Answer:
Exit velocity 
m/s.
Explanation:
Given:
At inlet:
Properties of steam at 100 bar and 600°C
At exit:Lets take exit velocity 
We know that if we know only one property inside the dome then we will find the other property by using steam property table.
Given that dryness or quality of steam at the exit of nozzle is 0.85 and pressure P=80 bar.So from steam table we can find the other properties.
Properties of saturated steam at 80 bar
So the enthalpy of steam at the exit of turbine
Now from first law for open system
In the case of adiabatic nozzle Q=0,W=0
m/s
So Exit velocity m/s.
Answer:
P ( 2.5 < X < 7.5 ) = 0.7251
Explanation:
Given:
- The pmf for normal distribution for random variable x is given:
f(x)=0.178 exp(-0.100(x-4.51)^2)
Find:
the fraction of individuals demonstrating a response in the range of 2.5 to 7.5.
Solution:
- The random variable X follows a normal distribution with mean u = 4.51, and standard deviation s.d as follows:
s.d = sqrt ( 1 / 0.1*2)
s.d = sqrt(5) =2.236067
- Hence, the normal distribution is as follows:
X ~ N(4.51 , 2.236)
- Compute the Z-score values of the end points 2.5 and 7.5:
P ( (2.5 - 4.51) / 2.236 < Z < (7.5 - 4.51 ) / 2.236 )
P ( -0.898899327 < Z < 1.337168651 )
- Use the Z-Table for the probability required:
P ( 2.5 < X < 7.5 ) = P ( -0.898899327 < Z < 1.337168651 ) = 0.7251
Answer:
(d) None. No provisions exist.
Explanation:
B&P Code § 6738 prohibits a non-licensed person from being the sole proprietor of an engineering business. The non-licensed can be a partner in an engineering business that offers civil, electrical, or mechanical services. It is mandatory that at least one licensed engineer must be a co-owner of the business.
Given Information:
Output power required = Pout = 2.80 MW
Efficiency = η = 30%
Intensity = I = 1180 W/m²
Required Information:
Effective area = A = ?
Answer:
Effective area = A = 7.907x10³ m²
Step-by-step explanation:
A community plans to build a facility to convert solar power into electrical power and this facility has an efficiency of 30%
As we know efficiency is given by
η = Pout/Pin
Where Pout is the output power and Pin is the input power.
Pin = Pout/η
Pin = 2.80x10⁶/0.30
Pin = 9.33x10⁶ W
The effective area of a perfectly absorbing surface used in such an installation can be found using
A = Pin/I
Where I is the in Intensity of the sunlight in W/m²
A = 9.33x10⁶/1180
A = 7.907x10³ m²
Therefore, the effective area of the absorbing surface would be 7.907x10³ m².
