Answer:74.75 kJ/kg
Explanation:
Given
mass of hot fluid
=5 kg/s
Energy associated with it=150 kJ/kg
mass of cold fluid
=15 kg/s
Energy associated with it=50 kJ/kg
Energy lost=5.5 kW
Let the Enthalpy of outlet fluid be h
mass at outlet
![m=m_l+m_h=20 kg/s](https://tex.z-dn.net/?f=m%3Dm_l%2Bm_h%3D20%20kg%2Fs)
Using Energy Conservation
![m_h\times E_h+m_l\times E_l+Q=m\times h](https://tex.z-dn.net/?f=m_h%5Ctimes%20E_h%2Bm_l%5Ctimes%20E_l%2BQ%3Dm%5Ctimes%20h)
![5\times 150+15\times 50-5.5=20\times h](https://tex.z-dn.net/?f=5%5Ctimes%20150%2B15%5Ctimes%2050-5.5%3D20%5Ctimes%20h)
h=74.75 kJ/kg
Answer:
A beam carries vertical loads
Explanation:
In bridge: Beam. The beam bridge is the most common bridge form. A beam carries vertical loads by bending. As the beam bridge bends, it undergoes horizontal compression on the top.
I hope this answered your question, if not let me know :)
Answer:
perform a full analysis to make sure there is no significant reason to change the design
Answer: the allowable load P is 242.7877 kips
Explanation:
Given that;
diameter bolts d = 1.83 in
ultimate shear strength of the bolts = 60 ksi
we know that
shear area = 2×(π/4)d²
= 2×(π/4)×(1.83)² = 5.2604 in²
so
p/3(5.2604) = 60000/3.9
p/15.7812 = 15384.6153
p = 15.7812 × 15384.6153
p = 242787.691 lb
p = 242.7877 kips
therefore the allowable load P is 242.7877 kips