Answer:
The level of service of of compound grade freeway is LOSB.
Explanation:
Find the provided attachments for explanation
A turbine produces shaft power from a gas that enters the turbine with a (static) temperature of 628 K, a velocity of 143 m/s an
d a stagnation pressure of 14 atm. The gas exits the turbine with a stagnation temperature of 275 K and a stagnation pressure of 0.95 atm. The gas passing through the turbine has a molecular weight of 16 and a specific heat ratio of 1.45. What is the stagnation temperature of the gas entering the turbine? Assuming the turbine is adiabatic, is it also reversible?
1 answer:
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Answer:
vB = - 0.176 m/s (↓-)
Explanation:
Given
(AB) = 0.75 m
(AB)' = 0.2 m/s
vA = 0.6 m/s
θ = 35°
vB = ?
We use the formulas
Sin θ = Sin 35° = (OA)/(AB) ⇒ (OA) = Sin 35°*(AB)
⇒ (OA) = Sin 35°*(0.75 m) = 0.43 m
Cos θ = Cos 35° = (OB)/(AB) ⇒ (OB) = Cos 35°*(AB)
⇒ (OB) = Cos 35°*(0.75 m) = 0.614 m
We apply Pythagoras' theorem as follows
(AB)² = (OA)² + (OB)²
We derive the equation
2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB
⇒ (AB)*(AB)' = (OA)*vA + (OB)*vB
⇒ vB = ((AB)*(AB)' - (OA)*vA) / (OB)
then we have
⇒ vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)
⇒ vB = - 0.176 m/s (↓-)
The pic can show the question.
Answer:
a). Work transfer = 527.2 kJ
b). Heat Transfer = 197.7 kJ
Explanation:
Given:
= 5 Mpa
= 1623°C
= 1896 K
= 0.05
Also given
Therefore, = 1
R = 0.27 kJ / kg-K
= 0.8 kJ / kg-K
Also given :
Therefore, =
= 0.1182 MPa
a). Work transfer, δW =
= 527200 J
= 527.200 kJ
b). From 1st law of thermodynamics,
Heat transfer, δQ = ΔU+δW
=
=
=
= 197.7 kJ