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NikAS [45]
3 years ago
7

Preheat and postheating are necessary when welding gray cast iron. * True False

Engineering
1 answer:
Zepler [3.9K]3 years ago
5 0
True will be your answer have a great day
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What best describes the relationship between missionaries and Native Americans?
Vlad [161]

Answer:

Missionaries taught the Native Americans to read but allowed them to keep their customs. Native Americans and missionaries fought in battles

Explanation:

sorry if this doesn't help u but i hope it does!

6 0
3 years ago
An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no
Stells [14]

Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus s_{1} =s_{2}

From the refrigerant table A11-A13

P_{1} =0.8MPa   \left \{ {{ {{v_{1}=v_{g}  @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g}  @0.8MPa =246.82 kJ/kg } -   also  {{s_{1}=s_{g}  @0.8MPa =0.91853 kJ/kgK } } \right.

sat vapor

m=\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa  \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339}   = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) =  38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}

T_{2} =T_{sat @ 0.2MPa} = -10.09^{o}  C

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

E_{in} - E_{out}  =ΔE

w_{b, out}  =ΔU

w_{b, out} =m([tex]u_{1} -u_{2)

w_{b, out}  = workdone during the isentropic process

=5.8491(246.82-219.9)

=5.8491(26.91)

=157.3993

=157.4kJ

4 0
3 years ago
Technician A states that if someone is asking a question of you, then he or she is showing interest. Technician B states that si
eimsori [14]

Answer:

They are both correct

Explanation:

When someone is asking a question of you, it means he or she has either been paying attention, and is interested in you, what you're doing, or what you're saying. Also, silence when used properly can be golden in the sense that it can prevent unnecessary problems from arising, and can save one from a lot of unforeseen problem. Whatever is said cannot be taken back again, and some things should never be said at all, especially in a professional setting.

8 0
4 years ago
Solve the problem with conditions if a wall has inner and outer surface temperatures of 16 and 6 C, respectively. The interior a
Angelina_Jolie [31]

Answer:

Heat flux is 20 W/m^2

Explanation:

Heat flux (Q) is computed as

Q = h \, \Delta T

where h is heat transfer coefficient and ΔT is the difference between body's temperature

From the interior air to the inner wall

Q = 5 \frac{W}{m^2 K} \, 4 K

Q = 20 \frac{W}{m^2}

From the the outer wall to the exterior air

Q = 20 \frac{W}{m^2 K} \, 1 K

Q = 20 \frac{W}{m^2}

The wall is under steady-state condition because heat flux is constant  

7 0
3 years ago
A condenser accepts steam from the turbine in problem 2 at a pressure of 2.34 kPa. Saturated water at the same pressure leaves t
vaieri [72.5K]

Answer:

The answer is "83.98, 1889.195, and 1889.195"

Explanation:

Given value:

\bold{P_{4}=2.34 \ kPa}

In point a:

The value of h_{f4}=83.915 \ \ \frac{Kj}{kg}\\

V_4=0.001002 \ \  \frac{Kj}{kg}\\\\U_4= 83.98 \ \ \frac{Kj}{kg}\\\\

In point b:

calculating heat leaves formula= h_3-h_{f4}

                                                      = 1973.11-83.915\\\\= 1889.195 \ \ \frac{KJ}{kg}

In point c:

calculating Heat transfer rate formula=m(h_3-h_4)

                                                              = 1(1889.195)\\\\                                     = 1889.19 \ \ kw.

7 0
3 years ago
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