Answer: preventive maintenance
Explanation:
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
Answer:
leggings
Explanation:
they allow the metal or sparks to not hurt you can the leggings can be easily and quickly removed.
Answer:
The answer is "+9.05 kw"
Explanation:
In the given question some information is missing which can be given in the following attachment.
The solution to this question can be defined as follows:
let assume that flow is from 1 to 2 then
Q= 1kw
m=0.1 kg/s
From the steady flow energy equation is:
![m\{n_1+ \frac{v^2_1}{z}+ gz_1 \}+Q= m \{h_2+ \frac{v^2_2}{2}+ gz_2\}+w\\\\\ change \ energy\\\\0.1[1.005 \times 800]-1= 0.01[1.005\times 700]+w\\\\w= +9.05 \ kw\\\\](https://tex.z-dn.net/?f=m%5C%7Bn_1%2B%20%5Cfrac%7Bv%5E2_1%7D%7Bz%7D%2B%20gz_1%20%5C%7D%2BQ%3D%20m%20%5C%7Bh_2%2B%20%5Cfrac%7Bv%5E2_2%7D%7B2%7D%2B%20gz_2%5C%7D%2Bw%5C%5C%5C%5C%5C%20change%20%5C%20energy%5C%5C%5C%5C0.1%5B1.005%20%5Ctimes%20800%5D-1%3D%200.01%5B1.005%5Ctimes%20700%5D%2Bw%5C%5C%5C%5Cw%3D%20%2B9.05%20%5C%20kw%5C%5C%5C%5C)
If the sign of the work performed is positive, it means the work is done on the surrounding so, that the expected direction of the flow is right.
