Answer:
4.modifying a building to achieve an entirely new purpose
Answer:
The reaction at support B
Rb= 235440N
The reaction at support C
RC= 29430N
Explanation : check attachment
Answer:
The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C
Explanation:
The properties of water at 100°C and 1 atm are:
pL = 957.9 kg/m³
pV = 0.596 kg/m³
ΔHL = 2257 kJ/kg
CpL = 4.217 kJ/kg K
uL = 279x10⁻⁶Ns/m²
KL = 0.68 W/m K
σ = 58.9x10³N/m
When the water boils on the surface its heat flux is:

For copper-water, the properties are:
Cfg = 0.0128
The heat flux is:
qn = 0.9 * 18703.42 = 16833.078 W/m²

The tube surface temperature immediately after installation is:
Tinst = 100 + 20.4 = 120.4°C
For rough surfaces, Cfg = 0.0068. Using the same equation:
ΔT = 10.8°C
The tube surface temperature after prolonged service is:
Tprolo = 100 + 10.8 = 110.8°C
Answer:
(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ
Explanation:
Solution
Recall that:
A 10 gr of air is compressed isentropically
The initial air is at = 27 °C, 110 kPa
After compression air is at = a450 °C
For air, R=287 J/kg.K
cv = 716.5 J/kg.K
y = 1.4
Now,
(a) W efind the pressure on [MPa]
Thus,
T₂/T₁ = (p₂/p₁)^r-1/r
=(450 + 273)/27 + 273) =
=(p₂/110) ^0.4/1.4
p₂ becomes 2390.3 kPa
So, p₂ = 2.39 MPa
(b) For the increase in total internal energy, is given below:
ΔU = mCv (T₂ - T₁)
=(10/100) (716.5) (450 -27)
ΔU =3030 J
ΔU =3.03 kJ
(c) The next step is to find the total work needed in kJ
ΔW = mR ( (T₂ - T₁) / k- 1
(10/100) (287) (450 -27)/1.4 -1
ΔW = 3035 J
Hence, the total work required is = 3.035 kJ