Answer:
15.64 MW
Explanation:
The computation of value of X that gives maximum profit is shown below:-
Profit = Revenue - Cost
= 15x - 0.2x 2 - 12 - 0.3x - 0.27x 2
= 14.7x - .47x^2 - 12
After solving the above equation we will get maximum differentiate for profit that is
14.7 - 0.94x = 0
So,
x = 15.64 MW
Therefore for computing the value of X that gives maximum profit we simply solve the above equation.
ANSWERS:
![-P_{2(a)} =15.6lbf/in^2\\-P_{2(b)} =30.146lbf/in^2\\ T_{2(a)} =0^oF\\T_{2(b)} =0^oF\\x_{2(b)} =49.87percent](https://tex.z-dn.net/?f=-P_%7B2%28a%29%7D%20%3D15.6lbf%2Fin%5E2%5C%5C-P_%7B2%28b%29%7D%20%3D30.146lbf%2Fin%5E2%5C%5C%20T_%7B2%28a%29%7D%20%3D0%5EoF%5C%5CT_%7B2%28b%29%7D%20%3D0%5EoF%5C%5Cx_%7B2%28b%29%7D%20%3D49.87percent)
Explanation:
Given:
Piston cylinder assembly which mean that the process is constant pressure process P=C.
<u>AMMONIA </u>
state(1)
saturated vapor ![x_{1} =1](https://tex.z-dn.net/?f=x_%7B1%7D%20%3D1)
The temperature ![T_{1} =0^0 F](https://tex.z-dn.net/?f=T_%7B1%7D%20%3D0%5E0%20F)
Isothermal process ![T=C](https://tex.z-dn.net/?f=T%3DC)
a)
( double)
b)
(reduced by half)
To find the final state by giving the quality in lbf/in we assume the friction is neglected and the system is in equilibrium.
state(1)
using PVT data for saturated ammonia
![-P_{1} =30.416 lbf/in^2\\-v_{1} =v_{g} =9.11ft^3/lb](https://tex.z-dn.net/?f=-P_%7B1%7D%20%3D30.416%20lbf%2Fin%5E2%5C%5C-v_%7B1%7D%20%3Dv_%7Bg%7D%20%3D9.11ft%5E3%2Flb)
then the state exists in the supper heated region.
a) from standard data
![-v_{1(a)} =2v_{1} =18.22ft^3/lb\\-T_{1} =0^oF](https://tex.z-dn.net/?f=-v_%7B1%28a%29%7D%20%3D2v_%7B1%7D%20%3D18.22ft%5E3%2Flb%5C%5C-T_%7B1%7D%20%3D0%5EoF)
![at\\P_{x} =14lbf/in^2\\-v_{x} =20.289 ft^3/kg](https://tex.z-dn.net/?f=at%5C%5CP_%7Bx%7D%20%3D14lbf%2Fin%5E2%5C%5C-v_%7Bx%7D%20%3D20.289%20ft%5E3%2Fkg)
![at\\P_{y} =16 lbf/in^2\\-v_{y} =17.701ft^3/kg](https://tex.z-dn.net/?f=at%5C%5CP_%7By%7D%20%3D16%20lbf%2Fin%5E2%5C%5C-v_%7By%7D%20%3D17.701ft%5E3%2Fkg)
assume linear interpolation
![\frac{P_{x}-P_{2(b)} }{P_{x}- P_{y} } =\frac{v_{x}-v_{1(a)} }{v_{x}-v_{y} }](https://tex.z-dn.net/?f=%5Cfrac%7BP_%7Bx%7D-P_%7B2%28b%29%7D%20%20%7D%7BP_%7Bx%7D-%20P_%7By%7D%20%7D%20%3D%5Cfrac%7Bv_%7Bx%7D-v_%7B1%28a%29%7D%20%20%7D%7Bv_%7Bx%7D-v_%7By%7D%20%20%7D)
![P_{1(b)}=P_{x} -(P_{x} -P_{y} )*\frac{v_{x}- v_{1(b)} }{v_{x}-v_{y} }\\ \\P_{1(b)} =14-(14-16)*\frac{20.289-18.22}{20.289-17.701} =15.6lbf/in^2](https://tex.z-dn.net/?f=P_%7B1%28b%29%7D%3DP_%7Bx%7D%20-%28P_%7Bx%7D%20-P_%7By%7D%20%29%2A%5Cfrac%7Bv_%7Bx%7D-%20v_%7B1%28b%29%7D%20%7D%7Bv_%7Bx%7D-v_%7By%7D%20%20%7D%5C%5C%20%5C%5CP_%7B1%28b%29%7D%20%3D14-%2814-16%29%2A%5Cfrac%7B20.289-18.22%7D%7B20.289-17.701%7D%20%3D15.6lbf%2Fin%5E2)
b)
