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nevsk [136]
3 years ago
9

Vector A extends from the origin to a point having polar coordinates (7, 70ᵒ ) and vector B extends from the origin to a point h

aving polar coordinates (4, 130ᵒ ). Find A · B.
Engineering
1 answer:
yaroslaw [1]3 years ago
8 0

Answer:

13.95

Explanation:

Given :

Vector A polar coordinates = ( 7, 70° )

Vector B polar coordinates = ( 4, 130° )

To find A . B we  will

A ( r , ∅ ) = ( 7, 70 )

A = rcos∅ + rsin∅

therefore ; A  = 2.394i + 6.57j

B ( r , ∅ ) = ( 4, 130° )

B = rcos∅ + rsin∅

therefore ;  B = -2.57i + 3.06j

Hence ; A .B

( 2.394 i + 6.57j ) . ( -2.57 + 3.06j ) = 13.95

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134a refrigerant enters an adiabatic compressor at 140kPa and -10C, the refrigerant is compressed at 0.5kW up to 700kPa and 60C.
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Answer:

(a) 65.04%

(b) 16.91%

Solution:

As per the question:

At inlet:

Pressure of the compressor, P = 140 kPa

Temperature, T = - 10^{\circ}C = 263 K

Isentropic work, W = 700 kPa

At outlet:

Pressure, P' = 700 kPa

Temperature, T' = 60^{\circ}C = 333 K

Now, from the steam table;

At the inlet , at a P = 700 kPa, T =60^{\circ}C:

h = 243.40 kJ/kg, s = 0.9606 kJ/kg.K

At outlet, at  P = 140 kPa, T =- 10^{\circ}C:

h' = 296.69 kJ/kg, s' = 1.0182 kJ/kg.K

Also in isentropic process, s = s'_{s} and h'_{s} = 278.06 kJ/kg.K at 700kPa

(a) Isentropic efficiency of the compressor, \eta_{s} = \frac{Work\ done\ in\ isentropic\ process}{Actual\ work\ done}

\eta_{s} = \frac{h'_{s} - h}{h' - h} = frac{278.06 - 243.40}{296.69 - 243.40} = 0.6504 = 65.04%

(b) The temperature of the environment, T_{e} = 27^{\circ}C = 273 + 27 = 300 K

Availability at state 1, \Psi = h - T_{e}s = 243.40 - 300\times 0.9606 = - 44.78 kJ/kg

Similarly for state 2, \Psi' = h' - T_{e}s' = 296.69 - 300\times 1.0182 = - 8.77 kJ/kg

Now, the efficiency of the compressor as per the second law;

\eta' = \frac{\Psi' - \Psi}{h' - h} = \frac{- 8.77 - (- 44.78)}{296.69 - 243.40} = 0.6757 = 67.57%

4 0
3 years ago
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