Answer:
Power output, ![P_{out} = 178.56 kW](https://tex.z-dn.net/?f=P_%7Bout%7D%20%3D%20178.56%20kW)
Given:
Pressure of steam, P = 1400 kPa
Temperature of steam, ![T = 350^{\circ}C](https://tex.z-dn.net/?f=T%20%3D%20350%5E%7B%5Ccirc%7DC)
Diameter of pipe, d = 8 cm = 0.08 m
Mass flow rate, ![\dot{m} = 0.1 kg.s^{- 1}](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D%20%3D%200.1%20kg.s%5E%7B-%201%7D)
Diameter of exhaust pipe, ![d_{h} = 15 cm = 0.15 m](https://tex.z-dn.net/?f=d_%7Bh%7D%20%3D%2015%20cm%20%3D%200.15%20m)
Pressure at exhaust, P' = 50 kPa
temperature, T' = ![100^{\circ}C](https://tex.z-dn.net/?f=100%5E%7B%5Ccirc%7DC)
Solution:
Now, calculation of the velocity of fluid at state 1 inlet:
![\dot{m} = \frac{Av_{i}}{V_{1}}](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D%20%3D%20%5Cfrac%7BAv_%7Bi%7D%7D%7BV_%7B1%7D%7D)
![0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}](https://tex.z-dn.net/?f=0.1%20%3D%20%5Cfrac%7B%5Cfrac%7B%5Cpi%20d%5E%7B2%7D%7D%7B4%7Dv_%7Bi%7D%7D%7B0.2004%7D)
![0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}](https://tex.z-dn.net/?f=0.1%20%3D%20%5Cfrac%7B%5Cfrac%7B%5Cpi%200.08%5E%7B2%7D%7D%7B4%7Dv_%7Bi%7D%7D%7B0.2004%7D)
![v_{i} = 3.986 m/s](https://tex.z-dn.net/?f=v_%7Bi%7D%20%3D%203.986%20m%2Fs)
Now, eqn for compressible fluid:
![\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}](https://tex.z-dn.net/?f=%5Crho_%7B1%7Dv_%7Bi%7DA_%7B1%7D%20%3D%20%5Crho_%7B2%7Dv_%7Be%7DA_%7B2%7D)
Now,
![\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}](https://tex.z-dn.net/?f=%5Cfrac%7BA_%7B1%7Dv_%7Bi%7D%7D%7BV_%7B1%7D%7D%20%3D%20%5Cfrac%7BA_%7B2%7Dv_%7Be%7D%7D%7BV_%7B2%7D%7D)
![\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cfrac%7B%5Cpi%20d_%7Bi%7D%5E%7B2%7D%7D%7B4%7Dv_%7Bi%7D%7D%7BV_%7B1%7D%7D%20%3D%20%5Cfrac%7B%5Cfrac%7B%5Cpi%20d_%7Be%7D%5E%7B2%7D%7D%7B4%7Dv_%7Be%7D%7D%7BV_%7B2%7D%7D)
![\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cfrac%7B%5Cpi%20%5Ctimes%200.08%5E%7B2%7D%7D%7B4%7D%5Ctimes%203.986%7D%7B0.2004%7D%20%3D%20%5Cfrac%7B%5Cfrac%7B%5Cpi%200.15%5E%7B2%7D%7D%7B4%7Dv_%7Be%7D%7D%7B3.418%7D)
![v_{e} = 19.33 m/s](https://tex.z-dn.net/?f=v_%7Be%7D%20%3D%2019.33%20m%2Fs)
Now, the power output can be calculated from the energy balance eqn:
![P_{out} = -\dot{m}W_{s}](https://tex.z-dn.net/?f=P_%7Bout%7D%20%3D%20-%5Cdot%7Bm%7DW_%7Bs%7D)
![P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}](https://tex.z-dn.net/?f=P_%7Bout%7D%20%3D%20-%5Cdot%7Bm%7D%28H_%7B2%7D%20-%20H_%7B1%7D%29%20%2B%20%5Cfrac%7Bv_%7Be%7D%5E%7B2%7D%20-%20v_%7Bi%7D%5E%7B2%7D%7D%7B2%7D)
![P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW](https://tex.z-dn.net/?f=P_%7Bout%7D%20%3D%20-%200.1%283.4181%20-%200.2004%29%20%2B%20%5Cfrac%7B19.33%5E%7B2%7D%20-%203.986%5E%7B2%7D%7D%7B2%7D%20%3D%20178.56%20kW)
Answer:
![\dot Q_{in} = 372.239\,MW](https://tex.z-dn.net/?f=%5Cdot%20Q_%7Bin%7D%20%3D%20372.239%5C%2CMW)
Explanation:
The water enters to the pump as saturated liquid and equation is modelled after the First Law of Thermodynamics:
![w_{in} + h_{in}- h_{out} = 0](https://tex.z-dn.net/?f=w_%7Bin%7D%20%2B%20h_%7Bin%7D-%20h_%7Bout%7D%20%3D%200)
