Answer:
The pressure upstream and downstream of a shock wave are related as
where,
= Specific Heat ratio of air
M = Mach number upstream
We know that 
Applying values we get
Similarly the temperature downstream is obtained by the relation
![\frac{T_{1}}{T_{o}}=\frac{[2\gamma M^{2}-(\gamma -1)][(\gamma -1)M^{2}+2]}{(\gamma +1)^{2}M^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7BT_%7B1%7D%7D%7BT_%7Bo%7D%7D%3D%5Cfrac%7B%5B2%5Cgamma%20M%5E%7B2%7D-%28%5Cgamma%20-1%29%5D%5B%28%5Cgamma%20-1%29M%5E%7B2%7D%2B2%5D%7D%7B%28%5Cgamma%20%2B1%29%5E%7B2%7DM%5E%7B2%7D%7D)
Applying values we get
![\frac{T_{1}}{423}=\frac{[2\times 1.4\times 1.8^{2}-(1.4-1)][(1.4-1)1.8^{2}+2]}{(1.4+1)^{2}\times 1.8^{2}}\\\\\therefore \frac{T_{1}}{423}=1.53\\\\\therefore T_{1}=647.85K=374.85^{o}C](https://tex.z-dn.net/?f=%5Cfrac%7BT_%7B1%7D%7D%7B423%7D%3D%5Cfrac%7B%5B2%5Ctimes%201.4%5Ctimes%201.8%5E%7B2%7D-%281.4-1%29%5D%5B%281.4-1%291.8%5E%7B2%7D%2B2%5D%7D%7B%281.4%2B1%29%5E%7B2%7D%5Ctimes%201.8%5E%7B2%7D%7D%5C%5C%5C%5C%5Ctherefore%20%5Cfrac%7BT_%7B1%7D%7D%7B423%7D%3D1.53%5C%5C%5C%5C%5Ctherefore%20T_%7B1%7D%3D647.85K%3D374.85%5E%7Bo%7DC)
The Mach number downstream is obtained by the relation
Answer:
Mechanic: a person who <u>repairs and maintains machinery</u>
Mechanical engineers: design <u>power-producing machines</u>
Explanation:
Answer:
the rate of heat loss from the steam turbine is Q = 200 kW
Explanation:
From the first law of thermodynamics applied to open systems
Q-W₀ = F*(ΔH + ΔK + ΔV)
where
Q= heat loss
W₀= power generated by the turbine
F= mass flow
ΔH = enthalpy change
ΔK = kinetic energy change
ΔV = potencial energy change
If we neglect the changes in potential and kinetic energy compared with the change in enthalpy , then
Q-W₀ = F*ΔH
Q = F*ΔH+ W₀
replacing values
Q = F*ΔH+ W₀ = 420 kg/min * (-600 kJ/kg) * 1 min/60 s * 1 MW/1000 kW + 4 MW = -0.2 MW = -200 kW (negative sign comes from outflow of energy)
Answer:
-Differential equation: d²T/dx² = 0
-The boundary conditions are;
1) Heat flux at bottom;
-KAdT(0)/dx = ηq_e
2) Heat flux at top surface;
-KdT(L)/dx = h(T(L) - T(water))
Explanation:
To solve this question, let's work with the following assumptions that we are given;
- Heat transfer is steady and one dimensional
- Thermal conductivity is constant.
- No heat generation exists in the medium
- The top surface which is at x = L will be subjected to convection while the bottom surface which is at x = 0 will be subjected to uniform heat flux.
Will all those assumptions given, the differential equation can be expressed as; d²T/dx² = 0
Now the boundary conditions are;
1) Heat flux at bottom;
q(at x = 0) is;
-KAdT(0)/dx = ηq_e
2) Heat flux at top surface;
q(at x = L):
-KdT(L)/dx = h(T(L) - T(water))
Answer:
-⅓ cos³ x + C
Explanation:
∫ cos² x sin x dx
If u = cos x, then du = -sin dx.
∫ -u² du
Integrate using power rule:
-⅓ u³ + C
Substitute back:
-⅓ cos³ x + C
