Answer:
Explanation:
Given that:
Superheated vapor enters the turbine at 10 MPa, 480°C,
From the tables of superheated steam tables; the following values are obtained

Also; from the system, the isentropic line is 1-2 in which s_2 is in wet state



From the values obtained;

Therefore;
6.52846 = 0.51624+7.82x
6.52846 - 0.51624 = 7.82 x
6.01222 = 7.82 x
x = 6.01222/7.82
x = 0.7688
The enthalpy for this process at state (s_2) can be determined as follows:

The actual enthalpy at s_2 by using the isentropic efficiency of the turbine can determined by using the expression:






kJ/kg
The work pump is calculated by applying the formula:




However;

From the process;





The actual enthalpy at s_4 by using the isentropic efficiency of the turbine can determined by using the expression:

Answer:
In engineering design, failure is expected. It helps you find the best solutions before implementing them in the “real world”. Having a prototype fail is a GOOD thing, because that means you have learned something new about the problem and potential solutions.
Explanation:
Answer:
0.00650 Ib s /ft^2
Explanation:
diameter ( D ) = 0.71 inches = 0.0591 ft
velocity = 0.90 ft/s ( V )
fluid specific gravity = 0.96 (62.4 ) ( x )
change in pressure ( P ) = 0 because pressure was constant
viscosity = (change in p - X sin∅ )
/ 32 V
= ( 0 - 0.96( 62.4) sin -90 ) * 0.0591 ^2 / 32 * 0.90
= - 59.904 sin (-90) * 0.0035 / 28.8
= 0.1874 / 28.8
viscosity = 0.00650 Ib s /ft^2
Answer:
a) benzene = 910 days
b) toluene = 1612.67 days
Explanation:
Given:
Kd = 1.8 L/kg (benzene)
Kd = 3.3 L/kg (toluene)
psolid = solids density = 2.6 kg/L
K = 2.9x10⁻⁵m/s
pores = n = 0.37
water table = 0.4 m
ground water = 15 m
u = K/n = (2.9x10⁻⁵ * (0.4/15)) / 0.37 = 2.09x10⁻⁶m/s
a) For benzene:

The time will take will be:

b) For toluene:


Answer:
the maximum length of the specimen before deformation is 0.4366 m
Explanation:
Given the data in the question;
Elastic modulus E = 124 GPa = 124 × 10⁹ Nm⁻²
cross-sectional diameter D = 4.2 mm = 4.2 × 10⁻³ m
tensile load F = 1810 N
maximum allowable elongation Δl = 0.46 mm = 0.46 × 10⁻³ m
Now to calculate the maximum length
for the deformation, we use the following relation;
= [ Δl × E × π × D² ] / 4F
so we substitute our values into the formula
= [ (0.46 × 10⁻³) × (124 × 10⁹) × π × (4.2 × 10⁻³)² ] / ( 4 × 1810 )
= 3161.025289 / 7240
= 0.4366 m
Therefore, the maximum length of the specimen before deformation is 0.4366 m