan to take: (1) on average our predictions are usually too high, and (2) when asked to give ranges for our predictions our ranges tend to be too narrow.

This gives a useful insight into some familiar mental outlooks that we might try to adopt:

Pessimism: Expecting unrealistically low net benefits from courses of action (i.e. poor results achieved, if at all, after a struggle), and being unrealistically sure that those poor benefits will be achieved.

Optimism: Expecting unrealistically high net benefits from courses of action (i.e. great results easily achieved), and being unrealistically sure that those great benefits will be achieved.

Both of these involve overly narrow predictions. If we correct that by being more open-minded then other possible outlooks emerge. The one I will focus on in this article is this:

Open-minded realism: Having a view of the results of courses of action that is not biased towards high or low net benefits, and is open to a range of possibilities in a way that is, again, rational and unbiased, rather than overly narrow.

Here's a picture that shows each of these outlooks. The horizontal axis represents the outcome of some course of action, ranging from terrible on the left to great on the right. The height of the graphs represents the person's belief that each level of outcome will occur. Pessimism is narrow and negative, represented by the red distribution. Optimism is narrow and positive, represented by the green distribution. Open-minded realism, represented by the black curve, is unbiased and more spread, reflecting an open mind about what might happen.

Explanation:

3 0
2 years ago
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
Nataly_w [17]

Answer:

The exit temperature of the gas = 32° C

Explanation:

Solution

Given that:

Inlet temperature T₁ = 27°C ≈ 300.15 K

Inlet pressure P₁ = 100 KPa = 100 * 10^3 Pa

Volume flow rate , V = 15 m/s³

Diameter of the deduct, D = 500 mm = 0.5 m

Electric heater power, W heater = 130 kW = 130 * 10^3 W

The heat lost Q = 80 kW =  80 * 10^3 W

Now,

From the ideal gas law, density of the air at the inlet is given as :

ρ₁ = P₁/RT₁ = 100 * 10^3/500 * 300

=0.6667 kg/m³

The mass flow rate through the duct is computed below:

m = ρ₁ V = 0.6667 * 15 = 10 kg/s

Thus

Applying the first law of thermodynamics to the process is shown below:

Q + m (h₁ + V₁²/2 + gz₁) = m (h₂ + V₂²/2 + gz₂) + W (Conservation energy)

So,

If we neglect the potential and kinetic energy changes of the air, the above equation can be written again as:

Q + m (h₁) = m (h₂) + W

or

Q - W heater =m (h₂ - h₁) or Q - W heater =m (T₂ - T₁)

Thus

h₂ - h₁ = Cp T₂ - T₁

Now by method of substitution the known values are:

(- 80 *10^3) - (-130 * 10^3) = 10 * 100 * (T₂ -27)

Note: The heat transfer is  taken as negative because the heat is lost by the gas and work done is also taken as negative because the work is done on the gas

So,

Solving for T₂,

T₂ = 32° C

Therefore the exit temperature of the gas = 32° C

7 0
4 years ago
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