Answer:
Explanation:
Given data:
flow rate = 10 gallon per minute = 0.0223 ft^3/sec
diameter = 0.75 inch
we know discharge is given as
Q = VA
solve for velocity V = \frac{Q}{A}[/tex]
V = 7.27 ft/sec
we know that Reynold number
calculate the ratio to determine the fanning friction f
from moody diagram f value corresonding to Re and is 0.037
for horizontal pipe
where 1.94 slug/ft^3is density of water
Answer:
I. Tension (cable A) ≈ 6939 lbf
II. Tension (cable B) ≈ 17199 lbf
Explanation:
Let's begin by listing out the data that we were given:
mass of beam (m) = 570 lb, deceleration (cable A) = -20 ft/s², deceleration (cable B) = -2 ft/s²,
g = 32.17405 ft/s²
The tension on an object is given by the product of mass of the object by gravitational force plus/minus the product of mass by acceleration.
Mathematically represented thus:
T = mg + ma
where:
T = tension, m = mass, g = gravitational force,
a = acceleration
I. For Cable A, we have:
T = mg + ma = (570 * 32.17405) + [570 * (-20)]
T = 18339.2085 - 11400 = 6939.2085
T ≈ 6939 lbf
II. For Cable B, we have:
T = mg + ma = (570 * 32.17405) + [570 * (-2)]
T = 18339.2085 - 1140 = 17199.2085
T ≈ 17199 lbf
Answer:
0.31
126.23 kg/s
Explanation:
Given:-
- Fluid: Water
- Turbine: P3 = 8MPa , P4 = 10 KPa , nt = 85%
- Pump: Isentropic
- Net cycle-work output, Wnet = 100 MW
Find:-
- The thermal efficiency of the cycle
- The mass flow rate of steam
Solution:-
- The best way to deal with questions related to power cycles is to determine the process and write down the requisite properties of the fluid at each state.
First process: Isentropic compression by pump
P1 = P4 = 10 KPa ( condenser and pump inlet is usually equal )
h1 = h-P1 = 191.81 KJ/kg ( saturated liquid assumption )
s1 = s-P1 = 0.6492 KJ/kg.K
v1 = v-P1 = 0.001010 m^3 / kg
P2 = P3 = 8 MPa( Boiler pressure - Turbine inlet )
s2 = s1 = 0.6492 KJ/kg.K .... ( compressed liquid )
- To determine the ( h2 ) at state point 2 : Pump exit. We need to determine the wok-done by pump on the water ( Wp ). So from work-done principle we have:
- From the following relation we can determine ( h2 ) as follows:
h2 = h1 + wp
h2 = 191.81 + 8.0699
h2 = 199.88 KJ/kg
Second Process: Boiler supplies heat to the fluid and vaporize
- We have already evaluated the inlet fluid properties to the boiler ( pump exit property ).
- To determine the exit property of the fluid when the fluid is vaporized to steam in boiler ( super-heated phase ).
P3 = 8 MPa
T3 = ? ( assume fluid exist in the saturated vapor phase )
h3 = hg-P3 = 2758.7 KJ/kg
s3 = sg-P3 = 5.7450 KJ/kg.K
- The amount of heat supplied by the boiler per kg of fluid to the water stream. ( qs ) is determined using the state points 2 and 3 as follows:
Third Process: The expansion ( actual case ). Turbine isentropic efficiency ( nt ).
- The saturated vapor steam is expanded by the turbine to the condenser pressure. The turbine inlet pressure conditions are similar to the boiler conditions.
- Under the isentropic conditions the steam exits the turbine at the following conditions:
P4 = 10 KPa
s4 = s3 = 5.7450 KJ/kg.K ... ( liquid - vapor mixture phase )
- Compute the quality of the mixture at condenser inlet by the following relation:
- Determine the isentropic ( h4s ) at this state as follows:
- Since, we know that the turbine is not 100% isentropic. We will use the working efficiency and determine the actual ( h4 ) at the condenser inlet state:
- We can now compute the work-produced ( wt ) due to the expansion of steam in turbine.
- The net power out-put from the plant is derived from the net work produced by the compression and expansion process in pump and turbine, respectively.
Answer: The mass flow rate of the steam would be 126.23 kg/s
- The thermal efficiency of the cycle ( nth ) is defined as the ratio of net work produced by the cycle ( Wnet ) and the heat supplied by the boiler to the water ( Qs ):
Answer: The thermal efficiency of the cycle is 0.31