The advantages that can be associated to
drawings and symbols over written descriptions in engineering design and prototyping process are;
Communicate design ideas as well as technical information to engineers.
Symbols and drawings can be universal which means it is easy to interpret any where by professionals.
- An engineering drawing serves as complex dimensional object and symbol use by engineer to communicate.
- Drawings and symbols makes it easier to communicate design ideas and technical information to engineers and and how the process will go.
Therefore, drawings and symbols is universal to all engineer unlike written one.
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Answer:
Correct Answer:
A. water pump
Explanation:
<em>Timing belt in a vehicle helps to ensure that crankshaft, pistons and valves operate together in proper sequence.</em> Timing belts are lighter, quieter and more efficient than chains that was previously used in vehicles.
<em>Most car manufacturers recommended that, when replacing timing belt, tension assembly, water pump, camshaft oil seal should also be replaced with it at same time. </em>
Explanation:
Look at the drawings and decide which view is missing. Front? Side? Top? Then draw it
Answer:
c
Explanation:
it's the only engineering career
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}](https://tex.z-dn.net/?f=s_%7B1%7D%20%3Ds_%7B2%7D)
From the refrigerant table A11-A13
![\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.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7B%20%7B%7Bv_%7B1%7D%3Dv_%7Bg%7D%20%20%400.8MPa%20%3D0.025645%20m%5E%7B3%2F%7D%2Fkg%20%7D%20%7D%20%5Catop%20%7B%20%7B%7Bu_%7B1%7D%3Du_%7Bg%7D%20%20%400.8MPa%20%3D246.82%20kJ%2Fkg%20%7D%20-%20%20%20also%20%20%7B%7Bs_%7B1%7D%3Ds_%7Bg%7D%20%20%400.8MPa%20%3D0.91853%20kJ%2FkgK%20%7D%20%7D%20%5Cright.)
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}](https://tex.z-dn.net/?f=%5Cfrac%7BV%7D%7Bv_%7B1%7D%20%7D%20%3D%5Cfrac%7B0.15%7D%7B0.025645%7D%20%3D5.8491%20kg%5C%5Cand%20%5C%5C%5C%5CP_%7B2%7D%20%3D0.2MPa%20%20%5Cleft%20%5C%7B%20%7B%7Bx_%7B2%7D%20%3D%5Cfrac%7Bs_%7B2%7D%20-s_%7Bf%7D%20%7D%7Bs_%7Bfg%20%7D%7D%3D%5Cfrac%7B0.91853-0.15449%7D%7B0.78339%7D%20%20%20%3D%200.9753%20%5Catop%20%7Bu_%7B2%7D%20%3Du_%7Bf%7D%20%2Bx_%7B2%7D%20%7D%28u_%7Bfg%7D%29%20%3D%20%2038.26%2B0.9753%28186.25%29%3D%2038.26%2B181.65%20%3D219.9kJ%2Fkg%20%5Cright.%20%5C%5Cs_%7B1%7D%20%3D%20s_%7B2%7D)
![T_{2} =T_{sat @ 0.2MPa} = -10.09^{o} C](https://tex.z-dn.net/?f=T_%7B2%7D%20%3DT_%7Bsat%20%40%200.2MPa%7D%20%3D%20-10.09%5E%7Bo%7D%20%20C)
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:
ΔE
ΔU
)
workdone during the isentropic process
=5.8491(246.82-219.9)
=5.8491(26.91)
=157.3993
=157.4kJ