![h_{out} = w_{in}+h_{in}](https://tex.z-dn.net/?f=h_%7Bout%7D%20%3D%20w_%7Bin%7D%2Bh_%7Bin%7D)
![h_{out} = 12\,\frac{kJ}{kg} + 191.81\,\frac{kJ}{kg}](https://tex.z-dn.net/?f=h_%7Bout%7D%20%3D%2012%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D%20%2B%20191.81%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D)
![h_{out} = 203.81\,\frac{kJ}{kg}](https://tex.z-dn.net/?f=h_%7Bout%7D%20%3D%20203.81%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D)
The boiler heats the water to the state of saturated vapor, whose specific enthalpy is:
![h_{out} = 2685.4\,\frac{kJ}{kg}](https://tex.z-dn.net/?f=h_%7Bout%7D%20%3D%202685.4%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D)
The rate of heat transfer in the boiler is:
![\dot Q_{in} = \left(150\,\frac{kg}{s}\right)\cdot \left(2685.4\,\frac{kJ}{kg}-203.81\,\frac{kJ}{kg} \right)\cdot \left(\frac{1\,MW}{1000\,kW} \right)](https://tex.z-dn.net/?f=%5Cdot%20Q_%7Bin%7D%20%3D%20%5Cleft%28150%5C%2C%5Cfrac%7Bkg%7D%7Bs%7D%5Cright%29%5Ccdot%20%5Cleft%282685.4%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D-203.81%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D%20%5Cright%29%5Ccdot%20%5Cleft%28%5Cfrac%7B1%5C%2CMW%7D%7B1000%5C%2CkW%7D%20%5Cright%29)
![\dot Q_{in} = 372.239\,MW](https://tex.z-dn.net/?f=%5Cdot%20Q_%7Bin%7D%20%3D%20372.239%5C%2CMW)
Answer:
R = ![\left[\begin{array}{ccc}1&0&0\\0&cos30&-sin30\\0&sin30&cos30\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%5C%5C0%26cos30%26-sin30%5C%5C0%26sin30%26cos30%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}cos 60&-sin60&0\\sin60&cos60&60\0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dcos%2060%26-sin60%260%5C%5Csin60%26cos60%2660%5C0%260%261%5Cend%7Barray%7D%5Cright%5D)
Explanation:
The mappings always involve a translation and a rotation of the matrix. Therefore, the rotation matrix will be given by:
Let
and
be the the angles 60⁰ and 30⁰ respectively
that is
= 60⁰ and
= 30⁰
The matrix is given by the following expression:
![\left[\begin{array}{ccc}1&0&0\\0&cos30&-sin30\\0&sin30&cos30\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%5C%5C0%26cos30%26-sin30%5C%5C0%26sin30%26cos30%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}cos 60&-sin60&0\\sin60&cos60&60\0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dcos%2060%26-sin60%260%5C%5Csin60%26cos60%2660%5C0%260%261%5Cend%7Barray%7D%5Cright%5D)
The angles can be evaluated and left in the surd form.
Answer:
The surface area of the primary settling tank is 0.0095 m^2.
The effective theoretical detention time is 0.05 s.
Explanation:
The surface area of the tank is calculated by dividing the volumetric flow rate by the overflow rate.
Volumetric flow rate = 0.570 m^3/s
Overflow rate = 60 m/s
Surface area = 0.570 m^3/s ÷ 60 m/s = 0.0095 m^2
Detention time is calculated by dividing the volume of the tank by the its volumetric flow rate
Volume of the tank = surface area × depth = 0.0095 m^2 × 3 m = 0.0285 m^3
Detention time = 0.0285 m^3 ÷ 0.570 m^3/s = 0.05 s
True will be your answer have a great day