CrazyEngineers
CrazyEngineers

  • Rate Of Heat Transfer Problem

    ISHAN TOPRE

    Member

    Updated: Oct 26, 2024
    Views: 1.1K
    2kg of Water at 1 atm boils in 30 minutes. Find the rate of heat transfer to the water.

    The answer should be -(2kg)*334.5(KJ/Kg)/(30*60s)=0.3716KW

    Is the solution correct? The given options are not having my answer

    a. 2.51KW b.2.32KW c.2.97KW d.0.47KW e.3.12KW
    0
    Replies
Howdy guest!
Dear guest, you must be logged-in to participate on CrazyEngineers. We would love to have you as a member of our community. Consider creating an account or login.
Replies

  • Krzysztof Pelc

    Member • Apr 2, 2014

    Answer is d. 0.47 kW
    because:
    Heat of vaporization is calculated from a temperature of 100 degrees at a 1 atm and yet you have warmed these 2 kg of water at 100 degrees .... and hence the difference
    Are you sure? This action cannot be undone.
    Cancel

  • ISHAN TOPRE

    Member • Apr 2, 2014

    hfg for water at 100 deg= nearly (2675-491.17)=2183.83

    2*2183.83/(30*60)=2.42

    No option. How come you are getting 0.47?
    Are you sure? This action cannot be undone.
    Cancel

  • ISHAN TOPRE

    Member • Apr 2, 2014

    334.5 KJ/Kg is latent heat of ice and not steam.
    Are you sure? This action cannot be undone.
    Cancel

  • Krzysztof Pelc

    Member • Apr 8, 2014

    Heat of vaporization is the difference enthalpy between the points on boundary curves
    h1 = 419.71 kJ / kg, h2 = 2675.72 kJ / kg

    h2-h1= 2256 kJ/kg !!!!!!!!!!!!!!!!!

    Q=m*dh/3600 /heat related to hours

    Q=2*2256/(30*60)=2,51 kW

    Answer a. is correct - first time i go after you suggest 😀)
    Are you sure? This action cannot be undone.
    Cancel
Home Channels Search Login Register