Explanation:
Conduction:
Heat transfer in the conduction occurs due to movement of molecule or we can say that due to movement of electrons in the two end of same the body. Generally, phenomenon of conduction happens in the case of solid . In conduction heat transfer takes places due to direct contact of two bodies.
Convection:
In convection heat transfer of fluid takes place due to density difference .In simple words we can say that heat transfer occur due to motion of fluid.
Answer:
5,4,1, this is a explication
So I’m thinking C because they both have a lot to do with design here is my evidence. Structural engineering is a component of civil engineering which focuses on the design and development of infrastructures such as bridges, skyscrapers, dams. Civil engineering is a professional engineering discipline that deals with the design, construction, and maintenance of the physical and naturally built environment. I may be wrong but hope this helped!
Answer:
Amount of air left in the cylinder=m
=0.357 Kg
The amount of heat transfer=Q=0
Explanation:
Given
Initial pressure=P1=300 KPa
Initial volume=V1=0.2![m^{3}](https://tex.z-dn.net/?f=m%5E%7B3%7D)
Initial temperature=T
=20 C
Final Volume=
=0.1 ![m^{3}](https://tex.z-dn.net/?f=m%5E%7B3%7D)
Using gas equation
![m_{1}=((P_{1}*V_{1})/(R*T_{1}))](https://tex.z-dn.net/?f=m_%7B1%7D%3D%28%28P_%7B1%7D%2AV_%7B1%7D%29%2F%28R%2AT_%7B1%7D%29%29)
m1==(300*0.2)/(.287*293)
m1=0.714 Kg
Similarly
m2=(P2*V2)/R*T2
m2=(300*0.1)/(0.287*293)
m2=0.357 Kg
Now calculate mass of air left,where me is the mass of air left.
me=m2-m1
me=0.715-0.357
mass of air left=me=0.357 Kg
To find heat transfer we need to apply energy balance equation.
![Q=(m_{e}*h_{e})+(m_{2}*h_{2})-(m_{1}*h_{1})](https://tex.z-dn.net/?f=Q%3D%28m_%7Be%7D%2Ah_%7Be%7D%29%2B%28m_%7B2%7D%2Ah_%7B2%7D%29-%28m_%7B1%7D%2Ah_%7B1%7D%29)
Where me=m1-m2
And as the temperature remains constant,hence the enthalpy also remains constant.
h1=h2=he=h
Q=(me-(m1-m2))*h
me=m1-me
Thus heat transfer=Q=0
Answer:
4.5kg/min
Explanation:
Given parameters
![T_1 = 32^0 C, m_1 = 3 kg/min, T_2 = 7^0 C ,T_3 = 17^0](https://tex.z-dn.net/?f=T_1%20%3D%2032%5E0%20C%2C%20%20m_1%20%3D%203%20kg%2Fmin%2C%20T_2%20%3D%207%5E0%20C%20%2CT_3%20%3D%2017%5E0)
if we take
The mass flow rate of the second stream = ![m_2(kg/min)](https://tex.z-dn.net/?f=m_2%28kg%2Fmin%29)
The mass flow rate of mixed exit stream = ![m_3 (kg/min)](https://tex.z-dn.net/?f=m_3%20%28kg%2Fmin%29)
Now from mass conservation
![m_3 = m_2 + m_1](https://tex.z-dn.net/?f=m_3%20%3D%20m_2%20%2B%20m_1)
![m_3 = m_2 + 3 (kg/min)](https://tex.z-dn.net/?f=m_3%20%3D%20m_2%20%2B%203%20%28kg%2Fmin%29)
The temperature of the mixed exit stream given as
![T_3m_3 = T_2m_2 +T_1m_1\\\\17 ( 3 + m_2) = 7 \times m_2 + 32 \times 3\\\\51 + 17 m_2 = 7 m_2 + 96\\\\10 m_2 = 96 - 51\\\\m_2 = 4.5 kg/min\\\\\\\\](https://tex.z-dn.net/?f=T_3m_3%20%3D%20T_2m_2%20%2BT_1m_1%5C%5C%5C%5C17%20%28%203%20%2B%20m_2%29%20%3D%207%20%5Ctimes%20m_2%20%2B%2032%20%5Ctimes%203%5C%5C%5C%5C51%20%2B%2017%20m_2%20%3D%207%20m_2%20%2B%2096%5C%5C%5C%5C10%20m_2%20%3D%2096%20-%2051%5C%5C%5C%5Cm_2%20%3D%204.5%20kg%2Fmin%5C%5C%5C%5C%5C%5C%5C%5C)
Therefore the mass flow rate of second stream will be 4.5 kg/min.