![-v_{2(a)} =2v_{1} =4.555ft^3/lb\\v_{g}](https://tex.z-dn.net/?f=-v_%7B2%28a%29%7D%20%3D2v_%7B1%7D%20%3D4.555ft%5E3%2Flb%5C%5Cv_%7Bg%7D%20%3Cv_%7B2%28a%29%7D)
from standard data
![-v_{f} =0.02419ft^3/kg\\-v_{g} =9.11ft^3/kg\\v_{f}](https://tex.z-dn.net/?f=-v_%7Bf%7D%20%3D0.02419ft%5E3%2Fkg%5C%5C-v_%7Bg%7D%20%3D9.11ft%5E3%2Fkg%5C%5Cv_%7Bf%7D%20%3Cv_%7B2%28a%29%7D%20%3Cv_%7Bg%7D)
then the state exist in the wet zone
![-P_{s} =30.146lbf/in^2\\v_{2(a)} =v_{f} +x(v_{g} -v_{f} )](https://tex.z-dn.net/?f=-P_%7Bs%7D%20%3D30.146lbf%2Fin%5E2%5C%5Cv_%7B2%28a%29%7D%20%3Dv_%7Bf%7D%20%2Bx%28v_%7Bg%7D%20-v_%7Bf%7D%20%29)
![x=\frac{v_{2(a)-v_{f} } }{v_{g} -v_{f} } \\x=\frac{4.555-0.02419}{9.11-0.02419} =49.87%](https://tex.z-dn.net/?f=x%3D%5Cfrac%7Bv_%7B2%28a%29-v_%7Bf%7D%20%7D%20%7D%7Bv_%7Bg%7D%20-v_%7Bf%7D%20%7D%20%5C%5Cx%3D%5Cfrac%7B4.555-0.02419%7D%7B9.11-0.02419%7D%20%3D49.87%25)
Gases, liquids and solids are all made up of atoms, molecules, and/or ions, but the behaviors of these particles differ in the three phases. ... gas are well separated with no regular arrangement. liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.
Answer and Explanation:
In any experiment, the observed values are the actual values obtained in any experiment.
The calculated values are the values that are measured by using the observed values in a formula.
The observed values are primary values whereas the calculated values are the secondary values as calaculations are made using observed values.
Yes, if the observed values are of low accuracy.
The values should be recorded with proper care and attention in order to avoid any error.
Answer:
The answer is "828.75"
Explanation:
Please find the correct question:
For W21x93 BEAM,
![Z_x = 221.00 in^3 \\\\\to \frac{b_t}{2t_f} =4.53\\\\\to \frac{h}{t_w}=32.3](https://tex.z-dn.net/?f=Z_x%20%3D%20221.00%20in%5E3%20%5C%5C%5C%5C%5Cto%20%5Cfrac%7Bb_t%7D%7B2t_f%7D%20%3D4.53%5C%5C%5C%5C%5Cto%20%5Cfrac%7Bh%7D%7Bt_w%7D%3D32.3)
For A992 STREL,
![F_y= 50\ ks](https://tex.z-dn.net/?f=F_y%3D%2050%5C%20%20ks)
Check for complete section:
![\to \frac{b_t}{2t_f} =4.53 < \frac{65}{\sqrt{f_y = 9.19}}\\\\\to \frac{h}{t_w} =32.3 < \frac{640}{\sqrt{f_y = 90.5}}](https://tex.z-dn.net/?f=%5Cto%20%5Cfrac%7Bb_t%7D%7B2t_f%7D%20%3D4.53%20%3C%20%5Cfrac%7B65%7D%7B%5Csqrt%7Bf_y%20%3D%209.19%7D%7D%5C%5C%5C%5C%5Cto%20%5Cfrac%7Bh%7D%7Bt_w%7D%20%3D32.3%20%3C%20%5Cfrac%7B640%7D%7B%5Csqrt%7Bf_y%20%3D%2090.5%7D%7D)
Design the strength of beam =![\phi_b Z_x F_y\\\\](https://tex.z-dn.net/?f=%5Cphi_b%20Z_x%20F_y%5C%5C%5C%5C)
![=0.9 \times 221 \times 50\\\\=9945 \ in \ \ kips\\\\=\frac{9945}{12}\\\\= 828.75 \ft \ kips \\](https://tex.z-dn.net/?f=%3D0.9%20%5Ctimes%20221%20%5Ctimes%2050%5C%5C%5C%5C%3D9945%20%5C%20in%20%5C%20%5C%20kips%5C%5C%5C%5C%3D%5Cfrac%7B9945%7D%7B12%7D%5C%5C%5C%5C%3D%20828.75%20%5Cft%20%5C%20kips%20%5C%5C)