Answer:
mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h
Explanation:
The properties of water at 1 atm and thus at the saturation temperature of 100C are hfg =2256.4 kJ/kg (Table A-4). The net rate of heat transfer to the water is ˙Q=0 . 60×3 kW=1 . 8 kWNoting that it takes 2256.4 kJ of energy to vaporize 1 kg of saturated liquid water, therate of evaporation of water is determined to be mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h
Answer:
the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm
Explanation:
Given the data in the question;
yield strength σ = 690 Mpa
plane strain fracture toughness K = 32 MPa-
minimum component thickness for which the condition of plane strain is valid = ?
Now, for plane strain conditions, the minimum thickness required is expressed as;
t ≥ 2.5( K / σ
)²
so we substitute our values into the formula
t ≥ 2.5( 32 / 690 )²
t ≥ 2.5( 0.0463768 )²
t ≥ 2.5 × 0.0021508
t ≥ 0.005377 m or 5.38 mm
Therefore, the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm
Answer:
Explanation:
Given
Required
Convert to standard form
From laws of indices
So, is equivalent to
Hence, the standard form of is