Ice at -50 oF and 1.00 atmis to be melted in a continuous flow process with saturated steam at 1 bar. Find the minimum quantity
of steam required per pound of ice.The heat capacity of ice is 0.45 Btu/lbmoF.
1 answer:
Answer:
The minimum quantity of steam required per pound of ice is -22.5 BTU/lbm
Explanation:
Q = mCT
Q/m = CT
Q/m is the quantity of steam required per pound of ice
C is the heat capacity of ice = 0.45 BTU/lbm°F
T is the temperature of ice = -50 °F
Q/m = 0.45 BTU/lbm°F × -50 °F = -22.5 BTU/lbm
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Its true hopefully this helps you.
Answer:
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Explanation:
Given:
Δx = 250 m
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t = 17 s
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Explanation:
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Momentum(M)=25kgm/s
Speed=v
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Ke=(m x v x v)/2
175=(mv^2)/2
Cross multiply
175 x 2=mv^2
350=mv^2
Momentum=mass x velocity
25=mv
m=25/v
Substitute m=25/v in 350=mv^2
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350=25v
v=350/25
v=14 m/s
