Answer: the answer will be d because it is the right one to be
Explanation:
Answer:

Explanation:
First, we will find actual properties at given inlet and outlet states by the use of steam tables:
AT INLET:
At 4MPa and 350°C, from the superheated table:
h₁ = 3093.3 KJ/kg
s₁ = 6.5843 KJ/kg.K
AT OUTLET:
At P₂ = 125 KPa and steam is saturated in vapor state:
h₂ =
= 2684.9 KJ/kg
Now, for the isentropic enthalpy, we have:
P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K
Since s₂ is less than
and greater than
at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

Now, we will find
(enthalpy at the outlet for the isentropic process):

Now, the isentropic efficiency of the turbine can be given as follows:

Answer: 150m
Explanation:
The following can be depicted from the question:
Dimensions of outer walls = 9.7m × 14.7m.
Thickness of the wall = 0.30 m
Therefore, the plinth area of the building will be:
= (9.7 + 0.30/2 + 0.30/2) × (14.7 × 0.30/2 + 0.30/2)
= 10 × 15
= 150m
Answer:
Explanation:
A smaller clearance volume means a higher compression. A higher compression means better thermal efficiency. However a compression ratio too high might be troublesome, as it can cause accidental ignition of the fuel-air mix. This is the reason why Otto cycle engines have lower compressions that Diesel engines. In a Diesel engine the mix ignites by compression instead of a spark.
Answer:
When the brakes are applied the in the typical double transverse wishbone front suspension, it "drives" the car ground due to the setting of the link-type system pivot points on the lower wishbone are have parallel alignment to the road
Explanation:
In order to minimize the car's reaction to the application of the brakes, the front and rear pivot are arranged with the lower wishbone's rear pivot made to be higher than the front pivot as such the inclined wishbone torque results in an opposing vertical force to the transferred extra weight from the back due to breaking.