The pressure of water is 7.3851 kPa
<u>Explanation:</u>
Given data,
V = 150×

m = 1 Kg
= 2 MPa
= 40°C
The waters specific volume is calculated:
= V/m
Here, the waters specific volume at initial condition is
, the containers volume is V, waters mass is m.
= 150×
/1
= 0.15
/ Kg
The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15
/ Kg and 0.13
/ Kg.
= 350+(400-350) 
= 395.17°C
Hence, the initial temperature is 395.17°C.
The volume is constant in the rigid container.
=
= 0.15
/ Kg
In saturated water labels for
= 40°C.
= 0.001008
/ Kg
= 19.515
/ Kg
The final state is two phase region
<
<
.
In saturated water labels for
= 40°C.
=
= 7.3851 kPa
= 7.3851 kPa
Answer:

Explanation:
The water enters to the pump as saturated liquid and equation is modelled after the First Law of Thermodynamics:




The boiler heats the water to the state of saturated vapor, whose specific enthalpy is:

The rate of heat transfer in the boiler is:


Answer:
M = 281.25 lb*ft
Explanation:
Given
W<em>man</em> = 150 lb
Weight per linear foot of the boat: q = 3 lb/ft
L = 15.00 m
M<em>max</em> = ?
Initially, we have to calculate the Buoyant Force per linear foot (due to the water exerts a uniform distributed load upward on the bottom of the boat):
∑ Fy = 0 (+↑) ⇒ q'*L - W - q*L = 0
⇒ q' = (W + q*L) / L
⇒ q' = (150 lb + 3 lb/ft*15 ft) / 15 ft
⇒ q' = 13 lb/ft (+↑)
The free body diagram of the boat is shown in the pic.
Then, we apply the following equation
q(x) = (13 - 3) = 10 (+↑)
V(x) = ∫q(x) dx = ∫10 dx = 10x (0 ≤ x ≤ 7.5)
M(x) = ∫10x dx = 5x² (0 ≤ x ≤ 7.5)
The maximum internal bending moment occurs when x = 7.5 ft
then
M(7.5) = 5(7.5)² = 281.25 lb*ft