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Galina-37 [17]
3 years ago
5

Proper handling of blueprints includes which of the following

Engineering
1 answer:
marta [7]3 years ago
3 0

Answer:

folding plans neatly after use

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Speed limit signs are classified as
luda_lava [24]

Answer:

B. Regulatory signs

Explanation:

traffic rules

5 0
3 years ago
Read 2 more answers
Consider 4.8 pounds per minute of water vapor at 100 lbf/in2, 500 oF, and a velocity of 100 ft/s entering a nozzle operating at
andriy [413]

Answer:

A) v_2 = 2016.80 ft/s

B) \Delta s = 0.006 Btu/lbm R  

Explanation:

Given data:

P-1 = 100 lbf/in^2

T_1 = 500 degree f

V_1 = 100 ft/s

P_2 = 40 lbf/inc^2

effeciency = 80%

from steady flow enerfy equation

h_1 +\frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}

where h1 and h2 are inlet and exit enthalpy

for P1 = 100 lbf/in^2 and T1 = 500 degree F

H_1 = 1278.8 Btu/lbm

s_1 = 1.708 Btu/lbm -R

for P1 = 40 lbf/in^2

H_1 = 1193.5 Btu/lbm

s_1 = 1.708 Btu/lbm -R

exit enthalapy h_2

\eta = \frac{h_1 - h'_2}{ h_1 - h_2}

0.80 = \frac{1278.8 - h'_2}{1278.8 -1193.5} = 1197.77 Btu/lbm

from above equation

1278.8 \times 25037 + \frac{100^2}{2} = 1197.77   \times 25037 + \frac{v_2^2}{2}                   [1 Btu/lbm = 25037 ft^2/s^2]

v_2 = 2016.80 ft/s

b) amount of entropy

\Delta s = s_2 - s_1

s_1 = 1.708 Btu/lbm -R

at h_2 = 1197.77 Btu/lbm [\tex]  and [tex]P_2 = 40 lbf/in^2

s_2 is 1.714 Btu/lbm -R

\Delta s = 1.714 - 1.708 = 0.006 Btu/lbm R

6 0
3 years ago
A car engine with a thermal efficiency of 33% drives the air-conditioner unit (a refrigerator) besides powering the car and othe
fgiga [73]

Answer:

The rate of fuel required to drive the air conditioner Q_h = 6.061 kW

The flow rate of the cold air is  \r m = 0.30765 kg/s

Explanation:

From this question we are told that

    The efficiency is \eta = 33% = 0.33

   Temperature for the hot day is  T_h = 35^oC = 308 K \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (35+273)

        Temperature after cooling is  T_c = 5^oC = 278K

      The input power is  P_{in} = 2kW

The rate of fuel required to drive the air conditioner can be mathematically represented as

              Q_h = \frac{P_{in}}{\eta}

                    = \frac{2}{0.33} = 6.061 kW

From the question the air condition is assumed to be half as a Carnot refrigeration unit

 This can be Mathematically interpreted in terms of COP(coefficient of performance) as

             \beta_{air} = 0.5 \beta

where \beta  denotes COP and is mathematically represented as

                     \beta = \frac{Q_c}{P_{in}}

= >              Q_c = \beta P_{in}

Where Q_c is the rate of flue being burned for cold air to flow

Now if  the COP of a Carnot refrigerator is having this value

                \beta_{Carnot } = \frac{T_c}{T_h - T_c}

                            = \frac{278}{308-278}

                            \beta_{Carnot} = 9.267\\

Then

     \beta_{air} = 0.5 * 9.2667

            = 4.6333

Now substituting the value of \beta to solve for Q_c

                             Q_c = \beta P_{in}

                                  = 4.6333 *2

                                  9.2667kW

The equation for the rate of fuel being burned for the cold air to flow

                       Q_c = \r mc_p \Delta T

Making the flow rate of the cold air

                       \r m = \frac{Q_c}{c_p \Delta T}

                            = \frac{9.2667}{1.004}* (308 - 278)

                            = 0.30765 kg/s

                         

                             

4 0
3 years ago
Are currently supporting communities affected by GBV in community violations​
White raven [17]

Answer:By obey peoples propertice

Explanation:

8 0
2 years ago
What the different methods to turn on thyrister and how can a thyrister turned off​
myrzilka [38]

Answer:

forward voltage triggering

temperature triggering

dv/dt triggering

light triggering

gate triggering

Then turning off;

Turn off is accomplished by a "negative voltage" pulse between the gate and cathode terminals

Explanation:

hope it helps

8 0
3 years ago
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