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stiv31 [10]
3 years ago
11

How to pass sharp edged tools to another student in welding

Engineering
1 answer:
lisov135 [29]3 years ago
3 0

Answer: When walking with a sharp tool, the tool should be carried with the blade down and away from the body. When climbing with a sharp tool, tool belts, or buckets with hand lines should be used so workers can have both hands to grip the ladder.

Explanation:

You might be interested in
Consider a thin suspended hotplate that measures 0.25 m × 0.25 m. The isothermal plate has a mass of 3.75 kg, a specific heat of
Orlov [11]

Answer:

Heat losses by convection, Qconv = 90W

Heat losses by radiation, Qrad = 5.814W

Explanation:

Heat transfer is defined as the transfer of heat from the heat surface to the object that needs to be heated. There are three types which are:

1. Radiation

2. Conduction

3. Convection

Convection is defined as the transfer of heat through the actual movement of the molecules.

Qconv = hA(Temp.final - Temp.surr)

Where h = 6.4KW/m2K

A, area of a square = L2

= (0.25)2

= 0.0625m2

Temp.final = 250°C

Temp.surr = 25°C

Q = 64 * 0.0625 * (250 - 25)

= 90W

Radiation is a heat transfer method that does not rely upon the contact between the initial heat source and the object to be heated, it can be called thermal radiation.

Qrad = E*S*(Temp.final4 - Temp.surr4)

Where E = emissivity of the surface

S = boltzmann constant

= 5.6703 x 10-8 W/m2K4

Qrad = 5.6703 x 10-8 * 0.42 * 0.0625 * ((250)4 - (25)4)

= 5.814 W

7 0
3 years ago
Saturated liquid-vapor mixture of water, called wet steam, in a steam line at 1500 kPa is throttled to 50 kPa and 100°C. What is
frozen [14]

Answer:

x = 0.944

Explanation:

Steam at outlet is an superheated steam, since T > T_{sat}. From steam tables, the specific enthalpy is:

h_{out}=2682.4\,\frac{kJ}{kg}

The throttle valve is modelled after the First Law of Thermodynamics:

h_{in} = h_{out}

Hence, specific enthalpy at inlet is:

h_{in}=2682.4\,\frac{kJ}{kg}

The quality in the steam line is:

x = \frac{2682.4\,\frac{kJ}{kg}-844.55\,\frac{kJ}{kg}}{2791.0\,\frac{kJ}{kg} - 844.55\,\frac{kJ}{kg} }

x = 0.944

6 0
3 years ago
Write a MATLAB program in a script file that calculate the average, standard
Kryger [21]

Answer:

Code in MATLab is given as  below:

Explanation:

grade = input('Enter the grades as elements of a vector ');

x1 = length(grade);

fprintf('There are %5.2f grades\n',x1);

x2 = mean(grade);

fprintf('The average grade is %5.2f \n',x2);

x3=std(grade);

fprintf('The standard deviation is %5.2f \n',x3);

x4 = median(grade);

fprintf('The median grade is %5.2f \n',x4);

7 0
3 years ago
Read 2 more answers
Evaluate the performance of the proposed heat pump for three locations Using R134a. Discuss the effect of outdoor temperature on
Phoenix [80]

Answer:Table 2.2: Differences in runstitching times (standard − ergonomic).

1.03 -.04 .26 .30 -.97 .04 -.57 1.75 .01 .42

.45 -.80 .39 .25 .18 .95 -.18 .71 .42 .43

-.48 -1.08 -.57 1.10 .27 -.45 .62 .21 -.21 .82

A paired t-test is the standard procedure for testing this null hypothesis.

We use a paired t-test because each worker was measured twice, once for Paired t-test for

each workplace, so the observations on the two workplaces are dependent. paired data

Fast workers are probably fast for both workplaces, and slow workers are

slow for both. Thus what we do is compute the difference (standard − er-

gonomic) for each worker, and test the null hypothesis that the average of

these differences is zero using a one sample t-test on the differences.

Table 2.2 gives the differences between standard and ergonomic times.

Recall the setup for a one sample t-test. Let d1, d2, . . ., dn be the n differ-

ences in the sample. We assume that these differences are independent sam-

ples from a normal distribution with mean µ and variance σ

2

, both unknown.

Our null hypothesis is that the mean µ equals prespecified value µ0 = 0

(H0 : µ = µ0 = 0), and our alternative is H1 : µ > 0 because we expect the

workers to be faster in the ergonomic workplace.

The formula for a one sample t-test is

t =

¯d − µ0

s/√

n

,

where ¯d is the mean of the data (here the differences d1, d2, . . ., dn), n is the The paired t-test

sample size, and s is the sample standard deviation (of the differences)

s =

vuut

1

n − 1

Xn

i=1

(di − ¯d )

2 .

If our null hypothesis is correct and our assumptions are true, then the t-

statistic follows a t-distribution with n − 1 degrees of freedom.

The p-value for a test is the probability, assuming that the null hypothesis

is true, of observing a test statistic as extreme or more extreme than the one The p-value

we did observe. “Extreme” means away from the the null hypothesis towards

the alternative hypothesis. Our alternative here is that the true average is

larger than the null hypothesis value, so larger values of the test statistic are

extreme. Thus the p-value is the area under the t-curve with n − 1 degrees of

freedom from the observed t-value to the right. (If the alternative had been

µ < µ0, then the p-value is the area under the curve to the left of our test

Explanation: The curve represents the sum total of the evaluation

4 0
3 years ago
Air enters a turbine operating at steady state at 440 K, 20 bar, with a mass flow rate of 6 kg/s, and exits at 290 K, 5 bar. The
marysya [2.9K]

Answer:

The rate of heat transfer is - 935.392 kW

Explanation:

Given;

inlet pressure, P₁ = 20 bar

inlet temperature, T₁ = 440 k

outlet pressure, P₂ = 5 bar

outlet temperature, T₂ = 290 k

inlet velocity, v₁ = 18 m/s

outlet velocity, v₂ = 30 m/s

mass flow rate, Q = 6kg/s

developed power, W = 815 kW

From steam table;

at P₁ and T₁, interpolating between T = 400 and T = 450, h₁ = 3335.52 kJ/kg

at P₂ and T₂, interpolating between T= 250 and T= 300, h₂ = 3043.5 kJ/kg

Applying steady state energy balance equation;

\frac{Q}{m} -\frac{W}{m} = h_2-h_1 +\frac{1}{2} (v_2{^2} -v_1{^2})\\\\\frac{Q}{m} = h_2-h_1 +\frac{1}{2} (v_2{^2} -v_1{^2}) +\frac{W}{m}

substitute values above into this equation and evaluate the rate of heat transfer

\frac{Q}{m} = h_2-h_1 +\frac{1}{2} (v_2{^2} -v_1{^2}) +\frac{W}{m}\\\\\frac{Q}{m}  = 3043.5 - 3335.52 +\frac{1}{2000} (30{^2} -18{^2}) + \frac{815}{m} \\\\\frac{Q}{m}  = -291.732 + \frac{815}{m}\\\\\frac{Q}{m}  = \frac{815}{m}  -291.732\\\\Q = 815 - m(291.732)\\Q = 815 - 6*291.732\\\\Q = 815-1750.392 = -935.392 \ kW

Therefore, the rate of heat transfer, in kW, for a control volume enclosing the turbine is - 935.392 kW

4 0
3 years ago
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