Question: Problem 4 ( = 0 . 3 3 mathrm { ~J } / mathrm { g } . { } ^ {

Problem 4\(=0.33\mathrm{~J}/\mathrm{g}.{}^{\circ}\mathrm{C}\), specific heat in the liquid stay (density is \(7800\mathrm{~kg}/\mathrm{m}^{3}\), melting point \(=800^{\circ}\mathrm{C}\); specific heat
[5 Points] from room temperature (\(25{}^{\circ}\mathrm{C}\)) to liquid state \(0.29\mathrm{~J}/\mathrm{g}\).\({}^{\circ}\mathrm{C}\); and heat of fusion=\(=160\mathrm{~J}/\mathrm{g}\)) will be hecated in a cant in the solid state has been designed to avoid aspiration and above its melting point in sand casting. A down-sprue shown in Fig. \(10{}^{\circ}\mathrm{C}\) below for the metal shrinkage, a cylindrical riser deliver the liquid alloy to the mold at a rate of e 10 kg sec . To . compensate part to be cast is shown in Fig. 2. The rer is designed, the length of the cylinder is to be 2 times its diameter. The sleeve line is located at the top of tha elo... The sleeve part cavity (made by pattem) is fully placed in the drag where the parting 0 mm diameter), compute the following.
Fig. 1
a) How much heat energy must be added to accomplish the heating (J), assuming no losses?
(2 points)
b) Determine the dimension of the riser that will ensure that the riser will take \(30\%\) longer time to solidify compared to the casting.
Problem 4 \ ( = 0 . 3 3 \ mathrm { ~J } / \

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