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Ganezh [65]
3 years ago
15

What is 7/8 + 1/2 on a ruler

Engineering
2 answers:
Ad libitum [116K]3 years ago
8 0

Answer:

11/8

Explanation:

7/8+4/8=11/8 if you want it on a ruler you need to go online and type fraction to decimal converter and put in 11/8 into a decimal

svp [43]3 years ago
3 0
11/8 or 1.375 as a decimal
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Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turb
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Answer:

\eta_{turbine} = 0.603 = 60.3\%

Explanation:

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AT INLET:

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h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

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h₂ = h_{g\ at\ 125KPa} = 2684.9 KJ/kg

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P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than s_g and greater than s_f 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

Now, we will find h_{2s}(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

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\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\%

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