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N76 [4]
3 years ago
14

A(n) 78-hp compressor in a facility that operates at full load for 2500 h a year is powered by an electric motor that has an eff

iciency of 93 percent. If the unit cost of electricity is $0.11/kWh, what is the annual electricity cost of this compressor
Engineering
1 answer:
Jet001 [13]3 years ago
5 0

Answer: $17,206.13

Explanation:

Hi, to answer this question we have to apply the next formula:  

Annual electricity cost = (P x 0.746 x Ckwh x h) /η  

P = compressor power = 78 hp  

0.746 kw/hp= constant (conversion to kw)

Ckwh = Cost per kilowatt hour = $0.11/kWh  

h = operating hours per year = 2500 h  

η = efficiency = 93% = 0.93 (decimal form)  

Replacing with the values given :  

C = ( 78 hp x 0.746 kw/hp x 0.11 $/kwh x 2500 h ) / 0.93 = $17,206.13  

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Complete Question

Complete Question is attached below.

Answer:

V'=5m/s

Explanation:

From the question we are told that:

Diameter d=0.10m

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Head loss \mu=10m

 \frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu

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Generally the equation for Power is mathematically given by

 P=\rho gQH_m

Therefore

 Q=\frac{P}{\rho g H_m}

 Q=\frac{4*10^4}{1000*9.81*10.9}

 Q=0.03935m^3/sec

Since

 Q=AV'

Where

 A=\pi r^2\\A=3.142 (0.05)^2

 A=7.85*10^{-3}

Therefore

 V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}

 V'=5m/s

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