Answer:
Option ‘a’ is the cheapest for this house.
Explanation:
Cheapest method of heating must have least cost per kj of energy. So, convert all the energy in the same unit (say kj) and take select the cheapest method to heat the house.
Given:
Three methods are given to heat a particular house are as follows:
Method (a)
Through Gas, this gives energy of amount $1.33/therm.
Method (b)
Through electric resistance, this gives energy of amount $0.12/KWh.
Method (c)
Through oil, this gives energy of amount $2.30/gallon.
Calculation:
Step1
Change therm to kj in method ‘a’ as follows:
![C_{1}=\frac{\$ 1.33}{therm}\times(\frac{1therm}{105500kj})](https://tex.z-dn.net/?f=C_%7B1%7D%3D%5Cfrac%7B%5C%24%201.33%7D%7Btherm%7D%5Ctimes%28%5Cfrac%7B1therm%7D%7B105500kj%7D%29)
$/kj.
Step2
Change kWh to kj in method ‘b’ as follows:
![C_{2}=\frac{\$ 0.12}{kWh}\times(\frac{1 kWh }{3600kj})](https://tex.z-dn.net/?f=C_%7B2%7D%3D%5Cfrac%7B%5C%24%200.12%7D%7BkWh%7D%5Ctimes%28%5Cfrac%7B1%20kWh%20%7D%7B3600kj%7D%29)
$/kj.
Step3
Change kWh to kj in method ‘c’ as follows:
![C_{3}=\frac{\$ 2.30}{gallon}\times(\frac{1 gallon }{138500kj})](https://tex.z-dn.net/?f=C_%7B3%7D%3D%5Cfrac%7B%5C%24%202.30%7D%7Bgallon%7D%5Ctimes%28%5Cfrac%7B1%20gallon%20%7D%7B138500kj%7D%29)
$/kj.
Thus, the method ‘a’ has least cost as compare to method b and c.
So, option ‘a’ is the cheapest for this house.
Water can freeze in cold weather and cause brake failure.
Answer:
![\eta_{turbine} = 0.603 = 60.3\%](https://tex.z-dn.net/?f=%5Ceta_%7Bturbine%7D%20%3D%200.603%20%3D%2060.3%5C%25)
Explanation:
First, we will find actual properties at given inlet and outlet states by the use of steam tables:
AT INLET:
At 4MPa and 350°C, from the superheated table:
h₁ = 3093.3 KJ/kg
s₁ = 6.5843 KJ/kg.K
AT OUTLET:
At P₂ = 125 KPa and steam is saturated in vapor state:
h₂ =
= 2684.9 KJ/kg
Now, for the isentropic enthalpy, we have:
P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K
Since s₂ is less than
and greater than
at 125 KPa. Therefore, the steam is in a saturated mixture state. So:
![x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7Bs_2-s_f%7D%7Bs_%7Bfg%7D%7D%20%5C%5C%5C%5Cx%20%3D%20%5Cfrac%7B6.5843%5C%20KJ%2Fkg.K%20-%201.3741%5C%20KJ%2Fkg.K%7D%7B5.91%5C%20KJ%2Fkg.K%7D%5C%5C%5C%5Cx%20%3D%200.88)
Now, we will find
(enthalpy at the outlet for the isentropic process):
![h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg](https://tex.z-dn.net/?f=h_%7B2s%7D%20%3D%20h_%7Bf%5C%20at%5C%20125KPa%7D%2Bxh_%7Bfg%5C%20at%5C%20125KPa%7D%5C%5C%5C%5Ch_%7B2s%7D%20%3D%20444.36%5C%20KJ%2Fkg%20%2B%20%280.88%29%282240.6%5C%20KJ%2Fkg%29%5C%5Ch_%7B2s%7D%20%3D%202416.088%5C%20KJ%2Fkg)
Now, the isentropic efficiency of the turbine can be given as follows:
![\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%](https://tex.z-dn.net/?f=%5Ceta_%7Bturbine%7D%20%3D%20%5Cfrac%7Bh_1-h_2%7D%7Bh_1-h_%7B2s%7D%7D%5C%5C%5C%5C%5Ceta_%7Bturbine%7D%20%3D%20%5Cfrac%7B3093.3%5C%20KJ%2Fkg-2684.9%5C%20KJ%2Fkg%7D%7B3093.3%5C%20KJ%2Fkg-2416.088%5C%20KJ%2Fkg%7D%5C%5C%5C%5C%5Ceta_%7Bturbine%7D%20%3D%20%5Cfrac%7B408.4%5C%20KJ%2Fkg%7D%7B677.212%5C%20KJ%2Fkg%7D%5C%5C%5C%5C%5Ceta_%7Bturbine%7D%20%3D%200.603%20%3D%2060.3%5C%25)
In order to understand a monomer let´s first see the structure of a polymer. As an example, in the first figure polyethylene (or polyethene) is shown. This polymer, like every other one, is composed of many repeated subunits, these subunits are called monomer. In the second figure, polyethylene's monomer is shown.
Answer:
A force must s applied to a wall or roof rafters to add strength and keep the building straight and plumb